{"id":1160,"date":"2023-03-18T01:34:57","date_gmt":"2023-03-18T01:34:57","guid":{"rendered":"https:\/\/vartmaaninstitutesirsa.com\/?page_id=1160"},"modified":"2024-03-13T12:29:34","modified_gmt":"2024-03-13T12:29:34","slug":"chapter-1-electric-charge-and-electric-field","status":"publish","type":"page","link":"https:\/\/vartmaaninstitutesirsa.com\/index.php\/unit-i-electrostatics\/chapter-1-electric-charge-and-electric-field\/","title":{"rendered":"chapter 1 Electric Charge and Electric Field"},"content":{"rendered":"
\n

chapter 1 Electric Charge and Electric Field<\/h1>\n

Chapter 1 Electric Charges and Fields :\u00a0 Electric charges, Conservation of charge, Coulomb\u2019s law-force between two point charges, forces between multiple charges; superposition principle and continuous charge distribution. Electric field, electric field due to a point charge, electric field lines, electric dipole, electric field due to a dipole, torque on a dipole in uniform electric field. Electric flux, statement of Gauss\u2019s theorem and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell (field inside and outside).<\/p>\n

\n

\u00a0<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Chapter\u20131: Electric Charges and Fields<\/span><\/strong><\/p>\n

1. Electrostatics:-\u00a0<\/span><\/strong><\/p>\n

The branch of physics which deals with the study of charges at rest & the forces & fields, potential of charges is called electrostatics.<\/span><\/em><\/strong><\/p>\n

* Electrostatics was discovered around 600Bc by a Greek philosopher\u00a0<\/span>Thales of Miletus<\/span><\/strong>.\u00a0<\/span><\/p>\n

He rubbed amber with a cloth, then cloths starts attracting small piece of paper.\u00a0<\/span><\/p>\n

In Greek amber is called electron so phenomenon was called electricity.\u00a0<\/span><\/p>\n

(Amber is a yellow gum like substance obtained from old plants).\u00a0<\/span><\/em><\/p>\n

* In 1544 \u2013 1603<\/span>\u00a0William Gilbert<\/span><\/strong>\u00a0found that a force was present after rubbing of amber or any other substance called electrostatic force.<\/span><\/p>\n

* In 1731\u00a0<\/span>Stephen Gray<\/span><\/strong>\u00a0found that charge can be moved through a metal for a long distance but not through a thread. This leads to two types of materials<\/span>\u00a0\u00a0\u00a0<\/span>(i) Conductor (ii) Insulator<\/span><\/p>\n

* In 1733\u00a0<\/span>Du Fay<\/span><\/strong>\u00a0discovered that charges are of two types + ve charge and \u2013 ve charge.<\/span><\/p>\n

* In 1750\u00a0<\/span>Benjamin Franklin<\/span><\/strong>\u00a0proposed one fluid theory. He believed that charge occurs due to transfer of electrons. The excess electrons means -vely charged body & deficiency of electrons means + vely charged body.\u00a0<\/span><\/p>\n

2. What is electric charge:-<\/span><\/strong>\u00a0<\/span><\/p>\n

\"C:\\Users\\vartmaan\\Downloads\\index.png\"The property of electrons due to which they experience force of interactions is called electric charge.\u00a0<\/span><\/p>\n

\uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>The electrostatic force is much larger than gravitational force.<\/span><\/em><\/p>\n

\uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>The charge on an electron is -1.6<\/span>\"\"10<\/span>-19<\/sup><\/span>C &\u00a0<\/span><\/p>\n

\uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Charge on a proton is=+ 1.6<\/span>\"\"10<\/span>-19<\/sup><\/span>C.<\/span><\/p>\n

\uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Neutrons have no charge.<\/span><\/p>\n

3. Two kinds of charge:-\u00a0<\/span><\/strong><\/p>\n

In 1733 Du Fay discovered that charges are of two types. This can be shown by the following simple experiments.<\/span>\u00a0<\/span><\/p>\n

\"C:\\Users\\vartmaan\\AppData\\Local\\Microsoft\\Windows\\TemporaryExp-1.<\/span><\/em><\/strong>\u00a0<\/span><\/p>\n

When two glass rods rubbed with silk are brought near each other then they started repel each other.<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

Exp.2<\/span><\/em><\/strong>\u00a0<\/span><\/p>\n

When two ebonite rods rubbed with cat\u2019s fur are also brought near each other, then they also repel each other.<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

\u00a0<\/span>Exp.3:-<\/span><\/em><\/strong>\u00a0<\/span><\/p>\n

But when a glass rod & ebonite rod are brought near each other then they started attracting each other.<\/span><\/p>\n

    \n
  • The polarity differentiates the two kinds of charges.<\/span><\/li>\n
  • \u00a0<\/span>The charge on glass rod is called\u00a0<\/span>vitreous charge\u00a0<\/span><\/strong>(Latin vitrum = glass) & the charge on amber when rubbed with wool is called\u00a0<\/span>resinous charge<\/span><\/strong>\u00a0(Amber is resin)<\/span><\/li>\n<\/ul>\n

    Now according to Benjamin frankly, charges may define as<\/span><\/p>\n

      \n
    • Positive charge<\/span><\/strong>:-<\/span><\/li>\n<\/ul>\n

      The charge developed on glass rod when rubbed with silk is called + ve charge. Or if numbers of p are greater than number of e<\/span>–\u00a0<\/sup><\/span>in a body, then body is called +vely charged.<\/span><\/p>\n

        \n
      • Negative charge:-<\/span><\/strong><\/li>\n<\/ul>\n

        The charge developed on plastic rod when rubbed with wool is called negative charge. Or if number of e<\/span>–<\/sup><\/span>\u00a0is more than number of p<\/span>+<\/sup><\/span>\u00a0in a body, and then body is called negative charged.<\/span><\/p>\n

        E, g:-Two kind of charge developed on rubbing\u00a0<\/span><\/p>\n

        4. Conductor, insulator & dielectrics:-<\/span><\/strong><\/p>\n

        \u00a0<\/span><\/strong>Conductor:-\u00a0<\/span><\/strong><\/p>\n

        A substance which allows charge to move from one place to another place is called conductor. Silver is the best conductor. Other examples are copper, aluminum, iron mercury, earth, human body etc.<\/span><\/p>\n

        Insulator: –\u00a0<\/span><\/strong><\/p>\n

        A<\/span><\/strong>\u00a0substance which cannot be used to conduct electric charge is called insulator\u00a0<\/span>e<\/span><\/strong>, g: – glass, rubber, plastic, ebonite, mica, wax etc.\u00a0<\/span><\/p>\n

        Dielectric:-<\/span><\/strong>\u00a0\u00a0<\/span><\/strong><\/p>\n

        \"F:\\unitDielectrics are that insulator which does not conduct electricity but on Appling external electric field charge induces on it e, g: – glass rod & paper acquire charge on rubbing.<\/span><\/p>\n

        5<\/span><\/em><\/strong>. Gold leaf electroscope ( GLE )<\/span><\/strong>\u00a0<\/span><\/p>\n

        It is consist of a vertical conducting rod passing through a rubber stopper fitted in the mouth of glass vessel. Two thin gold leafs are attached to the lower end of the rod. When a charged object touches the metal knob at the outer end of the rod, the charge flow down throw the leaves. The leaves diverge (moves away) due to repulsion of the like charge. The degree of divergence of the leaves gives measure of the amount of charge.<\/span><\/p>\n

        \u00a0<\/span><\/p>\n

        6.\u00a0<\/span>Method of charging\u00a0<\/span><\/strong>imp<\/sup><\/span><\/strong><\/p>\n

        \u00a0\u00a0\u00a0<\/span>(i)<\/span>\u00a0\u00a0<\/span>Charging by friction:-<\/span><\/strong><\/p>\n

        When a body having loosely bounded electrons, is rubbed with a body having strongly bounded electrons, then both the bodies becomes charged by transfer of e<\/span>–<\/sup><\/span>.<\/span><\/p>\n

        E,g:- when glass rod is rubbed with silk then glass rod become +vely charged by transferring electron & silk become<\/span>\u00a0\u00a0<\/span>-vely charged by acquiring electron.<\/span><\/p>\n

        \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>By loosing electron mass of the body decreases & by gaining e<\/span><\/em>–<\/sup><\/span><\/em>\u00a0mass of body increases. As electron have mass 9.1<\/span><\/em>\"\"10<\/span><\/em>-32<\/sup><\/span><\/em>kg.<\/span><\/em><\/p>\n

        \u00a0\u00a0<\/span>(ii)\u00a0<\/span>Charging by electrostatic induction:-<\/span><\/strong><\/p>\n

        \"F:\\unitThe phenomenon of charging a neutral body by placing it in the neighboring of a charge body is called electrostatic<\/span><\/em>\u00a0induction.<\/span><\/p>\n

        When a positively charge body is bring toward a neutral body then charge separation takes place in neutral body which remains till the + vely charged body remains near to the neutral body. These induced charges may be explained in following ways.<\/span><\/p>\n

        (a) Charging by induction ( by earthing a conductor);-<\/span><\/strong><\/p>\n

        Suppose a neutral ball on an insulating stand & a positively charged glass rod is bringing toward it. Due to + ve charge on glass rod, the \u2013ve charge induce toward the glass rod in the ball &+ ve charge induce on opposite side of the<\/span><\/p>\n

        \"C:\\Users\\vartmaan\\Downloads\\Figure_19_02_03a.jpg\"<\/p>\n

        \u00a0ball. By earthing the + ve charge of the ball, we get \u2013 vely charged ball. Hence charges can by induced by earthing.<\/span><\/p>\n

        (b) Charging by induction ( By separating conductors)<\/span><\/strong><\/p>\n

        Suppose two metallic balls mounted on insulating stands are placed in contact with each other. When a + vely charged glass rod is bring toward the balls then the face of first ball toward rod become \u2013 ve& face of second ball away from the rod becomes + vely charged. After separating balls, the ball A acquire -ve charge & ball B acquire +ve charge as shown in fig.<\/span><\/p>\n

        Q.1<\/strong>\u00a0\u00a0\u00a0<\/strong>how can you charge a metal sphere positively without touching it?<\/strong><\/p>\n

        \"F:\\unitAns: By bringing a negatively charged body towared metal sphere.<\/span><\/p>\n

        (<\/span><\/strong>iii)<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>Charging by conduction (contact):-<\/span><\/strong>\u00a0<\/span><\/p>\n

        Charging by conduction requires the actual contact between the two bodies.<\/span><\/p>\n

        In case of gold leaf electroscope when glass rod rubbed with silk is touched to the knob of leaf, than leaf diverse from their actual position, which remains separated even after removal of glass rod. Thus charging may be done by conduction.<\/span><\/p>\n

        \"F:\\unit7.<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>Polar & Non Polar Bodies:-<\/span><\/strong><\/p>\n

          \n
        • Polar body:-\u00a0<\/span><\/strong><\/li>\n<\/ul>\n

          A body having different center of +ve& -ve charge is called polar body<\/span><\/em>.\u00a0<\/span><\/p>\n

          E, g: – HCl , H<\/span>2<\/sub><\/span>O etc.<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

            \n
          • Non polar body:-<\/span><\/strong><\/li>\n<\/ul>\n

            \u00a0<\/span>A body having same center of +ve& -ve charge is called non polar body<\/span><\/em>.<\/span>\u00a0\u00a0<\/span><\/p>\n

            E, g: – H<\/span>2<\/sub><\/span>, O<\/span>2<\/sub><\/span>, N<\/span>2<\/sub><\/span>, etc<\/span><\/p>\n

            \"F:\\57.jpg\"<\/p>\n

            \u00a0<\/span><\/p>\n

            \u00a0<\/span><\/p>\n

             <\/p>\n

             <\/p>\n

            Q.2. (a) A comb run through one\u2019s dry hair attracts small bits of paper. Why? What happens if the hair is wet or if it is a rainy day? (Remember, a paper does not conduct electricity.)\u00a0<\/span><\/strong><\/p>\n

            (b)Ordinary rubber is an insulator. But special rubber tyres of aircraft are made slightly conducting. Why is this necessary?\u00a0<\/span><\/strong><\/p>\n

            \u00a9 Vehicles carrying inflammable materials usually have metallic ropes touching the ground during motion. Why?<\/span><\/strong><\/p>\n

            (d) A bird perches on a bare high power line, and nothing happens to the bird. A man standing on the ground touches the same line and gets a fatal shock. Why?\u00a0<\/span><\/strong><\/p>\n

            8.<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>Some Basic Properties & Electric Charge:-<\/span><\/strong>\u00a0imp<\/sup><\/span><\/strong>\u00a0<\/span><\/strong><\/p>\n

            Charge is a fundamental & intrinsic property of matter. There are three basic properties of charge.<\/span><\/p>\n

            (i)<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>Quantization Of Charge:- (Discrete Nature of charge )<\/span><\/strong><\/p>\n

            According to quantization nature of charge, the charge on a body is always whole number multiple of charge on an electron.\u00a0<\/span><\/p>\n

            I,e<\/span> \u00a0<\/span>charge on a body<\/span><\/p>\n

            \u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0<\/span>n= 1, 2, 3, 4, \u2026\u2026\u2026but<\/span>\u00a0\u00a0<\/span>n\u2260 \u00bd, 3\/2, 1.5 etc.\u00a0<\/span><\/p>\n

            Where e = 1.6<\/span>\"\"10<\/span>\u201319<\/sup><\/span>C (charge on an electron) & n is a whole number\u00a0<\/span><\/p>\n

            \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>During rubbing, electron can transfer from one body to another body in a whole number. So charge is quantized as half electron cannot be transferred.<\/span><\/em><\/p>\n

            \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>An elementary particle quart of very small life time have fraction of charge.<\/span><\/em><\/p>\n

            \u00b1\u00a0<\/span><\/em>\"\"e & \u00b1\u00a0<\/span><\/em>\"\"\u00a0e. (it is exception of quantization of charge)<\/span><\/em><\/p>\n

            Q.3.<\/strong>\u00a0\u00a0<\/strong>If 10<\/strong>9<\/sup><\/span><\/strong>\u00a0electrons move out of a body to another body every second, how much time is required to get a total charge of 1 C on the other body?\u00a0<\/strong><\/p>\n

            Solution:\u00a0 charge given out in one second is= 1.6 \u00d7 10\u201319<\/sup><\/span> \u00d7 109<\/sup><\/span> C = 1.6 \u00d7 10\u201310<\/sup><\/span> C.<\/p>\n

            The time required to have a charge of 1 C is<\/p>\n

            \"\"= 6.25 \u00d7 109<\/sup><\/span> s = \"\"\u00a0years = 198 years.<\/p>\n

            \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>One coulomb is, therefore, a very large unit for many practical purposes.<\/p>\n

            \u00a0 Q.4. how much positive and negative charge is there in a cup of water?\u00a0<\/strong><\/p>\n

            Solution:\u00a0 assume that the mass of one cup of water is 250 g.<\/p>\n

            The molecular mass of water is 18g.\u00a0\u00a0\u00a0 so 18g of water contain 6.022×1023<\/sup><\/span> molecules<\/p>\n

            \u00a01g of water contain =\"\"\u00a0molecules.\u00a0 Therefore total molecules \"\" \u00d7 6.02 \u00d7 1023<\/sup><\/span><\/p>\n

            Each molecule of water contains 10 electrons and 10 protons.<\/p>\n

            \u00a0Hence the total positive or total negative charge = \"\" \u00d7 6.02 \u00d7 1023<\/sup><\/span> \u00d7 10 \u00d7 1.6 \u00d7 10\u201319<\/sup><\/span> C = 1.34 \u00d7 107<\/sup><\/span> C.<\/p>\n

            Q.5. A polythene piece rubbed with wool is found to have a negative charge of 3 \u00d7 10<\/strong>\u22127<\/sup><\/span><\/strong>\u00a0C. (a) Estimate the number of electrons transferred (from which to which?) (b) Is there a transfer of mass from wool to polythene?<\/strong><\/p>\n

            (a) Amount of charge on the polythene piece,<\/p>\n

            q = \u22123 \u00d7 10\u22127<\/sup><\/span> C\u00a0 And\u00a0 e = \u22121.6 \u00d7 10\u221219<\/sup><\/span> C<\/p>\n

            \u00a0 Number of electrons transferred from wool to polythene = n<\/p>\n

            \u00a0 As q = ne Therefore, the number of electrons transferred from wool to polythene is 1.87 \u00d7 1012<\/sup><\/span>.<\/p>\n

            \u00a0\u00a0\u00a0\u00a0<\/span>Yes. Because an electron has mass, me = 9.1 \u00d7 10\u22123<\/sup><\/span> kg.<\/p>\n

            Total mass transferred to polythene from wool,<\/p>\n

            m = me<\/sub><\/span> \u00d7 n = 9.1 \u00d7 10\u221231<\/sup><\/span> \u00d7 1.85 \u00d7 1012<\/sup><\/span> = 1.706 \u00d7 10\u221218<\/sup><\/span> kg.<\/p>\n

            (ii) Conservative nature of electric charge:-<\/span><\/strong><\/p>\n

            The total charge on an isolated system remains constant; it can neither be created nor be destroyed & can only be transferred from one body to another body.<\/span>\u00a0\u00a0<\/span>E,g:-<\/span><\/p>\n

            (i)<\/span>\"\"\u00a0Here charge on reactant is zero & charge on product = +1-1=0<\/span><\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>So law of conservation of charge hold good.<\/span><\/p>\n

            \u00a0\u00a0<\/span>(ii)<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0 <\/span>(Zero charge)<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>(-1)<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>+<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>(+1)<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>= zero charge\u00a0<\/span><\/p>\n

            (iii) Addition nature of electric charge:-<\/span><\/strong><\/p>\n

            The net charge on an isolated system can be simply obtained by adding all the charges scalarly.<\/span><\/p>\n

            If a system having charge<\/span><\/p>\n

            \"\",<\/span>\"\",\u00a0<\/span>\"\"\u00a0<\/sub><\/span>\u2026<\/span>\"\"\u00a0.<\/sub><\/span><\/p>\n

            Then total charge on the system<\/span><\/p>\n

            \"\".<\/sub><\/span><\/p>\n

            E.g. If system have four charges<\/span>\u00a0\u00a0<\/span><\/p>\n

            \u00a0<\/span>\"\"<\/p>\n

            Then total charge<\/span>\u00a0<\/span>\"\"<\/p>\n

            \u00a0<\/span><\/strong><\/p>\n

            9. Comparison of charge & mass:-<\/span><\/strong><\/p>\n

              \n
            • Electric Charge<\/span><\/strong><\/li>\n<\/ul>\n

              \u00a0<\/span>1. Electric charge may be + ve, -ve or zero.<\/span>\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

              2. Electric charge is always quantized.<\/span><\/p>\n

              3. Charge on a body does not depend on its speed.<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

              4. Charge is strictly conserved.<\/span><\/p>\n

              5. The electrostatic force may be attractive or repulsive.<\/span><\/p>\n

              6. Electrostatic force may cancel out.<\/span><\/p>\n

              7. A charged body always has some mass.<\/span><\/p>\n

              \u00a0<\/span><\/p>\n

                \n
              • Mass\u00a0<\/span><\/strong><\/li>\n<\/ul>\n

                1. The mass of a body is always a positive.<\/span><\/p>\n

                2. Quantization of mass is yet not obeyed.\u00a0<\/span><\/p>\n

                3. Mass of the body increase with its velocity.<\/span><\/p>\n

                4. Mass is not conserved as it may convert into energy.<\/span><\/p>\n

                5. A gravitational force is always attractive.<\/span><\/p>\n

                6. Gravitational forces never cancel out.<\/span><\/p>\n

                7. A body having mass may not have any charge.<\/span><\/p>\n

                10. Coulombs Law In Electrostatics \ud83d\ude41 Scalar Form)<\/span><\/strong>\u00a0M.Imp<\/sup><\/span><\/strong><\/p>\n

                In 1785, the French physicist Coulomb measured the electric force between small charged spheres by using a torsion balance & gave a law which is as below.<\/span><\/p>\n

                Coulombs law states that the force of interaction between two stationary charge is directly proportion to the product of magnitude of charges & inversely proportional to the square of distance between the charges.\u00a0<\/span><\/p>\n

                This force acts along the line joining among the charges.<\/span><\/p>\n

                \"F:\\58.jpg\"I, e<\/span>\u00a0\u00a0<\/span>F<\/span>\"\"\u00a0,.1<\/span><\/p>\n

                F<\/span>\"\"\u00a0\u00a0<\/span>…2<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\u00a0<\/span><\/p>\n

                Combining equations 1& 2\u00a0<\/span><\/p>\n

                We get F<\/span>\"\"<\/p>\n

                Or<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                Where k is constant of proportionality & its value is<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

                \u00a0<\/span>k =<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0= 9<\/span>\"\"10<\/span>9<\/sup><\/span>\u00a0Nm<\/span>2<\/sup><\/span>C<\/span>-2<\/sup><\/span><\/p>\n

                Here\u00a0<\/span>\"\"o\u00a0<\/sub><\/span>is called permittivity of free space & its value is<\/span><\/p>\n

                \u00a0<\/span>\"\"o<\/sub><\/span>= 8.854<\/span>\"\"10<\/span>-12<\/sup><\/span>C<\/span>2<\/sup><\/span>N<\/span>-1<\/sup><\/span>m<\/span>-2\u00a0<\/sup><\/span><\/p>\n

                So coulomb\u2019s law becomes<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

                F=<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>\"\"\u00a0\u00a0<\/span><\/strong><\/p>\n

                =<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>\"\"\u00a0<\/span><\/strong><\/p>\n

                =\u00a0<\/span><\/strong>9<\/span><\/strong>\"\"10<\/span><\/strong>9<\/sup><\/span><\/strong>\"\"<\/p>\n

                Unite of charge:-<\/span><\/strong><\/p>\n

                If<\/span>\u00a0\u00a0<\/span>F= 9<\/span>\"\"10<\/span>9<\/sup><\/span>N,\u00a0<\/span>\"\"\u00a0(say), and<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

                As we know F= 9<\/span>\"\"10<\/span>9<\/sup><\/span>\u00a0\u00a0<\/sup><\/span>\"\"<\/p>\n

                Or 9<\/span>\"\"10<\/span>9 =<\/sup><\/span>\u00a09<\/span>\"\"10<\/span>9\u00a0<\/sup><\/span>q<\/span>2<\/sup><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/strong><\/p>\n

                \u21d2<\/span>\"\"<\/p>\n

                Or<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                If two equal charges repel each other with a force of 9<\/span><\/em>\"\"10<\/span><\/em>9<\/sup><\/span><\/em>N when placed one meter distance apart then charge is said to be one coulomb.<\/span><\/em><\/p>\n

                \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\u00a0The c, g, s unit of charge is Stat Coulomb ( stat C ) or electrostatic unit of charge (emu )\u00a0<\/span><\/p>\n

                One stat coulomb is that charge which repel an identical charge in vacuum at distance of 1 cm with a force of 1 dyne<\/span><\/em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/em>\u00a0<\/span><\/p>\n

                1<\/span><\/strong>\u00a0Coulomb =3<\/span><\/strong>\"\"\u00a010<\/span><\/strong>9<\/sup><\/span><\/strong>\u00a0stat Coulomb (e m u)<\/span><\/strong><\/p>\n

                \uf0d8<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>In electromagnetic c, g, s unit of charge is abcoulomb’s or electromagnetic unit of charge\u00a0<\/span><\/p>\n

                (e m u)<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                1 coulomb =\u00a0<\/span>\"\" abcoulomb’s =\u00a0<\/span>\"\"\u00a0emu of charge.<\/span><\/p>\n

                  \n
                • Significance of coulombs law:-<\/span><\/strong><\/li>\n<\/ul>\n

                  \u00a0<\/span>(i) It tells us about the force which bounds the electrons around the nucleus to form a atom.<\/span><\/p>\n

                  \u00a0<\/span>(ii) It tells us about the force which binds the molecule to form solids & liquids\u00a0<\/span><\/p>\n

                    \n
                  • Limitations of coulomb\u2019s law:-<\/span><\/strong><\/li>\n<\/ul>\n

                    (i) Coulombs law is applicable only on point charges.<\/span><\/p>\n

                    \u00a0<\/span>(ii) It holds good only when charges are in rest only.<\/span><\/p>\n

                    11. Comparison between coulomb\u2019s force &Newton\u2019s gravitational force:-<\/span><\/strong><\/p>\n

                    Similarity<\/span><\/strong>\u00a0<\/span><\/em><\/strong><\/p>\n

                    \u00a0<\/span>(i) Both obey inverse square law.<\/span><\/p>\n

                    \u00a0<\/span>(ii) Both forces are central forces.<\/span><\/p>\n

                    \u00a0<\/span>(iii) Both forces are conservative forces.\u00a0<\/span><\/p>\n

                    \u00a0<\/span>(iv) Both forces are directly proportional to product of interacting Patrick<\/span><\/p>\n

                    Dissimilarity:-<\/span><\/strong><\/p>\n

                    \u00a0<\/span>(i)\u00a0<\/span>Coulomb\u2019s force is attractive as well as repulsive while Gravitational force is always attractive in nature.<\/span><\/p>\n

                    \u00a0<\/span>(ii) Coulomb\u2019s force is much stronger than gravitational force (10<\/span>36<\/sup><\/span>times)\u00a0<\/span><\/p>\n

                    \u00a0<\/span>(iii) Coulombs force depends upon the medium in which charges are placed while gravitational force does not depends upon medium.<\/span><\/p>\n

                    Q.6 : What is the force between two small charged spheres having charges of 2 \u00d7 10<\/span><\/strong>\u22127<\/sup><\/span><\/strong>\u00a0C and 3 \u00d7 10<\/span><\/strong>\u22127<\/sup><\/span><\/strong>\u00a0C placed 30 cm apart in air?\u00a0<\/span><\/strong><\/p>\n

                    Ans: F= 9<\/span>\"\"10<\/span>9<\/sup><\/span>\u00a0\u00a0<\/sup><\/span>\"\"<\/p>\n

                    \"\"<\/p>\n

                    \"\"<\/p>\n

                    Q.7 How much stronger is coulombs force from gravitational force between a electron & proton separated by r distance<\/span><\/strong>?<\/span><\/p>\n

                    \"\"<\/span>I.e .<\/span>\"\"<\/p>\n

                    \u21d2<\/span>\u00a0\u00a0<\/span>\"\"=\u00a0<\/span>\"\"\u00a0\u00a0\u00a0<\/span><\/p>\n

                    Or<\/span>\"\"\u00a0= 1.27<\/span>\"\"10<\/span>36<\/sup><\/span>\u00a0\u00a0<\/span><\/p>\n

                    Hence Fe is 10<\/span>36<\/sup><\/span>\u00a0time stronger then F<\/span>G<\/sub><\/span><\/p>\n

                    Q.8: The electrostatic force on a small sphere of charge 0.4 \u00b5C due to another small sphere of charge \u2212 0.8 \u00b5C in air is 0.2 N.\u00a0<\/span><\/strong><\/p>\n

                    (a) What is the distance between the two spheres?<\/span><\/strong><\/p>\n

                    \u00a0<\/span><\/strong>(b) What is the force on the second sphere due to the first?\u00a0<\/span><\/strong><\/p>\n

                    Ans: (a)<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                    (b) Both have equal force so F=0.2N<\/span><\/p>\n

                    Q.9 Two large conducting spheres carrying charges\u00a0<\/span><\/strong>\"\"\u00a0are kept with their centers r distance apart. The magnitude of electrostatic force between them is not exactly<\/span><\/strong>\u00a0<\/span>F=\u00a0<\/span><\/strong>\"\"\u00a0because:<\/span><\/strong><\/p>\n

                    A. are not point charges.\u00a0<\/span><\/p>\n

                    B. Charge distribution on the spheres is not uniform*.<\/span><\/p>\n

                    C. Charges on spheres will shift towards the centers of their respective spheres.<\/span><\/p>\n

                    D. Charges will shift towards the portions of the spheres which are closer and facing towards each other.<\/span><\/p>\n

                    12. Coulomb\u2019s Law in vector from: –<\/span><\/strong>\u00a0imp<\/sup><\/span><\/strong><\/p>\n

                    Suppose two charge q<\/span>1\u00a0<\/sub><\/span>& q<\/span>2<\/sub><\/span>\u00a0placed at A & B having position vectors<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0<\/span>&<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0<\/span>from origin O and charge q<\/span>1<\/sub><\/span>\u00a0exert\u00a0<\/span>\"\"\u00a0force on q<\/span>2\u00a0<\/sub><\/span>& q<\/span>2<\/sub><\/span>\u00a0exerts\u00a0<\/span>\"\"\u00a0force on q<\/span>1<\/sub><\/span>.\u00a0<\/span><\/p>\n

                    \"F:\\unitThen from coulomb\u2019s Law\u00a0<\/span><\/p>\n

                    \"\"\u00a0=<\/span>\u00a0\u00a0<\/span>\"\".<\/span>\"\"\u2026 1<\/span><\/p>\n

                    Where<\/span>\"\"\u00a0is a position vector acting from q<\/span>2<\/sub><\/span>\u00a0to q<\/span>1<\/sub><\/span><\/p>\n

                    Again from Coulomb\u2019s Law force on q<\/span>2<\/sub><\/span>\u00a0to q<\/span>1<\/sub><\/span><\/p>\n

                    \"\"\u00a0\u00a0<\/span>=\u00a0<\/span>\"\".\u00a0<\/span>\"\"\u20262<\/span><\/p>\n

                    Where\u00a0<\/span>\"\"is a position vector, which acts along q<\/span>1<\/sub><\/span>\u00a0from q<\/span>2<\/sub><\/span>.<\/span><\/p>\n

                    Clearly\u00a0<\/span>\"\"=- \"\"<\/p>\n

                    So from eq<\/span>n<\/sup><\/span>\u00a02 we get<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                    \"\"=\u00a0<\/span>\"\"<\/p>\n

                    Comparing\u00a0eq<\/span>n<\/sup><\/span>\u00a01 & 3 we get<\/span>\"\"<\/p>\n

                    \"\"\u00a0= \"\"<\/p>\n

                    Hence Coulomb\u2019s Law in vector from obeys Newton\u2019s third Law of motion.<\/p>\n

                    \u00a0Also from diagram\u00a0 \"\"\u00a0 (from\u00a0 \"\"\u00a0law of vector addition)<\/p>\n

                    &\"\"(from\u00a0 \"\"\u00a0law of vector addition)<\/p>\n

                    \u00a0So eqn<\/sup><\/span> 1 & 2 becomes<\/p>\n

                    \u00a0\"\"\u00a0= \"\"=\"\"\u00a0)<\/p>\n

                    Or\"\"=\u00a0<\/strong>\"\"\u00a0\u20264<\/p>\n

                    & similarly\u00a0 \"\"\u00a0=\u00a0<\/strong>\"\"\u00a0\u2026 5<\/p>\n

                    Equation 4 and 5 represents Coulombs Law in Position Vector Form<\/strong>.<\/p>\n

                    Q .10: Four point charges q<\/span><\/strong>A<\/sub><\/span><\/strong>\u00a0= 2 \u00b5C, q<\/span><\/strong>B<\/sub><\/span><\/strong>\u00a0= \u22125 \u00b5C, q<\/span><\/strong>C<\/sub><\/span><\/strong>\u00a0= 2 \u00b5C, and q<\/span><\/strong>D<\/sub><\/span><\/strong>\u00a0= \u22125 \u00b5C are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 \u00b5C placed at the centre of the square?<\/span><\/strong>\u00a0\u00a0<\/span><\/strong><\/p>\n

                    Ans: 0<\/span><\/p>\n

                    Q.11. (a) two small insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 \u00d7 10<\/strong>\u22127<\/sup><\/span><\/strong>\u00a0C?\u00a0<\/strong><\/p>\n

                    \u00a0<\/strong>(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?<\/strong><\/p>\n

                    \u00a0<\/strong>Ans:<\/strong>\u00a0\u00a0\u00a0<\/strong>(a)\u00a0<\/strong>force between the two spheres is 1.52 \u00d7 10\u22122<\/sup><\/span> N.<\/p>\n

                    (b) 0.243 N.<\/p>\n

                    12. Three equal charges,\u00a0<\/span><\/strong>\"\"\u00a0each, are held fixed at the three corners of an equilateral triangle of side 5 cm.\u00a0<\/span><\/strong><\/p>\n

                    Find the Coulomb force experienced by one of the charges due to the rest two.<\/span><\/strong><\/p>\n

                    Ans<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>24.9 N at 30\u00b0 with the extended sides from the charge<\/span>\u00a0<\/span><\/strong>under consideration<\/span><\/p>\n

                    13. Four equal charges\u00a0<\/span><\/strong>\"\" each are fixed at the four corners of a square of side 5 cm.\u00a0<\/span><\/strong><\/p>\n

                    Find the Coulomb force experienced by one of the charges due to the rest three.<\/span><\/strong><\/p>\n

                    Ans<\/span><\/strong>\u00a0\u00a0\u00a0\u00a0<\/span><\/strong>27.5 N at 45\u00b0 with the extended sides of the square from the charge under consideration<\/span><\/p>\n

                    13. Dielectric Constant or Relative Permittivity: –<\/span><\/strong>\u00a0M.Imp<\/sup><\/span><\/strong><\/p>\n

                    Permittivity is a property of the medium which determines the electric force between the charges situated in that medium<\/span><\/em>.<\/span><\/p>\n

                    \u00a0<\/span><\/strong>Relative permittivity:-<\/span><\/strong><\/p>\n

                    As we know coulomb\u2019s force between the two charge placed in vacuum is\u00a0<\/span><\/p>\n

                    \"\"=<\/span>\"\"\u00a0\u20261<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

                    \u00a0<\/span>Where \"\"o<\/sub><\/span> is the permittivity in free space<\/p>\n

                    Again force between the same charges, when placed in a medium is\u00a0<\/span><\/p>\n

                    \"\"=<\/span>\"\"\u00a0. \u20262<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

                    Here<\/span>\"\" is the permittivity in the medium<\/span><\/p>\n

                    Dividing eq<\/span>n<\/sup><\/span>\u00a01 & 2 we get<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

                    \"\"\u00a0\u00a0<\/span>=\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

                    =\u00a0<\/span>\"\"<\/p>\n

                    \"\"\u00a0or K<\/span>\u00a0\u00a0<\/span>(note: k=<\/span>\"\")<\/span><\/p>\n

                    Where<\/span>\u00a0\u00a0<\/span>\"\" is called relative permittivity.<\/span><\/p>\n

                    Hence\u00a0<\/span>relative permittivity may be defined as the ratio of force between two charges in vacuum to the force between the charges in medium<\/span><\/em>.<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Or<\/span><\/p>\n

                    \u00a0<\/span>Relative permittivity may be defined as the ratio of permittivity in the medium to the permittivity in free space.<\/span><\/p>\n

                    \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Hence coulombs law for material medium becomes<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                    F<\/span><\/strong>m<\/sub><\/span><\/strong>=<\/span><\/strong>\"\"<\/p>\n

                    Q.14. what will be the Coulombs force if two charges are placed in water?<\/span><\/strong>\u00a0\u00a0\u00a0\u00a0<\/span><\/strong><\/p>\n

                    Clearly<\/span>\u00a0\u00a0\u00a0<\/span>F<\/span>m<\/sub><\/span>=<\/span>\"\"\u00a0<\/span><\/p>\n

                    If charges are placed in water (<\/span>\"\"=80) then force between charges becomes\u00a0<\/span><\/p>\n

                    F<\/span>w<\/sub><\/span>=<\/span>\u00a0\u00a0<\/span>\"\"\u00a0.\u00a0<\/span><\/p>\n

                    Here we can see that force reduces by 80 in water as compare to vacuum<\/span><\/p>\n

                    14. Force between Multiple Charges:\u00a0<\/span><\/strong><\/p>\n

                    The superposition Principle:-<\/span><\/strong><\/p>\n

                    If there are a number of charges exerting force on a single charge, then total force on single charge will be equal to sum of all the forces exerted by individual charge on the charge.<\/span><\/em>\u00a0<\/span><\/p>\n

                    Suppose there are n charges q<\/span>1<\/sub><\/span>, q<\/span>2<\/sub><\/span>, q<\/span>3<\/sub><\/span>\u2026q<\/span>n<\/sub><\/span>\u00a0exerting force on a charge q<\/span>0<\/sub><\/span>.\u00a0<\/span><\/p>\n

                    Then from coulomb\u2019s law force on q<\/span>0<\/sub><\/span>\u00a0due to q<\/span>1<\/sub><\/span>\u00a0<\/span>\"F:\\unitis<\/span>\u00a0\u00a0<\/span><\/p>\n

                    \u00a0<\/span>\"\"\u00a0=<\/span>\"\"\u20261<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                    Again force on q<\/span>0<\/sub><\/span>\u00a0due to q<\/span>2<\/sub><\/span>\u00a0is<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\u00a0<\/span><\/p>\n

                    \"\"\u00a0=<\/span>\"\"\u20262<\/span><\/p>\n

                    Similarly<\/span><\/p>\n

                    \"\"\u00a0=<\/span>\"\"\u00a0\u00a0\u00a0\u00a0<\/span>\u2026<\/span>\u00a0\u00a0<\/span>3<\/span><\/p>\n

                    Similarly<\/span><\/p>\n

                    \"\"\u00a0\u00a0\u00a0<\/span>=<\/span>\"\"\u00a0\u00a0<\/span>\u2026.n<\/span><\/p>\n

                    Adding all the eq<\/span>n<\/sup><\/span>\u00a0s we get\u00a0<\/span><\/p>\n

                    F =\u00a0<\/span>\"\"\u00a0\u2026\u00a0<\/span>\"\"<\/p>\n

                    =<\/span>\"\"\u00a0<\/span><\/p>\n

                    \"\"<\/p>\n

                    \u00a0 So total force on a charge<\/p>\n

                    \u00a0\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>\"\"<\/p>\n

                    \uf076<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>In scalar form this expression may be written as<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

                    \"\"\u00a0\u00a0\u00a0\u00a0\u00a0 \"\"<\/p>\n

                    \uf076<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>In vector form<\/p>\n

                    \"\"\u00a0\u00a0\u00a0\u00a0\u00a0 \"\"<\/p>\n

                    15. Force on a point charge due to continuous charge distribution:-<\/span><\/strong><\/p>\n

                    \"F:\\unitA<\/span><\/em>\u00a0\u00a0<\/span><\/em>continuous charge distribution is a system of charge lying at infinitelly small distances from each other. There are three types of continuous charge distribution.<\/span><\/em><\/p>\n

                    (a) Linear or line charge distribution:-<\/span><\/strong>\u00a0<\/span><\/p>\n

                    If charges are arranged in such a way that they seem to be like a line then the distribution of charges is called line charge distribution.<\/span><\/p>\n

                    The line charge density\u00a0<\/span>\"\"\u00a0may be defined as the charge per unit length.<\/span><\/p>\n

                    I, e<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0<\/span><\/strong>=\u00a0<\/span><\/strong>\"\"\u00a0=\u00a0<\/span><\/strong>\"\"<\/p>\n

                    (b) Surface or area charge distribution:-\u00a0<\/span><\/strong><\/p>\n

                    \"F:\\unitIf charge are arranged in such a way that they seems to be a surface (plane) then the distribution of charge are called surface distribution of charge<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                    The surface charge density\u00a0<\/span>\"\"\u00a0may be defined as then charge per unit area of the conductor.<\/span><\/p>\n

                    I, e<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0=<\/span><\/em>\u00a0\u00a0<\/span><\/em>\"\"\u00a0\u00a0\u00a0<\/span><\/em>=\u00a0<\/em>\"\"<\/p>\n

                      \n
                    1. \u00a0\u00a0<\/span>Volume Charge distribution:-<\/li>\n<\/ol>\n

                      \"F:\\unitIf charges are arranged in such a way that they seems to be like a volume, then the distribution of the charges is called\u00a0<\/span>volume distribution of charge<\/span><\/em>.<\/span><\/p>\n

                      The volume charge density\u00a0<\/span>\u03c1 may be defined as the charge per unit volume of the conductor.<\/span><\/p>\n

                      i,e<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\u03c1<\/span>\u00a0\u00a0<\/span>=\u00a0<\/span>\"\"\u00a0<\/span> = \"\"<\/p>\n

                      16 force due to continuous distribution of charge:<\/span><\/strong><\/p>\n

                      (i)<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>Forces at a point due to continuous line distribution of charges<\/span><\/strong>:-<\/span><\/p>\n

                      Suppose a point P having charge\u00a0<\/span>\"\"\u00a0and r distance from a point O on continuous line distribution of charge. Than small force at placed at P due to small charge dq on small surface ds may be given by\u00a0<\/span><\/p>\n

                      \"F:\\unit\"\"\u00a0=\u00a0<\/em>\"\"\u00a0<\/span><\/em>\u00a0<\/span><\/em><\/p>\n

                      But<\/span>\u00a0\u00a0\u00a0<\/span>dq =\u00a0<\/span>\"\"d<\/span>l<\/span><\/em>\u00a0where\u00a0<\/span>\"\"\u00a0is line charge density.<\/span>\u00a0\u00a0<\/span>So<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      d<\/span>\"\"\u00a0=\u00a0<\/span>\"\"<\/p>\n

                      For complete length\u00a0<\/span><\/p>\n

                      \u00a0<\/span>\"\"=\u00a0<\/strong>\"\"\u00a0<\/span><\/p>\n

                      (ii)<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>Force due to continuous surface distribution of charge:-<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>\u00a0<\/span><\/p>\n

                      \"F:\\unitSuppose a test charge q<\/span>0<\/sub><\/span>\u00a0is placed at point p having r distance from small surface ds having charge dq. Then force on\u00a0<\/span>\"\"\u00a0is\u00a0<\/span><\/p>\n

                      \"\"\u00a0= \"\"\u00a0<\/span>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/span><\/p>\n

                      Where surface charge density<\/span><\/p>\n

                      \"\"=\u00a0<\/span>\"\"\u00a0\u00a0<\/span>dq =\u00a0<\/span>\"\"\u00a0ds<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\u00a0<\/span>\u00a0<\/span>\u00a0<\/span>\u00a0<\/span><\/p>\n

                      So<\/span>\u00a0\u00a0<\/span>\"\"\u00a0= \"\"\u00a0<\/span>\u00a0<\/span>\u00a0<\/span>\u00a0<\/span><\/p>\n

                      So the force at p due to complete surface is<\/span>\u00a0\u00a0\u00a0<\/span>\u00a0\u00a0<\/strong><\/p>\n

                      \"\"=\u00a0<\/strong>\"\"<\/p>\n

                      (iii) Force due to continuous volume distribution of charges.<\/span><\/strong><\/p>\n

                      \"F:\\unitSuppose a test charge\u00a0<\/span>\"\"is placed at point p having distance r from small volume dv having charge dq.<\/span>\u00a0<\/span><\/em><\/strong>Then force on\u00a0<\/span>\"\"due to dq<\/span>\"\"<\/p>\n

                      \"\"\u00a0= \"\"<\/p>\n

                      where volume charge density\u00a0<\/span><\/p>\n

                      \"\"=\u00a0<\/span>\"\"\u00a0\u00a0<\/span>dq =\u00a0<\/span>\"\"dv<\/span>\u00a0<\/span><\/p>\n

                      \"\"\u00a0= \"\"<\/p>\n

                      \u00a0<\/span>Hence force at p due to complete surface is\u00a0<\/span>\u00a0\u00a0<\/strong><\/p>\n

                      \u00a0<\/strong>\"\"=\u00a0<\/strong>\"\"<\/p>\n

                      Q.15 A uniformly charged conducting sphere of 2.5 m in diameter has a surface charge density of 100 \u00b5C\/m. Calculate the charge on the sphere.<\/span><\/strong><\/p>\n

                      \"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      Electric Field<\/span><\/strong><\/p>\n

                      17. Electric Field:-<\/span><\/strong><\/p>\n

                      The space around a positive charge in which its force of interaction may be experienced is called electric field<\/span><\/em>.<\/span><\/p>\n

                      Electric field intensity:-<\/span><\/strong>\u00a0M.Imp<\/sup><\/span><\/strong><\/p>\n

                      The force experienced by a unit + ve charge in the field of another charge is called electric field intensity. It is a vector quantity & denoted by\"\".<\/p>\n

                      I,e<\/span>\"\"<\/p>\n

                      Here the charge q<\/span>0\u00a0<\/sub><\/span>should be so small, that it does not disturb the position of original charge.\u00a0<\/span><\/p>\n

                      Also<\/span>\u00a0\u00a0<\/span>\"\"\u00a0 =\u00a0 \"\"\u00a0\u00a0<\/span><\/p>\n

                      =<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                      Or E\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/strong>=<\/span><\/strong>\"\"<\/p>\n

                      Unit &Dimensions:-<\/span><\/em><\/p>\n

                      As E=<\/span>\u00a0\u00a0\u00a0<\/span>\"\"=\u00a0<\/span>\"\"\u00a0=<\/span>\"\"<\/p>\n

                      \"\"& unit of E =\u00a0<\/span>\"\"\u00a0or\u00a0<\/span>volt per meter<\/span><\/em><\/strong>.<\/span><\/p>\n

                      \uf076<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>electric field is stronger near the charge & decreases exponentially with distance & becomes zero at<\/span><\/em>\"\"\u00a0<\/span><\/em><\/p>\n

                      Physical Significance Of Electric Field:-<\/span><\/strong>\u00a0imp<\/sup><\/span><\/strong>\u00a0<\/span><\/strong><\/p>\n

                      By knowing electric field at any point, we can determine force on a charge placed at that point.<\/span><\/p>\n

                      I.e<\/span>\"\"<\/p>\n

                      Thus electric field experience a intermediate role between two charge<\/span><\/p>\n

                      It is a characteristic of the system of charges & is independent of the test Charge that we place to determine the field.<\/span><\/em>\u00a0\u00a0\u00a0<\/span><\/em><\/p>\n

                      Q.16 A particle of charge\u00a0<\/span><\/strong>\"\"\u00a0and mass 1.6g is moving with a velocity\u00a0<\/span><\/strong>\"\"\u00a0.<\/span><\/strong><\/p>\n

                      At t =0 the particle enters in a region having an electric field<\/span><\/strong>\"\"\u00a0.\u00a0<\/span><\/strong><\/p>\n

                      Find velocity of the particle at t = 5 s.<\/span><\/strong><\/p>\n

                      Ans:\u00a0<\/span>\"\"<\/p>\n

                      As =<\/span>\"\"<\/p>\n

                      \"\"<\/p>\n

                      \"\"<\/p>\n

                      \"\"\u00a0<\/span><\/p>\n

                      And\u00a0<\/span>\"\"\u00a0So\u00a0<\/span>\"\"<\/p>\n

                      \"\"<\/p>\n

                      \"\"<\/p>\n

                      \"\"Q<\/span>.17: shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?<\/span><\/strong>\u00a0<\/span><\/p>\n

                      Answer: The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity.\u00a0<\/span><\/p>\n

                      Since the Deflection of particle 3 is the maximum, it has the highest charge to mass ratio<\/span><\/p>\n

                      Question 18: Two points charges of\u00a0<\/span><\/strong>\"\"and<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>\"\"\u00a0are kept 30cm apart. How far from the Charge\u00a0<\/span><\/strong>\"\"\u00a0on the line joining the two charges, will the net electric field be zero?<\/span><\/strong><\/p>\n

                      Ans: 10cm.<\/span><\/p>\n

                      Q. 19: Two point charges\u00a0<\/span><\/strong>\"\"\u00a0and\u00a0<\/span><\/strong>\"\"\u00a0are located 20 cm apart in vacuum.\u00a0<\/span><\/strong><\/p>\n

                      (a) What is the electric field at the midpoint O of the line AB joining the two charges?\u00a0<\/span><\/strong><\/p>\n

                      (b) If a negative test charge of magnitude 1.5 \u00d7 10<\/span><\/strong>\u22129<\/sup><\/span><\/strong>\u00a0C is placed at this point, what is the force experienced by the test charge?\u00a0<\/span><\/strong><\/p>\n

                      Ans:<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>(a)\u00a0<\/span><\/strong>The electric field at mid-point O is 5.4 \u00d7 10<\/span>6<\/sup><\/span>\u00a0N C<\/span>\u22121<\/sup><\/span>\u00a0along OB.\u00a0<\/span><\/p>\n

                      (b) A test charge of amount 1.5 \u00d7 10<\/span>\u22129<\/sup><\/span>\u00a0C is placed at mid-point O.<\/span>\u00a0\u00a0<\/span><\/p>\n

                      q = 1.5 \u00d7 10<\/span>\u22129<\/sup><\/span>\u00a0C\u00a0<\/span><\/p>\n

                      \"\"\u00a0= 1.5 \u00d7 10<\/span>\u22129\u00a0<\/sup><\/span>\u00d7 5.4 \u00d7 10<\/span>6<\/sup><\/span>\u00a0<\/span><\/p>\n

                      = 8.1 \u00d7 10<\/span>\u22123<\/sup><\/span>\u00a0N.\u00a0<\/span><\/p>\n

                      The force is directed along line OA.<\/span><\/p>\n

                      Question 20: An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 \u00d7 10<\/span><\/strong>4<\/sup><\/span><\/strong>\u00a0N C<\/span><\/strong>\u22121<\/sup><\/span><\/strong>\u00a0in Millikan\u2019s oil drop experiment. The density of the oil is<\/span><\/strong>\"\". Estimate the radius of the drop. (g = 9.81 m s<\/span><\/strong>\u22122<\/sup><\/span><\/strong>; e = 1.60 \u00d7 10<\/span><\/strong>\u221219<\/sup><\/span><\/strong>\u00a0C).<\/span><\/strong>\u00a0<\/span><\/p>\n

                      Ans:<\/span>\u00a0\u00a0<\/span>Excess electrons n = 12.<\/span>\u00a0\u00a0\u00a0<\/span>E = 2.55 \u00d7 10<\/span>4<\/sup><\/span>\u00a0N C<\/span>\u22121<\/sup><\/span>\u00a0<\/span><\/p>\n

                      And \u03c1 = 1.26 gm\/cm<\/span>3<\/sup><\/span>\u00a0= 1.26 \u00d7 10<\/span>3<\/sup><\/span>\u00a0kg\/m<\/span>3<\/sup><\/span>.g =9.81ms<\/span>\u22122<\/sup><\/span>.<\/span>\u00a0\u00a0<\/span>e = 1.6 \u00d7 10<\/span>\u221219<\/sup><\/span>\u00a0C .Radius of the oil drop = r<\/span><\/p>\n

                      Force (F) due to electric field E is equal to the weight of the oil drop F = W<\/span><\/p>\n

                      \"\"<\/p>\n

                      Where, q = Net charge on the oil drop = ne\u00a0<\/span><\/p>\n

                      m = Mass of the oil drop = Volume of the oil drop \u00d7 Density of oil\u00a0<\/span><\/p>\n

                      = 9.82 \u00d7 10<\/span>\u22124<\/sup><\/span>\u00a0kg\u00a0<\/span><\/p>\n

                      Therefore, the radius of the oil drop is<\/span><\/p>\n

                      \u00a0<\/span>\"\"= 9.82 \u00d7 10<\/span>\u22124<\/sup><\/span>\u00a0mm.<\/span><\/p>\n

                      18.<\/span>\u00a0Electric Field due to a point charge:-<\/span><\/strong><\/p>\n

                      Suppose a point charge q<\/span>0<\/sub><\/span>is placed at point p having r distance from a charge q placed at point o.<\/span><\/p>\n

                      Then according to coulomb\u2019s law the force of interaction between<\/span><\/p>\n

                      \"\"The charge is F=<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                      & the electric field at point p is\u00a0<\/span><\/p>\n

                      \"\"<\/p>\n

                      = \"\"\u00a0\u00a0<\/span><\/p>\n

                      =\u00a0<\/span>\"\"<\/p>\n

                      The magnitude of electric field is<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      E=<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>\"\"<\/p>\n

                      * Clearly E<\/span>\"\"\u00a0\u00a0<\/span>,\u00a0<\/span>it mesas E is same in all direction & does not depends upon direction of r.\u00a0<\/span><\/em><\/p>\n

                      Such a field is called\u00a0<\/span><\/em>spherically symmetric or radial field<\/span><\/em><\/strong>.<\/span><\/strong><\/p>\n

                      19. Electric field due to a system of point charges:-<\/span><\/strong><\/p>\n

                      Suppose q<\/span>0<\/sub><\/span>\u00a0charge is placed at point p having Distances<\/span>\"\"\u00a0,\u00a0<\/span>\"\"\u2026\u2026..<\/span>\"\". From charges q<\/span>1,\u00a0<\/sub><\/span>q<\/span>2<\/sub><\/span>\u2026\u2026\u2026. q<\/span>n..<\/sub><\/span>\u00a0Respectively.\u00a0<\/span><\/p>\n

                      Then total electric field at q<\/span>0<\/sub><\/span>\u00a0due to all charges may be calculated as given below.<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      \"F:\\unitElectric field at q<\/span>0<\/sub><\/span>\u00a0due to q<\/span>1<\/sub><\/span>\u00a0is\u00a0<\/span><\/p>\n

                      \"\"\u2026 1<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      Again electric field at q<\/span>0\u00a0<\/sub><\/span>due to q<\/span>2\u00a0<\/sub><\/span>charge is<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      \"\"\u2026 2<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      & Electric field at q<\/span>0<\/sub><\/span>\u00a0due to q<\/span>3<\/sub><\/span>\u00a0charge is<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      \"\"\u20263<\/span>\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      Similarly electric field at q<\/span>0\u00a0<\/sub><\/span>due to q<\/span>n\u00a0<\/sub><\/span>charge will be<\/span><\/p>\n

                      \"\"\u2026 n<\/span><\/p>\n

                      Now the total electric field at q<\/span>0\u00a0<\/sub><\/span>due to all charges will be<\/span><\/p>\n

                      E=<\/span>\"\"<\/p>\n

                      \"\"+\u2026\u2026.<\/span>\"\"<\/p>\n

                      Or<\/span>\u00a0\u00a0<\/span>E =\u00a0<\/span>\"\"<\/p>\n

                      Or<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>E<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>=<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>\"\"<\/p>\n

                      28. Electric field at a point due to continuous distributions of charge:-<\/span><\/strong><\/p>\n

                      (i)<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>Electric field due to continuous line distribution of charge:<\/span><\/strong><\/p>\n

                      \"F:\\unitThan small force at charge\u00a0<\/span>\"\"placed at P due to small charge dq on small surface ds may be given by coulomb\u2019s law,<\/span><\/p>\n

                      \u00a0<\/span>\"\"\u00a0\u00a0\u00a0 =\u00a0 \"\"\u00a0<\/span>\u00a0<\/span><\/p>\n

                      But<\/span>\u00a0\u00a0\u00a0<\/span>dq =\u00a0<\/span>\"\"dl where\u00a0<\/span>\"\"\u00a0is line charge density.\u00a0<\/span><\/p>\n

                      So<\/span>\u00a0\u00a0<\/span>d<\/span>\"\"\u00a0\u00a0<\/span>=<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                      For complete length<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      \"\"\u00a0\u00a0\u00a0<\/span>=\u00a0 \"\"<\/p>\n

                      So electric Field at point P may be as given as<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      \"\"\u00a0\u00a0<\/span>=<\/span>\"\"<\/p>\n

                      Q.28 : An infinite line charge produces a field of 9 \u00d7 10<\/strong>4<\/sup><\/span><\/strong>\u00a0N\/C at a distance of 2 cm. Calculate the linear charge density<\/strong>?<\/p>\n

                      Ans: the linear charge density is 10 \u00b5C\/m.<\/p>\n

                      (ii)<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>Electric field due to continuous surface distribution of charge:-<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>\u00a0<\/span><\/p>\n

                      \"F:\\unitSuppose a test charge q<\/span>0<\/sub><\/span>\u00a0is placed at point p having r distance from small surface ds having charge dq. Then force on\u00a0<\/span>\"\"\u00a0is<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                      \"\"\u00a0\u00a0\u00a0 = \"\"\u00a0<\/span>\u00a0<\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span>Where surface charge density<\/span>\"\"<\/p>\n

                      \"\"\u00a0\u00a0\u00a0<\/span>=<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0<\/span>dq =\u00a0<\/span>\"\"\u00a0ds<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\u00a0<\/span>\u00a0<\/span>\u00a0<\/span>\u00a0<\/span>\u00a0<\/span>\u21d2<\/span>\u00a0\u00a0<\/span>\"\"\u00a0= \"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      So the force at p due to complete surface is<\/span>\"\"<\/p>\n

                      \"\"=<\/strong>\u00a0\u00a0\u00a0<\/strong>\"\"<\/p>\n

                      Now electric field at point p is<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      \"\"\u00a0\u00a0<\/span>=<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                      (iii)<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>Electric field due to continuous volume distribution of charges<\/span><\/strong>.<\/span><\/strong><\/p>\n

                      Suppose a test charge\u00a0<\/span>\"\"is placed at point p having distance r from small volume dv having charge dq.<\/span><\/p>\n

                      \"F:\\unitThen force on\u00a0<\/span>\"\"due to dq<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      \"\"\u00a0\u00a0\u00a0 = \"\"<\/p>\n

                      Where volume charge density<\/span>\u00a0\u00a0<\/span><\/p>\n

                      \"\"=\u00a0<\/span>\"\"\u00a0\u00a0<\/span>dq =\u00a0<\/span>\"\"dv<\/span>\u00a0<\/span><\/p>\n

                      \"\"=\"\"<\/p>\n

                      Hence force at p due to complete surface is<\/span>\u00a0\u00a0<\/span><\/p>\n

                      F<\/strong>\u00a0\u00a0\u00a0<\/strong>=<\/strong>\u00a0\u00a0\u00a0<\/strong>\"\"<\/p>\n

                      So the electric field at point P will be<\/span><\/p>\n

                      \"\"\u00a0=<\/span>\"\"<\/p>\n

                      20.\u00a0<\/span>Electric field lines: –<\/span><\/strong>\u00a0M.Imp<\/sup><\/span><\/strong><\/p>\n

                      The electric field lines are the path followed by a unit positive charge when placed in the field of a charge. These lines are imaginary lines & tangent drawn on these lines gives the direction of electric field at that point<\/span><\/em>.<\/span><\/p>\n

                       <\/p>\n

                       <\/p>\n

                      Properties of electric field lines:-<\/span><\/strong><\/p>\n

                      \u00a0<\/span><\/p>\n

                      1. The electric lines of forces are continuous curves having no any breakages.<\/span><\/p>\n<\/div>\n

                      2. These lines originating from a +ve charge in Outward direction & originating from +ve charge is in inward direction.\u00a0<\/span><\/p>\n

                      \"\"3. The tangent drawn on an electric line of force gives the direction of electric field at that point.\u00a0<\/span><\/p>\n

                      \"C:\\Users\\vartmaan\\Downloads\\index.png\"4. No two electric lines of forces can intersect each other. Because if they do so, it means there are two direction of a single field at same point. This is not possible.\u00a0<\/span><\/p>\n

                      \"\"5. The electric field lines are always normal to the surface from which they originate or end.<\/span><\/p>\n

                      6. Larger the field lines in a region means more is electric field in that region & vice versa.<\/span><\/p>\n

                      \"\"7. These lines do not pass through a conductor because electric field inside the conductor is always zero.<\/span><\/p>\n

                      8. Electric field lines emits from a +ve charge & ends at \u2013ve charge.<\/span><\/p>\n

                      9. Electric lines of forces of two positive charges repel each other.<\/span><\/p>\n

                      21. Electric dipole:-<\/span><\/strong>\u00a0M.Imp<\/sup><\/span><\/strong>\u00a0<\/span><\/strong><\/p>\n

                      An electric dipole is consist of two equal & opposite Charge \u2013q & +q separated by a small distance 2a.<\/span><\/em><\/p>\n

                      Electric dipole moment:-<\/span><\/strong><\/p>\n

                      \"\"The product of either charge of electric dipole & the distance between the charges is called electric dipole moment. It is denoted by<\/span><\/em>\"\".\u00a0<\/span><\/em><\/p>\n

                      I.e.<\/span>\"\"<\/p>\n

                      \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Electric dipole moment is a vector quantity having direction from\u00a0<\/span><\/em>\u2013ve to +ve.<\/span><\/em><\/strong><\/p>\n

                      Unit &Dimensional formula:-<\/span><\/strong><\/p>\n

                      (i) The S.I unit of electric dipole is Coulomb metre (Cm)<\/span><\/p>\n

                      (ii) The practical unit of electric dipole is Debye (D)<\/span><\/p>\n

                      1D = 10<\/span>-10\u00a0<\/sup><\/span>Franklin<\/span>\"\"\u00a01A<\/span>\"\"<\/p>\n

                      =<\/span>\u00a0\u00a0<\/span>10<\/span>-10<\/sup><\/span>\"\"10<\/span><\/strong>-9<\/sup><\/span><\/strong>\"\"10<\/span>-10<\/sup><\/span>\u00a0\u00a0<\/span>(<\/span>\"\"\u00a01C = 3\u00a0<\/span>\"\"10<\/span>9\u00a0<\/sup><\/span>Fr)<\/span><\/p>\n

                      Or 1D = 3.33\u00a0<\/span>\"\"\u00a010<\/span>-30\u00a0<\/sup><\/span>Cm\u00a0<\/span><\/p>\n

                      Dimensional formula: – As<\/span>\u00a0\u00a0<\/span>P = q.2a = [AT] [L] = [M<\/span>0<\/sup><\/span>\"\"\u00a0TA]<\/span><\/p>\n

                      \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>An\u00a0<\/span>ideal dipole\u00a0<\/span><\/strong>has very small distance between the large charges.<\/span><\/p>\n

                      \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>An electric field produced by an electric dipole is called dipole is called\u00a0<\/span>dipole field.<\/span><\/strong><\/p>\n

                      Physical Significance of Electric dipole:<\/span><\/strong><\/p>\n

                      An electric dipole has a common existence in Nature. A molecule consist of a +ve& -ve charge is an electric dipole. It is used in study of effect of electric field on insulators & in study of radiations of energy from an antenna.<\/span><\/p>\n

                      Question: 21 Two particles with equal charge magnitudes\u00a0<\/span><\/strong>\"\"\u00a0\u00a0<\/span><\/strong>but opposite in signs are held 15 cm apart. What are the magnitude and direction of the Electric Field E at the point midway between the charges?<\/span><\/strong><\/p>\n

                      Ans<\/span>\u00a0\u00a0\u00a0<\/span>\"\" here<\/span>\"\"\u00a0\u00a0\u00a0<\/span><\/p>\n

                      =<\/span>\"\"\u00a0\u00a0<\/span>toward left<\/span><\/p>\n

                      Question 22: A system has two charges\u00a0<\/span><\/strong>\"\"\u00a0and\u00a0<\/span><\/strong>\"\"\u00a0<\/span><\/strong>\"\"\u00a0\u00a0<\/span><\/strong>C located at points A: (0, 0, \u2212 15 cm) and B: (0, 0, + 15 cm), respectively.\u00a0<\/span><\/strong><\/p>\n

                      What are the total charge and electric dipole moment of the system?<\/span><\/strong>\u00a0<\/span><\/p>\n

                      Answer: Distance between two charges at points A and B,<\/span><\/p>\n

                      \u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

                      Electric dipole moment of the system is given by,<\/span><\/p>\n

                      \"\"<\/p>\n

                      \"\"<\/p>\n

                      \"\"<\/p>\n

                      \"\"\u00a0<\/span><\/p>\n

                      along positive z-axis. Therefore, the electric dipole moment of the system is\u00a0<\/span>\"\"\u00a0along positive z\u2212axis.<\/span><\/p>\n

                      22. Electric Field at Axial Point of an electric dipole:-<\/span><\/strong>\u00a0imp<\/sup><\/span><\/strong>\u00a0<\/span><\/strong><\/p>\n

                      Suppose a test charge q<\/span>0<\/sub><\/span> is placed at point p having r distance from center of the dipole<\/span>\"\".\u00a0<\/span><\/p>\n

                      Now electric field at point p from charge \u2013q at A is<\/span><\/p>\n

                      \"C:\\Users\\vartmaan\\Downloads\\e2ddd556630163421024eb36883c7fe8a51aca7e.png\"\"\"\u00a0along\u00a0<\/span>\"\"\u2026.1<\/span><\/p>\n

                      & Electric field at P due to q charge at B is<\/span><\/p>\n

                      \"\"\u00a0\u00a0\u00a0<\/span><\/em>=\u00a0<\/span><\/em>\"\"\u2026<\/span><\/em>\u00a0along\u00a0<\/span>\"\"\u2026.2<\/span><\/em><\/p>\n

                      So the net electric field at P will be\u00a0<\/span><\/p>\n

                      \"\"\u00a0=<\/span><\/em>\u00a0\u00a0<\/span><\/em>\"\"\u00a0+\u00a0<\/span><\/em>\"\"<\/p>\n

                      =<\/span><\/em>\u00a0\u00a0<\/span><\/em>\"\"\u00a0\u00a0\u00a0<\/span><\/em><\/p>\n

                      =<\/span><\/em>\"\"\u00a0\u00a0<\/span><\/em>[<\/span>\"\"\u00a0–<\/span><\/em>\"\"]<\/span>\u00a0<\/span><\/em><\/p>\n

                      =<\/span><\/em>\"\"\u00a0<\/span><\/em><\/p>\n

                      =\u00a0<\/span><\/em>\"\"[<\/span>\"\"<\/p>\n

                      Or<\/span>\u00a0\u00a0\u00a0<\/span><\/em>E =<\/span><\/em>\"\"<\/p>\n

                      Or\u00a0<\/span>\u00a0\u00a0<\/span><\/em>E =<\/span><\/em>\u00a0\u00a0<\/span><\/em>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>(<\/span>\u2235<\/span><\/em>\"\")<\/span><\/p>\n

                      Or<\/span>\u00a0\u00a0<\/span>\"\"(<\/span><\/em>\u2235<\/span><\/em>\u00a0q.2a=P)<\/span><\/em><\/p>\n

                      In vector form<\/span>\"\"\u00a0\u00a0<\/span><\/em>=<\/span><\/em>\u00a0\u00a0<\/span><\/em>\"\"
                      \nI,e electric field is in direction of electric dipole moment.<\/span><\/p>\n

                      Special cases<\/span><\/strong><\/p>\n

                      \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>If dipole is small then\u00a0<\/span>\"\". Here a<\/span>2<\/sup><\/span>\u00a0can be neglected.\u00a0<\/span><\/p>\n

                      In this case\u00a0<\/span>\"\"\u00a0=\u00a0<\/span>\"\"\u00a0\u00a0<\/span>=<\/span>\"\"<\/p>\n

                      \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Electric field varies as\u00a0<\/span><\/em>\"\"\u00a0in case of electric dipole.<\/span><\/em><\/p>\n

                      \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>If point p lies near the dipole then we can take\u00a0<\/span><\/em>\"\"\u00a0=<\/span><\/em>\u00a0\u00a0<\/span><\/em>\"\"<\/p>\n

                      23<\/span><\/strong>.\u00a0<\/sub><\/span><\/strong>Electric field at an equatorial point of a dipole:-<\/span><\/strong>\u00a0M.Imp<\/sup><\/span><\/strong><\/p>\n

                      Suppose a test charge q<\/span>0\u00a0<\/sub><\/span>is placed at point p equatorial to electric dipole \u00b1 q.2a.<\/span><\/p>\n

                      \"\"The distance of p from Centre of the dipole o is r. As shown in fig. Now electric field E<\/span>A<\/sub><\/span>\u00a0at p due to \u2013q charge at A is toward\u00a0<\/span>\"\"& electric field at p due to +q charge at B is toward<\/span>\"\".<\/span><\/p>\n

                      Such that\u00a0<\/span>\"\"<\/p>\n

                      \"\"\u00a0\u00a0<\/span>Toward\u00a0<\/span>\"\"\u20261<\/span><\/p>\n

                      &<\/span>\"\"<\/p>\n

                      \"\"\u00a0toward\u00a0<\/span>\"\"..2<\/span><\/p>\n

                      Cleary from equation (1) & (2)<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      \"\"<\/p>\n

                      Now resoling<\/span>\"\"\u00a0<\/span>into rectangular components,\u00a0<\/span><\/p>\n

                      We can see that\u00a0<\/span>\"\"sin<\/span>\u03b8\u00a0<\/span>&\u00a0<\/span>\"\"\u00a0sin<\/span>\"\" are equal and opposite so cancel out.\u00a0<\/span><\/p>\n

                      Net electric field is<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      E =<\/span>\"\"\u00a0cos\u03b8 +\u00a0<\/span>\"\"\u00a0cos\u03b8<\/span>\u00a0\u00a0<\/span><\/p>\n

                      Or E =<\/span>\"\"\u00a0cos\u03b8…3<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>(<\/span>\u2235<\/span>\"\"\u00a0\u00a0<\/span>)<\/span><\/p>\n

                      Here<\/span>\"\"\u00a0=\u00a0<\/span>\"\"\u00a0\u00a0\u00a0<\/sub><\/span>& cos\u03b8 =<\/span>\"\"<\/p>\n

                      So from equation (3)<\/span><\/p>\n

                      \u00a0<\/span>E<\/span>\u00a0\u00a0\u00a0<\/span>=<\/span>\"\"\u00a0\u00a0<\/span>cos\u03b8\u00a0<\/span><\/p>\n

                      = 2<\/span>\"\"<\/p>\n

                      \"\"\u00a0<\/sub><\/span><\/p>\n

                      Or<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                      In vector from\u00a0<\/span>\"\"<\/p>\n

                      Here \u2013ve sign indicate that electric field is in opposite direction to<\/span><\/em>\"\".\u00a0<\/span><\/em><\/p>\n

                      *<\/span>If electric dipole is small then a<<r.\u00a0<\/span><\/em><\/p>\n

                      Then electric field becomes<\/span><\/em>\"\"<\/p>\n

                      \"\"<\/p>\n

                      Clearly<\/span>\u00a0\u00a0<\/span>E\u00a0<\/span>\u221d<\/span>\"\"<\/p>\n

                      *Relation between\u00a0<\/span><\/strong>\"\"\u00a0axial &<\/span><\/strong>\"\"\u00a0for a short dipole<\/span><\/strong><\/p>\n

                      \"\"\u00a0\u00a0<\/span>=\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0<\/sub><\/span>= 2\u00a0<\/span><\/p>\n

                      Or<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                      \n

                      *<\/span>In case of point p lying nears the dipole we can take<\/span><\/em><\/p>\n

                      \u00a0<\/span><\/em>\"\"<\/p>\n

                      24. Electric Field at any Point of electric dipole: –<\/span><\/strong>\u00a0imp<\/sup><\/span><\/strong>\u00a0\u00a0<\/span><\/strong><\/p>\n

                      Suppose a test charge lying at any point P having distance r from the Centre of the dipole.\u00b1 q.2a inclined at an angle \u03b8.<\/span><\/p>\n

                      \u00a0<\/span>Now to calculate electric field at point P due to dipole, resolve dipole into two components, such that point p lies at axial &equatorial points of the components. <\/span>Again suppose that point P lies at axial point of dipole A\u2019B\u2019 &equatorial point of dipole A\u2019\u2019B\u2019\u2019.<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      Then Net electric field at point P is\u00a0<\/span><\/p>\n

                      \"D:\\New\u00a0<\/span>\"\"=<\/span>\u00a0\u00a0<\/span>\"\"\u2026..1<\/span><\/p>\n

                      Here for sort dipole<\/span><\/p>\n

                      \u00a0<\/span>\"\"\u00a0=<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                      \u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0<\/span>=\u00a0<\/span>\"\"<\/p>\n

                      So<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0=\u00a0<\/span>\"\"<\/p>\n

                      Or\u00a0<\/span>\"\"=<\/span>\u00a0\u00a0\u00a0<\/span>\"\".(<\/span>\"\")<\/span><\/p>\n

                      Or<\/span>\u00a0\u00a0<\/span>\"\"\u00a0=\u00a0<\/span>\"\"(<\/span>\"\"<\/p>\n

                      Or<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0<\/span>=<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                      Special cases<\/span><\/strong><\/p>\n

                      \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>If\u00a0<\/span>point P lies at axial point of electric dipole<\/span><\/em>\u00a0\u00a0<\/span><\/em>then \u03b8 =0<\/span><\/em>0\u00a0<\/sup><\/span><\/em>\u21d2<\/span>\u00a0\u00a0\u00a0<\/span>cos0<\/span><\/em>0<\/sup><\/span><\/em>=1<\/span><\/em><\/p>\n

                      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\u21d2<\/span>\u00a0\u00a0<\/span>\u00a0<\/span>E<\/span><\/strong>\u00a0\u00a0\u00a0\u00a0<\/span><\/strong>=<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>\"\"<\/p>\n

                      \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>If point P lies at equatorial point of electric dipole then \u03b8 =90<\/span><\/em>0<\/sup><\/span><\/em>\u00a0\u00a0<\/sup><\/span><\/em>\u21d2<\/span><\/em>\u00a0\u00a0\u00a0\u00a0<\/span><\/em>cos90<\/span><\/em>0<\/sup><\/span><\/em>= 0<\/span><\/em>\u00a0\u00a0<\/span><\/em><\/p>\n

                      \u21d2\u00a0<\/span>\u00a0<\/span>E<\/span><\/strong>\u00a0\u00a0\u00a0\u00a0<\/span><\/strong>=<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>\"\"<\/p>\n

                      Q23.A point charge is situated at an axial point of a small electric dipole at a large distance from it. The charge experiences a force F. If the distance of the charge is doubled, the force acting on the charge will become:<\/span><\/strong><\/p>\n

                        \n
                      1. \"\"<\/li>\n<\/ol>\n

                        \"\"Q 24<\/strong>: Two charges 10 \u00b5C and \u221210 \u00b5C are placed at points A and B separated by a distance of 10 cm. Find the electric field at a point P on the perpendicular bisector of AB at a distance of 12 cm from its middle point.<\/span><\/strong><\/p>\n

                        Ans:<\/span><\/em>\u00a0\u00a0<\/span><\/em>\"\"<\/p>\n

                        \u00a0<\/span><\/strong><\/p>\n

                        Q 25: Two charges 10 \u00b5C and \u221210 \u00b5C are placed at points A and B separated by a distance of 10 cm. Find the electric field at a point P on the perpendicular bisector of AB at a distance of 12 cm from its middle .point.<\/span><\/strong><\/p>\n

                        Solution:\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

                        25. Torque on a electric dipole placed in uniform two dimensional electric field:-<\/span><\/strong>\u00a0imp<\/sup><\/span><\/strong><\/p>\n

                        Suppose a electric dipole \u00b1q.2a is placed in a uniform two dimensional electric field E. Then dipole experience two equal & opposite forces in two directions which results into torque on the dipole.<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                        \"\"\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                        \"\"\u00a0<\/span><\/p>\n

                        Where<\/span>\"\"<\/p>\n

                        So<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\u03c4 = qE. 2a\u00a0<\/span>\"\"<\/p>\n

                        Or<\/span>\u00a0\u00a0<\/span>\u03c4<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>=<\/span>\u00a0\u00a0\u00a0<\/span>PE<\/span>\"\"<\/p>\n

                        In vector from<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0<\/span>=\u00a0<\/span>\"\"\u00d7<\/span>\"\"<\/p>\n

                        The direction of<\/span>\"\"\u00a0\u00a0\u00a0<\/span>can be giving by\u00a0<\/span>right hand screw rule<\/span><\/em><\/strong>.\u00a0<\/span><\/p>\n

                          \n
                        • Torque<\/span><\/em>\u00a0has direction\u00a0<\/span>\u22a5<\/span>\u00a0to place containing<\/span>\"\".<\/span><\/li>\n<\/ul>\n

                          \u00a0<\/span>When dipole align to direction of electric field\u00a0<\/span><\/p>\n

                          then\u00a0<\/span>\"\"\u00a0= 0<\/span><\/p>\n

                          \u21d2<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0\u00a0<\/span><\/p>\n

                          So<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\u03c4 = o\u00a0<\/span><\/p>\n

                            \n
                          • When dipole lins<\/span>\u22a5<\/span>\u00a0to direction field\u00a0<\/span><\/li>\n<\/ul>\n

                            then<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\"\"= 90\u00b0<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                            \u27f9<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                            \u21d2<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>\u03c4 = PE (max)\u00a0<\/span><\/p>\n

                              \n
                            • If\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0<\/span>&<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/li>\n<\/ul>\n

                              Then<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\u03c4 = P<\/span><\/p>\n

                              Hence dipole moment may be defined as torque acting on an electric dipole placed\u00a0<\/span><\/em>\u22a5<\/span><\/em>\u00a0to electric field of unit strength<\/span><\/em>.<\/span><\/p>\n

                              \uf076<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>Net force on electric dipole in uniform electric field is zero but<\/span><\/em>\u00a0\u00a0<\/span><\/em>\u03c4\u22600<\/span><\/em><\/p>\n

                              26. Potential Energy of Electric Dipole in Uniform Two dimensional Electric field:-<\/span><\/strong>\u00a0imp<\/sup><\/span><\/strong>\u00a0<\/span><\/strong><\/p>\n

                              Suppose an electric dipole is placed at an angle<\/span>\"\"\u00a0in a uniform two dimensional Electric field. Then torque experienced\u00a0<\/span>by electric dipole is<\/span><\/p>\n

                              \u03c4<\/span>\u00a0\u00a0<\/span>=<\/span>\u00a0\u00a0<\/span>PE<\/span>\"\"<\/p>\n

                              Now suppose we want to rotate the dipole through small angle d<\/span>\"\". Then amount of work done to rotate the dipole is<\/span><\/p>\n

                              \"\"<\/p>\n

                              Or\u00a0<\/span>\"\"<\/p>\n

                              Now to calculate total work done to rotate dipole from\u00a0<\/span>\"\"1\u00a0<\/sub><\/span>to\u00a0<\/span>\"\"2<\/sub><\/span>\u00a0is\u00a0<\/span><\/p>\n

                              \"\"W =\u00a0<\/span>\"\"d<\/span>\"\"\u00a0<\/span><\/p>\n

                              = PE<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

                              = PE<\/span>\"\"<\/p>\n

                              = – PE\u00a0<\/span>\"\"<\/p>\n

                              W = PE<\/span>\"\"<\/p>\n

                              \u00a0<\/span>This work is store in the form of potential energy\u00a0<\/span><\/p>\n

                              so<\/span>\u00a0\u00a0<\/span>\u00a0<\/span><\/strong>U = PE\u00a0<\/span><\/strong>\"\"<\/p>\n

                              27. Stable and Unstable Equilibrium:-\u00a0<\/span><\/strong><\/p>\n

                              In case (a):<\/span><\/strong>When the angle between the dipole and electric field will be zero Then, the potential energy of the dipole will be equal to the<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                              \u00a0<\/span>\"\"\u00a0<\/span><\/p>\n

                              \"\"<\/p>\n

                              \"https:\/\/byjus-answer-creation.s3.amazonaws.com\/uploads\/9569_Physics_62a482045eb502df78a587a8_EL_Dipole_And_Dipole_Moment-22_022089.jpg_img_upload_solution_2022-08-04%2007:44:40.469116.png\"And\u00a0\u00a0 \u03c4 =\u00a0<\/em>\"\"<\/p>\n

                              \"\"<\/p>\n

                              We can say that the dipole will be in stable equilibrium.\u00a0<\/span><\/em><\/p>\n

                              In case (b):<\/span><\/strong>\u00a0<\/span>When the angle between the dipole and electric field will be<\/span>\u00a0<\/span>180 Then<\/span>, the potential energy of the dipole will be equal to the<\/p>\n

                              \"\" \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                              And\u00a0\u00a0\u00a0 \u03c4\u00a0\u00a0\u00a0<\/em>=<\/em>\u00a0\u00a0\u00a0<\/em>\"\"<\/p>\n

                              \"\"<\/p>\n

                              We can say that the dipole will be in an unstable equilibrium.<\/em><\/p>\n

                                \n
                              • a dipole is said to be in stable equilibrium when the angle between the electric field and dipole moment is zero<\/span><\/li>\n<\/ul>\n
                                  \n
                                • The dipole’s moment is directed between the -ve and +ve charges. In equilibrium, the system’s net force and torque should be zero.<\/span><\/li>\n
                                • As a result, if potential energy is high, the equilibrium will be unstable, whereas if potential energy is low, the equilibrium will be stable.<\/span><\/li>\n<\/ul>\n

                                  Question 26: An electric dipole with dipole moment 4 \u00d7 10<\/strong>\u22129<\/sup><\/span><\/strong>\u00a0C m is aligned at 30\u00b0 with the direction of a uniform electric field of magnitude 5 \u00d7 10<\/strong>4<\/sup><\/span><\/strong>\u00a0N C<\/strong>\u22121<\/sup><\/span><\/strong>.\u00a0<\/strong><\/p>\n

                                  Calculate the magnitude of the torque acting on the dipole?<\/strong><\/p>\n

                                  Ans:<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>\u03c4<\/span>\u00a0\u00a0\u00a0<\/span>=\u00a0<\/span>10\u22124<\/sup><\/span> N m.<\/p>\n

                                  Question 27:\u00a0<\/strong>An electric dipole of length 4 cm, when placed with its axis making an angle of 60 with a uniform electric field, experiences a torque of\u00a0<\/span><\/strong>\"\"\u00a0Nm. Calculate the potential energy of the dipole, if it has charge\u00a0<\/span><\/strong>\"\"<\/p>\n

                                  Ans:<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>\u03c4 =<\/span>\u00a0\u00a0<\/span>pE<\/span>\"\"<\/p>\n

                                  \"\"<\/p>\n

                                  \"\"<\/p>\n

                                  29.<\/span>\u00a0Electric Field at a point on the axis of charged circular ring:-<\/span><\/strong>\u00a0\u00a0<\/span><\/strong><\/p>\n

                                  Suppose a point P having x distance from centre of charged circular ring. Now small electric field dE at point P, due to small charge dq of length d<\/span>l \u2018<\/span>is as shown in fig.<\/span><\/p>\n

                                  \"C:\\Users\\pc\\Desktop\\download.png\"Here we can see that the components\u00a0<\/span>\"\"are canceled out.\u00a0<\/span><\/p>\n

                                  The net electric field at P is\u00a0<\/span><\/p>\n

                                  E =\u00a0<\/span>\"\"\u2026\u2026\u2026\u2026(1)<\/span><\/p>\n

                                  Here<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                                  \u00a0<\/span>dE =\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                                  &<\/span>\u00a0\u00a0\u00a0<\/span>cos\u03b8<\/span>\u00a0\u00a0<\/span>=\u00a0<\/span>\"\"<\/p>\n

                                  So from (1)\u00a0<\/span><\/p>\n

                                  E<\/span>\u00a0\u00a0<\/span>=<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                                  =\u00a0<\/span>\"\"<\/p>\n

                                  As q is uniformly distribution.\u00a0<\/span><\/p>\n

                                  The charge\u00a0<\/span>\"\"= +<\/span>\"\".\u00a0<\/span>\"\"<\/p>\n

                                  E =\u00a0<\/span>\"\"<\/p>\n

                                  Or E<\/span>\u00a0\u00a0\u00a0<\/span>=<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00a0\u00a0\u00a0<\/span>(<\/span>\u2235<\/span>\"\"= 2<\/span>\u03c0<\/span>a)<\/span><\/p>\n

                                  \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>If p lies for away then<\/span><\/em>\u00a0\u00a0<\/span><\/em>a<\/span><\/em>\u00a0\u00a0<\/span><\/em><<<\/span><\/em>\u00a0\u00a0<\/span><\/em>r\u00a0<\/span><\/em><\/p>\n

                                  Then E =<\/span><\/em>\u00a0\u00a0<\/span><\/em>\"\"<\/p>\n

                                  This is same as in case of electric field due to point charge.<\/span><\/em><\/p>\n

                                  30. Area vector:-<\/span><\/strong><\/p>\n

                                  \"F:\\unitIn physics; we need to know not only the magnitude of a surface area but also its direction. The area vector is taken always perpendicular to the surface.\u00a0<\/span><\/p>\n

                                  As\u00a0<\/span>\"\"\u00a0=\u00a0<\/span><\/em>\"\"\u00a0ds<\/span><\/em><\/p>\n

                                  Here<\/span>\"\"\u00a0is a unit vector\u00a0<\/span>\u22a5\u00a0<\/span>to the surface.<\/span><\/p>\n

                                  31. Electric Flux:-<\/span><\/strong>\u00a0imp<\/sup><\/span><\/strong><\/p>\n

                                  The word flux comes from the Latin word \u2018fluere\u2019 means \u2018to flow\u2019.<\/span>\u00a0<\/span><\/em><\/p>\n

                                  Thus electric flux is a measure of flow of electric field through a surface. Or<\/span><\/p>\n

                                  \u00a0<\/span>Electric flux may be defined as the number of electric lines passing through a given area. It is a scalar quantity & denoted by<\/span><\/em>\"\".<\/span><\/em><\/p>\n

                                  \"\"Suppose a area element<\/span>\"\"\u00a0is placed in a uniform electric field\u00a0<\/span>\"\"\u00a0making at angle<\/span>\"\".<\/span><\/p>\n

                                  Then<\/span>\"\"\u00a0=\u00a0<\/span><\/em>\"\"\u00a0\u00a0<\/span><\/em><\/p>\n

                                  or\u00a0<\/span><\/em>\u00a0<\/span>\"\"<\/p>\n

                                  Unit of\u00a0<\/span><\/em>\"\"\u00a0= NC<\/span>-1<\/sup><\/span>\u00d7 m<\/span>2 =<\/sup><\/span>\u00a0Nm<\/span>2<\/sup><\/span>C<\/span>-1<\/sup><\/span><\/p>\n

                                  Special Cases:-\u00a0<\/span><\/strong><\/p>\n

                                    \n
                                  • \u00a0<\/span><\/em>If area element is placed in plane of electric field then<\/span><\/em>\u00a0area vector become perpendicular to electric field.<\/span><\/em>\u00a0\u00a0<\/span><\/em><\/li>\n<\/ul>\n

                                    \u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/em>So\u00a0<\/span><\/em>\"\"<\/p>\n

                                      \n
                                    • If area element is placed\u00a0<\/span><\/em>\u22a5 to plane containing E. Then area vector become parallel to electric field.\u00a0<\/span><\/em><\/li>\n<\/ul>\n

                                      \u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/em>\"\"\u00a0= 0<\/span><\/em>0<\/sup><\/span><\/em>\"\"\u00a0cos 0 = max =1\u00a0<\/span><\/em><\/p>\n

                                      \u00a0\u00a0\u00a0\u00a0<\/span><\/em>\"\"\u00a0= Eds (max)<\/span><\/em><\/p>\n

                                      Question 29:\u00a0<\/strong>If the electric flux entering and leaving a closed surface in air<\/span><\/strong>\"\"are<\/span><\/strong>\"\"\u00a0respectively, the net electric charge enclosed within the surface is ________.<\/span><\/strong><\/p>\n

                                      Ans: If the electric flux entering and leaving a closed surface in air are\u00a0<\/span>\"\"\u00a0respectively, the net electric charge enclosed within the surface is\u00a0<\/span>\"\"<\/p>\n

                                      Question 30 :\u00a0<\/strong>Given a uniform electric field:<\/span><\/strong>\"\", find the flux of this field through a square of 5cm on a side whose plane is parallel to the y-z plane. What would be the flux through the same square if the plane makes a<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>\"\"\u00a0angle with the x-axis?<\/span><\/strong><\/p>\n

                                      Ans:<\/span><\/strong>\u00a0\u00a0\u00a0<\/span><\/strong>\"\"<\/p>\n

                                      \"\"<\/p>\n

                                      Q.31 : Consider a uniform electric field E=3\u00d710<\/span><\/strong>3<\/sup><\/span><\/strong>\u00eeN\/C.\u00a0<\/span><\/strong><\/p>\n

                                      (a) What is the flux of this field through a square of 10cm on a side whose plane is parallel to the yz plane?<\/span><\/strong><\/p>\n

                                      \u00a0<\/span><\/strong>(b) What is the flux through the same<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>square if the normal to its plane makes a 60\u00b0 angle with the x; axis?<\/span><\/strong><\/p>\n

                                      Ans (a) \u03a6 = = 3 \u00d7 10<\/span>3<\/sup><\/span>\u00a0\u00d7 0.01 \u00d7 cos0\u00b0 = 30 N m<\/span>2<\/sup><\/span>\u00a0\/C<\/span><\/p>\n

                                      \u00a0<\/span>(b) Plane makes an angle of 60\u00b0 with the x; axis. Hence, \u03b8=60\u00b0<\/span>\u00a0\u00a0\u00a0<\/span><\/p>\n

                                      Flux, \u03a6=3\u00d710<\/span>3<\/sup><\/span>\u00a0\u00d70.01\u00d7cos60\u00b0=15Nm<\/span>2<\/sup><\/span>\/C<\/span><\/p>\n

                                      Question 32: What is the net flux of the uniform electric field of above question through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?<\/span><\/strong><\/p>\n

                                      Answer: All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero.<\/span><\/p>\n

                                      Question 33:<\/em><\/strong> In a region of space the electric field is given by<\/strong>\"\". Calculate the<\/strong>\u00a0\u00a0<\/strong><\/p>\n

                                      electric flux<\/strong>\u00a0\u00a0<\/strong>through<\/strong>\u00a0\u00a0<\/strong>a Surface of area 100 units in\u00a0<\/strong>x-y<\/em><\/strong>\u00a0plane.\u00a0<\/strong><\/p>\n

                                      Solution:<\/em><\/strong><\/p>\n

                                      A surface of area 100 units in the x<\/span><\/em>y<\/em> plane is represented by an area vector \"\"\u00a0=100 \"\"<\/p>\n

                                      (Direction along the Normal to the area)<\/p>\n

                                      \u00a0The electric flux through the surface is given by<\/p>\n

                                      \u00a0<\/strong>\"\"\u00a0<\/em><\/strong>=<\/em>\u00a0<\/em><\/strong>\"\"\u00a0= 300 units\u00a0<\/strong><\/p>\n

                                      Question 34:<\/em><\/strong> Calculate the electric flux through a cube of side\u2018<\/strong>a<\/em><\/strong>\u2019 as shown, where\u00a0<\/strong>E<\/em><\/strong>x<\/sub><\/span><\/em><\/strong>\u00a0=\u00a0<\/strong><\/p>\n

                                      bx<\/em><\/strong>1\/2<\/sup><\/span><\/strong>;\u00a0<\/strong>E<\/em><\/strong>y<\/sub><\/span><\/em><\/strong>\u00a0=\u00a0<\/strong>E<\/em><\/strong>z<\/sub><\/span><\/em><\/strong>\u00a0= 0,\u00a0<\/strong>a<\/em><\/strong>\u00a0= 10 cm and\u00a0<\/strong>b<\/em><\/strong>\u00a0= 800 N\/C-m<\/strong>1\/2<\/sup><\/span><\/strong>. (in Nm<\/strong>2<\/sup><\/span><\/strong>\/C)<\/strong>\u00a0\u00a0\u00a0<\/strong><\/p>\n

                                      Solution:<\/strong>\u00a0\u00a0\u00a0<\/em><\/strong>E<\/em>x<\/sub><\/span><\/em>\u00a0<\/em>= bx<\/em>1\/2<\/sup><\/span>, where b<\/em> = 800 N\/C1\/2<\/sup><\/span>.<\/p>\n

                                      For the left face perpendicular to the x<\/em>-axis,<\/p>\n

                                      we have x<\/em> = a<\/em> = 10 cm, while for the right face x<\/em> = 2a<\/em> = 20cm.<\/p>\n

                                      Hence for the left face, the x<\/em>-component of the field is<\/p>\n

                                      \u00a0E<\/em>x<\/sub><\/span><\/em>\u00a0<\/sub><\/span><\/em>= 800 \u00d7 (10 \u00d7 10-2<\/sup><\/span> m)1\/2<\/sup><\/span> = 253 N\/c<\/p>\n

                                      \"\"For the right face,<\/p>\n

                                      we have E<\/em>x<\/sub><\/span><\/em>\u00a0<\/em>\uf0a2<\/span> = 800 \u00d7 (20 \u00d7 10\u20132<\/sup><\/span> )1\/2<\/sup><\/span> = 358 N\/C<\/p>\n

                                      The area of each face is<\/p>\n

                                      \u00a0S<\/em> = 100 cm2<\/sup><\/span> = 10\u20132<\/sup><\/span>m<\/em>2\u00a0<\/sup><\/span><\/p>\n

                                      Hence, the flux through the left face<\/p>\n

                                      \u00a0= \u2013E<\/em>x<\/sub><\/span><\/em>S<\/em> = (253) (10-2<\/sup><\/span>) = \u2013 2.53 N-m2<\/sup><\/span>\/C<\/p>\n

                                      The flux through the right face<\/p>\n

                                      = E<\/em>x<\/sub><\/span><\/em> \uf0a2<\/span>S<\/em> = (358) (10\u20132<\/sup><\/span>) = 3.58 N-m2<\/sup><\/span>\/C<\/p>\n

                                      The net flux through the other faces is zero, because<\/p>\n

                                      E<\/em>y<\/sub><\/span><\/em> = E<\/em>z<\/sub><\/span><\/em> = 0<\/p>\n

                                      Hence, the net flux through the cube<\/p>\n

                                      \u00a0\uf066<\/span>E<\/sub><\/span><\/em> = 3.58 \u2013 2.53 = 1\u00a0<\/strong> (approx)<\/p>\n

                                      \"\"32. Gaussian\u00a0<\/span><\/strong>Surface:<\/span><\/strong><\/p>\n

                                      Any hypothetical closed surface around a charge having same electric field at all points is called Gaussian Surface<\/span><\/em>. There are three types of Gaussian surface.<\/span><\/p>\n

                                        \n
                                      1. Spherical Gaussian surface:-<\/li>\n<\/ol>\n

                                        A Gaussian surface around a point charge is called spherical Gaussian surface. As shown in fig (a)<\/span><\/p>\n

                                          \n
                                        1. Cylindrical Gaussian Surface:-<\/li>\n<\/ol>\n

                                          A Gaussian surface around a line charge is called cylindrical Gaussian Surface. As shown in fig (b)<\/span><\/p>\n

                                            \n
                                          1. Plane Gaussian surface:-<\/strong><\/li>\n<\/ol>\n

                                            An infinite small part of spherical or cylindrical Gaussian surface is called plane Gaussians surface. As shown in fig (c)\u00a0<\/span><\/p>\n

                                            Given the electric field in the<\/span><\/strong>\"\"\u00a0\u00a0<\/span><\/strong>, region; find the net electric flux through the cube and the charge enclosed by it.<\/span><\/strong><\/p>\n

                                            Ans:<\/span><\/strong> Only<\/span>\u00a0\u00a0<\/span>the<\/span>\u00a0\u00a0<\/span>faces<\/span>\u00a0\u00a0<\/span>perpendicular<\/span>\u00a0\u00a0<\/span>to<\/span>\u00a0\u00a0<\/span>the<\/span>\u00a0\u00a0<\/span>direction<\/span>\u00a0\u00a0<\/span>of<\/span>\u00a0\u00a0<\/span>x-axis,<\/span>\u00a0\u00a0<\/span>contribute<\/span>\u00a0\u00a0<\/span>to<\/span>\u00a0\u00a0<\/span>the<\/span>\u00a0\u00a0<\/span>electric<\/span>\u00a0\u00a0<\/span>flux.<\/span>\u00a0\u00a0<\/span>The remaining faces of the cube give zero contribution<\/span>.\u00a0<\/span><\/strong><\/p>\n

                                            \"\"\u00a0=\u00a0<\/span><\/em>\"\"\u00a0\u00a0<\/span><\/em><\/p>\n

                                            Q. 35: A point charge causes an electric flux of \u22121.0 \u00d7 10<\/strong>3\u00a0<\/sup><\/span><\/strong>Nm<\/strong>2<\/sup><\/span><\/strong>\u00a0\/C to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge.\u00a0<\/strong><\/p>\n

                                            (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?\u00a0<\/strong><\/p>\n

                                            (b) What is the value of the point charge?\u00a0<\/strong><\/p>\n

                                            Answer: (a) Electric flux, \u03a6 =\u00a0 \u22121.0 \u00d7 103<\/sup><\/span> N m2\u00a0<\/sup><\/span>\/C and, r = 10.0 cm<\/p>\n

                                            Electric flux does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same.<\/p>\n

                                            i.e., \u22121.0 \u00d7 103<\/sup><\/span> N m2\u00a0<\/sup><\/span>\/C<\/p>\n

                                            (b)\u00a0\u00a0 \"\",<\/p>\n

                                            Where, q = Net charge enclosed and \u2208<\/span>0\u00a0<\/sub><\/span>= 8.854 \u00d7 10\u221212\u00a0<\/sup><\/span>N\u22121<\/sup><\/span>C\u00a02<\/sup><\/span> m\u22122<\/sup><\/span><\/p>\n

                                            \u2234<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>q =<\/span> \u00a0 \u22121.0 \u00d7 103<\/sup><\/span> \u00d7 8.854 \u00d7 10\u221212\u00a0<\/sup><\/span><\/p>\n

                                            = \u22128.854 \u00d7 10\u22129<\/sup><\/span> C\u00a0\u00a0 = \u22128.854 nC<\/p>\n

                                            33. Gauss Theorem:-<\/span><\/strong>\u00a0M.Imp<\/sup><\/span><\/strong>\u00a0<\/span><\/strong><\/p>\n

                                            According to Gauss theorem, electric flux or surface integral of electrical field over a closed surface is equal to 1\/<\/span>\"\"\u00a0times the total charge enclosed by the surface.<\/span><\/p>\n

                                            I,e<\/span>\"\"\u00a0\u00a0<\/span>=\u00a0<\/span>\"\"\u00a0\u00a0<\/span>=<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                                            \"F:\\unitProof:-<\/span><\/strong>\u00a0\u00a0<\/span>Suppose a point p having r distance from +q charge on the Gaussians Surface of small area ds.\u00a0<\/span><\/p>\n

                                            Then electric field at point p is\u00a0<\/span><\/p>\n

                                            \"\"\u00a0\u00a0<\/span>=<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                                            & electric flux at point p is\u00a0<\/span><\/p>\n

                                            \"\"\u00a0=<\/span>\"\"\u00a0<\/span><\/p>\n

                                            =<\/span>\"\"<\/p>\n

                                            \"\"cos0\u00b0<\/span><\/p>\n

                                            \"\"\u00a0= E\u00a0<\/span>\"\"<\/p>\n

                                            =<\/span>\u00a0\u00a0<\/span>\"\"\u00a0\u00d74\u03c0\u00a0<\/span>\"\"(Here\u00a0<\/span>\"\"\u00a0= 4\u03c0<\/span>\"\"\u00a0is surface area of sphere)<\/span>\"\"<\/p>\n

                                            Or<\/span>\u00a0\u00a0<\/span>\"\"\u00a0=<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>\"\"\u00a0=<\/span><\/strong>\u00a0\u00a0<\/span><\/strong>\"\"<\/p>\n

                                            Hence Gauss theorem is proved\u00a0<\/span><\/em><\/p>\n

                                            Question 36:\u00a0<\/strong>A charge ‘q’ is placed at the centre of a cube of side l. What is the electric flux passing through each face of the cube?<\/span><\/strong><\/p>\n

                                            Ans:<\/span>\u00a0\u00a0\u00a0\u00a0<\/span>\"\"\u00a0=\u00a0<\/span>\"\"<\/p>\n

                                            Question 37:\u00a0<\/strong>What is the electric flux through a cube of side 1 cm which encloses an electric dipole?\u00a0<\/span><\/strong><\/p>\n

                                            Ans:<\/span>\u00a0\u00a0<\/span>Zero.<\/span><\/p>\n

                                            Question 38 :\u00a0<\/strong>The electric flux through a closed Gaussian surface depends upon:<\/span><\/strong><\/p>\n

                                            A Net charge enclosed and permittivity of the medium*.<\/span><\/p>\n

                                            B Net charge enclosed permittivity of the medium and the size of the Gaussian surface.<\/span><\/p>\n

                                            C Net charge enclosed only.<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                                            D Permittivity of the medium only.<\/span><\/p>\n

                                            \"\"Question 39 :\u00a0<\/strong>Electric flux through a spherical surface shown in the figure is ________.<\/span><\/strong><\/p>\n

                                            Ans:\u00a0<\/span><\/strong>\"\"\u00a0=\u00a0<\/span>\"\"<\/p>\n

                                            \u00a0<\/strong><\/p>\n

                                            Question 40 :\u00a0<\/strong>If the net electric flux through a closed surface is zero, and then we can infer:<\/span><\/strong><\/p>\n

                                            A No net charge is enclosed by the surface*.\u00a0<\/span><\/p>\n

                                            B Uniform electric field exists within the surface.<\/span><\/p>\n

                                            C Electric potential varies from point to point inside the surface.\u00a0<\/span><\/p>\n

                                            D Charge is present inside the surface.<\/span><\/p>\n

                                            34.\u00a0<\/span>Deduction of Coulomb\u2019s law from Gauss theorem:- Or\u00a0<\/span><\/strong><\/p>\n

                                            Electric field at a point due to point charge:-<\/span><\/strong>\u00a0<\/span><\/p>\n

                                            Suppose a point p on a Gaussian surface\u00a0<\/span>\"\"having charge q<\/span>0<\/sub><\/span>around a charge q. If r is the distance between q & q<\/span>0\u00a0<\/sub><\/span>then electric flux may be given by gauss theorem is<\/span><\/p>\n

                                            \"F:\\unit\"\"\u00a0=<\/span>\"\"\u00a0=<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                                            \"\"\u00a0=<\/span>\u00a0\u00a0\u00a0<\/span>\"\"<\/p>\n

                                            =<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0\u00a0<\/span>=\u00a0<\/span>\"\"<\/p>\n

                                            \"\"cos0\u00b0=\u00a0<\/span>\"\"<\/p>\n

                                            \"\"\u00a0<\/span><\/p>\n

                                            = E\u00a0<\/span>\"\"\u00a0\u00a0<\/span>=\u00a0<\/span>\"\"<\/p>\n

                                            E\u00a0<\/span>\"\"\u00a0=\u00a0<\/span>\"\"<\/p>\n

                                            Here<\/span>\u00a0\u00a0<\/span>\"\"\u00a0= area of Gaussian sphere<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                                            = 4\u03c0<\/span>\"\"\u00a0\u00a0\u00a0<\/span><\/p>\n

                                            \u21d2<\/span>\u00a0E \u00d7 4\u03c0<\/span>\"\"\u00a0=\u00a0<\/span>\"\"<\/p>\n

                                            Or<\/span>\u00a0\u00a0\u00a0<\/span>E =\u00a0<\/span>\"\"<\/p>\n

                                            Now force may be given as\u00a0<\/span><\/p>\n

                                            F<\/span>\u00a0\u00a0<\/span>=<\/span>\u00a0\u00a0<\/span>q<\/span>0<\/sub><\/span>E<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n

                                            So<\/span>\u00a0\u00a0\u00a0<\/span>F =<\/span>\"\"<\/p>\n

                                            This is coulomb\u2019s law. Hence coulombs law can be obtained by Gauss theorem.<\/span><\/p>\n

                                            \uf0fc<\/span>\u00a0\u00a0\u00a0\u00a0\u00a0<\/span>Gauss theorem is used to calculate electric field at a point, but Coulomb\u2019s cannot simplify problems related to electric field, Gauss\u2019s law is more easily, suitable & more useful in situations involving symmetry<\/span><\/em>.<\/span><\/p>\n

                                            35 APPLICATION OF GAUSS THEOREM<\/span><\/strong>\u00a0M.Imp<\/sup><\/span><\/strong><\/p>\n

                                            (I).\u00a0<\/span>Electric field due to an infinitely long charged wire:-<\/span><\/strong><\/p>\n

                                            Suppose a cylindrical Gaussian surface of radius r around a infinite long charged wire of length\u00a0<\/span>l\u00a0<\/span>& charge q having three small area elements<\/span>\"\",\u00a0<\/span>\"\"&\u00a0<\/span>\"\"\u00a0<\/sub><\/span>as shown in fig.<\/span><\/p>\n

                                            Now According to Gauss theorem\u00a0<\/span><\/p>\n

                                            \u2205<\/span>e<\/sub><\/span>=<\/span>\"\"\u00a0=\u00a0<\/span>\"\"<\/p>\n

                                            =\u00a0<\/span>\"\"+\u00a0<\/span>\"\"+\u00a0<\/span>\"\"=\u00a0<\/span>\"\"<\/p>\n

                                            \"F:\\59.jpg\"\u00a0<\/span>\u2205<\/span>e<\/sub><\/span>\"\"\u00a0\u00a0\u00a0<\/span>=\u00a0<\/span>\"\"=<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                                            \u2205<\/span>e<\/sub><\/span>\u00a0= E<\/span>\"\"=\u00a0<\/span>\"\"<\/p>\n

                                            Here<\/span>\u00a0\u00a0\u00a0<\/span>\"\"\u00a0= area of the cylinder =<\/span>\u00a0\u00a0\u00a0<\/span>2<\/span>\"\"r<\/span>l<\/span><\/p>\n

                                            So<\/span>\u00a0\u00a0\u00a0<\/span>\u2205<\/span>e<\/sub><\/span>\u00a0\u00a0<\/span>= E\u00a0<\/span>\"\"\u00a02<\/span>\"\"r<\/span>l<\/span>\u00a0\u00a0\u00a0<\/span>=\u00a0<\/span>\"\"<\/p>\n

                                            Or<\/span>\u00a0\u00a0\u00a0<\/span>E=<\/span>\u00a0\u00a0<\/span>\"\"<\/p>\n

                                            Or<\/span>\u00a0\u00a0<\/span>E=\u00a0<\/span>\"\"\u00a0<\/span>\"\"<\/p>\n