Chapter–1: Electric Charges and Fields

1. Electrostatics:- 

The branch of physics which deals with the study of charges at rest & the forces & fields, potential of charges is called electrostatics.

* Electrostatics was discovered around 600Bc by a Greek philosopher Thales of Miletus. 

He rubbed amber with a cloth, then cloths starts attracting small piece of paper. 

In Greek amber is called electron so phenomenon was called electricity. 

(Amber is a yellow gum like substance obtained from old plants). 

* In 1544 – 1603 William Gilbert found that a force was present after rubbing of amber or any other substance called electrostatic force.

* In 1731 Stephen Gray found that charge can be moved through a metal for a long distance but not through a thread. This leads to two types of materials   (i) Conductor (ii) Insulator

* In 1733 Du Fay discovered that charges are of two types + ve charge and – ve charge.

* In 1750 Benjamin Franklin proposed 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. 

2. What is electric charge:- 

The property of electrons due to which they experience force of interactions is called electric charge. 

     The electrostatic force is much larger than gravitational force.

     The charge on an electron is -1.610^-19C & 

     Charge on a proton is=+ 1.610^-19C.

     Neutrons have no charge.

3. Two kinds of charge:- 

In 1733 Du Fay discovered that charges are of two types. This can be shown by the following simple experiments. 

Exp-1. 

When two glass rods rubbed with silk are brought near each other then they started repel each other.                                  

Exp.2 

When two ebonite rods rubbed with cat’s fur are also brought near each other, then they also repel each other.                                                                       

 Exp.3:- 

But when a glass rod & ebonite rod are brought near each other then they started attracting each other.

• The polarity differentiates the two kinds of charges.
•  The charge on glass rod is called vitreous charge (Latin vitrum = glass) & the charge on amber when rubbed with wool is called resinous charge (Amber is resin)

Now according to Benjamin frankly, charges may define as

• Positive charge:-

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^– in a body, then body is called +vely charged.

• Negative charge:-

The charge developed on plastic rod when rubbed with wool is called negative charge. Or if number of e^– is more than number of p⁺ in a body, and then body is called negative charged.

E, g:-Two kind of charge developed on rubbing 

4. Conductor, insulator & dielectrics:-

 Conductor:- 

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.

Insulator: – 

A substance which cannot be used to conduct electric charge is called insulator e, g: – glass, rubber, plastic, ebonite, mica, wax etc. 

Dielectric:-  

Dielectrics 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.

5. Gold leaf electroscope ( GLE ) 

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.

 

6. Method of charging ^imp

   (i)  Charging by friction:-

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^–.

E,g:- when glass rod is rubbed with silk then glass rod become +vely charged by transferring electron & silk become  -vely charged by acquiring electron.

     By loosing electron mass of the body decreases & by gaining e^– mass of body increases. As electron have mass 9.110^-32kg.

  (ii) Charging by electrostatic induction:-

The phenomenon of charging a neutral body by placing it in the neighboring of a charge body is called electrostatic induction.

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.

(a) Charging by induction ( by earthing a conductor);-

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 –ve charge induce toward the glass rod in the ball &+ ve charge induce on opposite side of the

 ball. By earthing the + ve charge of the ball, we get – vely charged ball. Hence charges can by induced by earthing.

(b) Charging by induction ( By separating conductors)

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 – 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.

Q.1   how can you charge a metal sphere positively without touching it?

Ans: By bringing a negatively charged body towared metal sphere.

(iii)  Charging by conduction (contact):- 

Charging by conduction requires the actual contact between the two bodies.

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.

7.  Polar & Non Polar Bodies:-

• Polar body:- 

A body having different center of +ve& -ve charge is called polar body. 

E, g: – HCl , H₂O etc.      

• Non polar body:-

 A body having same center of +ve& -ve charge is called non polar body.  

E, g: – H₂, O₂, N₂, etc

 

 

 

 

Q.2. (a) A comb run through one’s 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.) 

(b)Ordinary rubber is an insulator. But special rubber tyres of aircraft are made slightly conducting. Why is this necessary? 

© Vehicles carrying inflammable materials usually have metallic ropes touching the ground during motion. Why?

(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? 

8.  Some Basic Properties & Electric Charge:- ^imp 

Charge is a fundamental & intrinsic property of matter. There are three basic properties of charge.

(i)  Quantization Of Charge:- (Discrete Nature of charge )

According to quantization nature of charge, the charge on a body is always whole number multiple of charge on an electron. 

I,e  charge on a body

     n= 1, 2, 3, 4, ………but  n≠ ½, 3/2, 1.5 etc. 

Where e = 1.610^–19C (charge on an electron) & n is a whole number 

      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.

      An elementary particle quart of very small life time have fraction of charge.

± e & ±  e. (it is exception of quantization of charge)

Q.3.  If 10⁹ electrons 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? 

Solution:  charge given out in one second is= 1.6 × 10^–19 × 10⁹ C = 1.6 × 10^–10 C.

The time required to have a charge of 1 C is

= 6.25 × 10⁹ s =  years = 198 years.

      One coulomb is, therefore, a very large unit for many practical purposes.

  Q.4. how much positive and negative charge is there in a cup of water? 

Solution:  assume that the mass of one cup of water is 250 g.

The molecular mass of water is 18g.    so 18g of water contain 6.022×10²³ molecules

 1g of water contain = molecules.  Therefore total molecules × 6.02 × 10²³

Each molecule of water contains 10 electrons and 10 protons.

 Hence the total positive or total negative charge = × 6.02 × 10²³ × 10 × 1.6 × 10^–19 C = 1.34 × 10⁷ C.

Q.5. A polythene piece rubbed with wool is found to have a negative charge of 3 × 10⁻⁷ C. (a) Estimate the number of electrons transferred (from which to which?) (b) Is there a transfer of mass from wool to polythene?

(a) Amount of charge on the polythene piece,

q = −3 × 10⁻⁷ C  And  e = −1.6 × 10⁻¹⁹ C

  Number of electrons transferred from wool to polythene = n

  As q = ne Therefore, the number of electrons transferred from wool to polythene is 1.87 × 10¹².

    Yes. Because an electron has mass, me = 9.1 × 10⁻³ kg.

Total mass transferred to polythene from wool,

m = m_e × n = 9.1 × 10⁻³¹ × 1.85 × 10¹² = 1.706 × 10⁻¹⁸ kg.

(ii) Conservative nature of electric charge:-

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.  E,g:-

(i) Here charge on reactant is zero & charge on product = +1-1=0

             So law of conservation of charge hold good.

  (ii)         

      (Zero charge)               (-1)    +        (+1)           = zero charge 

(iii) Addition nature of electric charge:-

The net charge on an isolated system can be simply obtained by adding all the charges scalarly.

If a system having charge

,,  … _.

Then total charge on the system

_.

E.g. If system have four charges  

 

Then total charge 

 

9. Comparison of charge & mass:-

• Electric Charge

 1. Electric charge may be + ve, -ve or zero.    

2. Electric charge is always quantized.

3. Charge on a body does not depend on its speed.   

4. Charge is strictly conserved.

5. The electrostatic force may be attractive or repulsive.

6. Electrostatic force may cancel out.

7. A charged body always has some mass.

 

• Mass 

1. The mass of a body is always a positive.

2. Quantization of mass is yet not obeyed. 

3. Mass of the body increase with its velocity.

4. Mass is not conserved as it may convert into energy.

5. A gravitational force is always attractive.

6. Gravitational forces never cancel out.

7. A body having mass may not have any charge.

10. Coulombs Law In Electrostatics 🙁 Scalar Form) ^M.Imp

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.

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. 

This force acts along the line joining among the charges.

I, e  F ,.1

F  …2                   

Combining equations 1& 2 

We get F

Or  

Where k is constant of proportionality & its value is   

 k =    = 910⁹ Nm²C^-2

Here _o is called permittivity of free space & its value is

 _o= 8.85410^-12C²N^-1m^-2 

So coulomb’s law becomes   

F=     

=   

= 910⁹

Unite of charge:-

If  F= 910⁹N,  (say), and    

As we know F= 910⁹  

Or 910^{9 =} 910⁹ q²      

⇒

Or   

If two equal charges repel each other with a force of 910⁹N when placed one meter distance apart then charge is said to be one coulomb.

       The c, g, s unit of charge is Stat Coulomb ( stat C ) or electrostatic unit of charge (emu ) 

One stat coulomb is that charge which repel an identical charge in vacuum at distance of 1 cm with a force of 1 dyne                 

1 Coulomb =3 10⁹ stat Coulomb (e m u)

      In electromagnetic c, g, s unit of charge is abcoulomb’s or electromagnetic unit of charge 

(e m u)          

1 coulomb =  abcoulomb’s =  emu of charge.

• Significance of coulombs law:-

 (i) It tells us about the force which bounds the electrons around the nucleus to form a atom.

 (ii) It tells us about the force which binds the molecule to form solids & liquids 

• Limitations of coulomb’s law:-

(i) Coulombs law is applicable only on point charges.

 (ii) It holds good only when charges are in rest only.

11. Comparison between coulomb’s force &Newton’s gravitational force:-

Similarity 

 (i) Both obey inverse square law.

 (ii) Both forces are central forces.

 (iii) Both forces are conservative forces. 

 (iv) Both forces are directly proportional to product of interacting Patrick

Dissimilarity:-

 (i) Coulomb’s force is attractive as well as repulsive while Gravitational force is always attractive in nature.

 (ii) Coulomb’s force is much stronger than gravitational force (10³⁶times) 

 (iii) Coulombs force depends upon the medium in which charges are placed while gravitational force does not depends upon medium.

Q.6 : What is the force between two small charged spheres having charges of 2 × 10⁻⁷ C and 3 × 10⁻⁷ C placed 30 cm apart in air? 

Ans: F= 910⁹  

Q.7 How much stronger is coulombs force from gravitational force between a electron & proton separated by r distance?

I.e .

⇒  =    

Or = 1.2710³⁶  

Hence Fe is 10³⁶ time stronger then F_G

Q.8: The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge − 0.8 µC in air is 0.2 N. 

(a) What is the distance between the two spheres?

 (b) What is the force on the second sphere due to the first? 

Ans: (a)        

(b) Both have equal force so F=0.2N

Q.9 Two large conducting spheres carrying charges  are kept with their centers r distance apart. The magnitude of electrostatic force between them is not exactly F=  because:

A. are not point charges. 

B. Charge distribution on the spheres is not uniform*.

C. Charges on spheres will shift towards the centers of their respective spheres.

D. Charges will shift towards the portions of the spheres which are closer and facing towards each other.

12. Coulomb’s Law in vector from: – ^imp

Suppose two charge q₁ & q₂ placed at A & B having position vectors    &    from origin O and charge q₁ exert  force on q₂ & q₂ exerts  force on q₁. 

Then from coulomb’s Law 

 =  .… 1

Where is a position vector acting from q₂ to q₁

Again from Coulomb’s Law force on q₂ to q₁

  = . …2

Where is a position vector, which acts along q₁ from q₂.

Clearly =-

So from eqⁿ 2 we get      

= 

Comparing eqⁿ 1 & 3 we get

 =

Hence Coulomb’s Law in vector from obeys Newton’s third Law of motion.

 Also from diagram    (from   law of vector addition)

&(from   law of vector addition)

 So eqⁿ 1 & 2 becomes

  = = )

Or=  …4

& similarly   =  … 5

Equation 4 and 5 represents Coulombs Law in Position Vector Form.

Q .10: Four point charges q_A = 2 µC, q_B = −5 µC, q_C = 2 µC, and q_D = −5 µC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 µC placed at the centre of the square?  

Ans: 0

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 × 10⁻⁷ C? 

 (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

 Ans:   (a) force between the two spheres is 1.52 × 10⁻² N.

(b) 0.243 N.

12. Three equal charges,  each, are held fixed at the three corners of an equilateral triangle of side 5 cm. 

Find the Coulomb force experienced by one of the charges due to the rest two.

Ans  24.9 N at 30° with the extended sides from the charge under consideration

13. Four equal charges  each are fixed at the four corners of a square of side 5 cm. 

Find the Coulomb force experienced by one of the charges due to the rest three.

Ans    27.5 N at 45° with the extended sides of the square from the charge under consideration

13. Dielectric Constant or Relative Permittivity: – ^M.Imp

Permittivity is a property of the medium which determines the electric force between the charges situated in that medium.

 Relative permittivity:-

As we know coulomb’s force between the two charge placed in vacuum is 

= …1   

 Where _o is the permittivity in free space

Again force between the same charges, when placed in a medium is 

= . …2   

Here is the permittivity in the medium

Dividing eqⁿ 1 & 2 we get   

  =  

= 

 or K  (note: k=)

Where   is called relative permittivity.

Hence relative permittivity may be defined as the ratio of force between two charges in vacuum to the force between the charges in medium.            Or

 Relative permittivity may be defined as the ratio of permittivity in the medium to the permittivity in free space.

     Hence coulombs law for material medium becomes       

F_m=

Q.14. what will be the Coulombs force if two charges are placed in water?    

Clearly   F_m= 

If charges are placed in water (=80) then force between charges becomes 

F_w=   . 

Here we can see that force reduces by 80 in water as compare to vacuum

14. Force between Multiple Charges: 

The superposition Principle:-

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. 

Suppose there are n charges q₁, q₂, q₃…q_n exerting force on a charge q₀. 

Then from coulomb’s law force on q₀ due to q₁ is  

  =…1                             

Again force on q₀ due to q₂ is                                      

 =…2

Similarly

 =    …  3

Similarly

   =  ….n

Adding all the eqⁿ s we get 

F =  … 

= 

  So total force on a charge

       

     In scalar form this expression may be written as   

     

     In vector form

     

15. Force on a point charge due to continuous charge distribution:-

A  continuous charge distribution is a system of charge lying at infinitelly small distances from each other. There are three types of continuous charge distribution.

(a) Linear or line charge distribution:- 

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.

The line charge density  may be defined as the charge per unit length.

I, e    =  = 

(b) Surface or area charge distribution:- 

If 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              

The surface charge density  may be defined as then charge per unit area of the conductor.

I, e    =     = 

1.   Volume Charge distribution:-

If charges are arranged in such a way that they seems to be like a volume, then the distribution of the charges is called volume distribution of charge.

The volume charge density ρ may be defined as the charge per unit volume of the conductor.

i,e    ρ  =   =

16 force due to continuous distribution of charge:

(i)  Forces at a point due to continuous line distribution of charges:-

Suppose a point P having charge  and 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 

 =   

But   dq = dl where  is line charge density.  So        

d = 

For complete length 

 =  

(ii)  Force due to continuous surface distribution of charge:-   

Suppose a test charge q₀ is placed at point p having r distance from small surface ds having charge dq. Then force on  is 

 =           

Where surface charge density

=   dq =  ds                                        

So   =     

So the force at p due to complete surface is     

= 

(iii) Force due to continuous volume distribution of charges.

Suppose a test charge is placed at point p having distance r from small volume dv having charge dq. Then force on due to dq

 =

where volume charge density 

=   dq = dv 

 =

 Hence force at p due to complete surface is   

 = 

Q.15 A uniformly charged conducting sphere of 2.5 m in diameter has a surface charge density of 100 µC/m. Calculate the charge on the sphere.

                                    

Electric Field

17. Electric Field:-

The space around a positive charge in which its force of interaction may be experienced is called electric field.

Electric field intensity:- ^M.Imp

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.

I,e

Here the charge q₀ should be so small, that it does not disturb the position of original charge. 

Also    =    

=   

Or E     =

Unit &Dimensions:-

As E=   =  =

& unit of E =  or volt per meter.

    electric field is stronger near the charge & decreases exponentially with distance & becomes zero at 

Physical Significance Of Electric Field:- ^imp 

By knowing electric field at any point, we can determine force on a charge placed at that point.

I.e

Thus electric field experience a intermediate role between two charge

It is a characteristic of the system of charges & is independent of the test Charge that we place to determine the field.   

Q.16 A particle of charge  and mass 1.6g is moving with a velocity  .

At t =0 the particle enters in a region having an electric field . 

Find velocity of the particle at t = 5 s.

Ans: 

As =

 

And  So 

Q.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? 

Answer: The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. 

Since the Deflection of particle 3 is the maximum, it has the highest charge to mass ratio

Question 18: Two points charges of and   are kept 30cm apart. How far from the Charge  on the line joining the two charges, will the net electric field be zero?

Ans: 10cm.

Q. 19: Two point charges  and  are located 20 cm apart in vacuum. 

(a) What is the electric field at the midpoint O of the line AB joining the two charges? 

(b) If a negative test charge of magnitude 1.5 × 10⁻⁹ C is placed at this point, what is the force experienced by the test charge? 

Ans:  (a) The electric field at mid-point O is 5.4 × 10⁶ N C⁻¹ along OB. 

(b) A test charge of amount 1.5 × 10⁻⁹ C is placed at mid-point O.  

q = 1.5 × 10⁻⁹ C 

 = 1.5 × 10⁻⁹ × 5.4 × 10⁶ 

= 8.1 × 10⁻³ N. 

The force is directed along line OA.

Question 20: An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10⁴ N C⁻¹ in Millikan’s oil drop experiment. The density of the oil is. Estimate the radius of the drop. (g = 9.81 m s⁻²; e = 1.60 × 10⁻¹⁹ C). 

Ans:  Excess electrons n = 12.   E = 2.55 × 10⁴ N C⁻¹ 

And ρ = 1.26 gm/cm³ = 1.26 × 10³ kg/m³.g =9.81ms⁻².  e = 1.6 × 10⁻¹⁹ C .Radius of the oil drop = r

Force (F) due to electric field E is equal to the weight of the oil drop F = W

Where, q = Net charge on the oil drop = ne 

m = Mass of the oil drop = Volume of the oil drop × Density of oil 

= 9.82 × 10⁻⁴ kg 

Therefore, the radius of the oil drop is

 = 9.82 × 10⁻⁴ mm.

18. Electric Field due to a point charge:-

Suppose a point charge q₀is placed at point p having r distance from a charge q placed at point o.

Then according to coulomb’s law the force of interaction between

The charge is F=  

& the electric field at point p is 

=   

= 

The magnitude of electric field is            

E=  

* Clearly E  , it mesas E is same in all direction & does not depends upon direction of r. 

Such a field is called spherically symmetric or radial field.

19. Electric field due to a system of point charges:-

Suppose q₀ charge is placed at point p having Distances , ……... From charges q_1, q₂………. q_n.. Respectively. 

Then total electric field at q₀ due to all charges may be calculated as given below.                                              

Electric field at q₀ due to q₁ is 

… 1             

Again electric field at q₀ due to q₂ charge is                                              

… 2                                 

& Electric field at q₀ due to q₃ charge is                        

…3    

Similarly electric field at q₀ due to q_n charge will be

… n

Now the total electric field at q₀ due to all charges will be

E=

+…….

Or  E = 

Or    E   =   

28. Electric field at a point due to continuous distributions of charge:-

(i)  Electric field due to continuous line distribution of charge:

Than small force at charge placed at P due to small charge dq on small surface ds may be given by coulomb’s law,

     =    

But   dq = dl where  is line charge density. 

So  d  =  

For complete length                 

   = 

So electric Field at point P may be as given as     

  =

Q.28 : An infinite line charge produces a field of 9 × 10⁴ N/C at a distance of 2 cm. Calculate the linear charge density?

Ans: the linear charge density is 10 µC/m.

(ii)  Electric field due to continuous surface distribution of charge:-   

Suppose a test charge q₀ is placed at point p having r distance from small surface ds having charge dq. Then force on  is                   

    =            Where surface charge density

   =    dq =  ds                               ⇒   =                                           

So the force at p due to complete surface is

=   

Now electric field at point p is       

  =  

(iii)  Electric field due to continuous volume distribution of charges.

Suppose a test charge is placed at point p having distance r from small volume dv having charge dq.

Then force on due to dq               

    =

Where volume charge density  

=   dq = dv 

=

Hence force at p due to complete surface is  

F   =   

So the electric field at point P will be

 =

20. Electric field lines: – ^M.Imp

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.

 

 

Properties of electric field lines:-

 

1. The electric lines of forces are continuous curves having no any breakages.

*In case of point p lying nears the dipole we can take

 

24. Electric Field at any Point of electric dipole: – ^imp  

Suppose a test charge lying at any point P having distance r from the Centre of the dipole.± q.2a inclined at an angle θ.

 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. Again suppose that point P lies at axial point of dipole A’B’ &equatorial point of dipole A’’B’’.      

Then Net electric field at point P is 

 =  …..1

Here for sort dipole

  =        

     = 

So    = 

Or =   .()

Or   = (

Or    =    

Special cases

     If point P lies at axial point of electric dipole  then θ =0⁰ ⇒   cos0⁰=1

      ⇒   E    =   

     If point P lies at equatorial point of electric dipole then θ =90⁰  ⇒    cos90⁰= 0  

⇒  E    =   

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:

1. 

Q 24: Two charges 10 µC and −10 µC 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.

Ans:  

 

Q 25: Two charges 10 µC and −10 µC 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.

Solution:  

25. Torque on a electric dipole placed in uniform two dimensional electric field:- ^imp

Suppose a electric dipole ±q.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.                                                                          

                                              

 

Where

So    τ = qE. 2a 

Or  τ      =   PE

In vector from    = ×

The direction of   can be giving by right hand screw rule. 

• Torque has direction ⊥ to place containing.

 When dipole align to direction of electric field 

then  = 0

⇒     

So    τ = o 

• When dipole lins⊥ to direction field 

then      = 90°     

⟹        

⇒     τ = PE (max) 

• If     &   

Then    τ = P

Hence dipole moment may be defined as torque acting on an electric dipole placed ⊥ to electric field of unit strength.

    Net force on electric dipole in uniform electric field is zero but  τ≠0

26. Potential Energy of Electric Dipole in Uniform Two dimensional Electric field:- ^imp 

Suppose an electric dipole is placed at an angle in a uniform two dimensional Electric field. Then torque experienced by electric dipole is

τ  =  PE

Now suppose we want to rotate the dipole through small angle d. Then amount of work done to rotate the dipole is

Or 

Now to calculate total work done to rotate dipole from ₁ to ₂ is 

W = d 

= PE    

= PE

= – PE 

W = PE

 This work is store in the form of potential energy 

so   U = PE 

27. Stable and Unstable Equilibrium:- 

In case (a):When the angle between the dipole and electric field will be zero Then, the potential energy of the dipole will be equal to the     

  

And   τ = 

We can say that the dipole will be in stable equilibrium. 

In case (b): When the angle between the dipole and electric field will be 180 Then, the potential energy of the dipole will be equal to the

                                     

And    τ   =   

We can say that the dipole will be in an unstable equilibrium.

• a dipole is said to be in stable equilibrium when the angle between the electric field and dipole moment is zero

• The dipole’s moment is directed between the -ve and +ve charges. In equilibrium, the system’s net force and torque should be zero.
• As a result, if potential energy is high, the equilibrium will be unstable, whereas if potential energy is low, the equilibrium will be stable.

Question 26: An electric dipole with dipole moment 4 × 10⁻⁹ C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 10⁴ N C⁻¹. 

Calculate the magnitude of the torque acting on the dipole?

Ans:       τ   = 10⁻⁴ N m.

Question 27: 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  Nm. Calculate the potential energy of the dipole, if it has charge 

Ans:  τ =  pE

29. Electric Field at a point on the axis of charged circular ring:-  

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 dl ‘is as shown in fig.

Here we can see that the components are canceled out. 

The net electric field at P is 

E = …………(1)

Here         

 dE =       

&   cosθ  = 

So from (1) 

E  =       

= 

As q is uniformly distribution. 

The charge = +. 

E = 

Or E   =      (∵= 2πa)

     If p lies for away then  a  <<  r 

Then E =  

This is same as in case of electric field due to point charge.

30. Area vector:-

In 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. 

As  =  ds

Here is a unit vector ⊥ to the surface.

31. Electric Flux:- ^imp

The word flux comes from the Latin word ‘fluere’ means ‘to flow’. 

Thus electric flux is a measure of flow of electric field through a surface. Or

 Electric flux may be defined as the number of electric lines passing through a given area. It is a scalar quantity & denoted by.

Suppose a area element is placed in a uniform electric field  making at angle.

Then =   

or  

Unit of  = NC^-1× m^{2 =} Nm²C^-1

Special Cases:- 

•  If area element is placed in plane of electric field then area vector become perpendicular to electric field.  

     So 

• If area element is placed ⊥ to plane containing E. Then area vector become parallel to electric field. 

      = 0⁰ cos 0 = max =1 

     = Eds (max)

Question 29: If the electric flux entering and leaving a closed surface in airare respectively, the net electric charge enclosed within the surface is ________.

Ans: If the electric flux entering and leaving a closed surface in air are  respectively, the net electric charge enclosed within the surface is 

Question 30 : Given a uniform electric field:, 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   angle with the x-axis?

Ans:   

Q.31 : Consider a uniform electric field E=3×10³îN/C. 

(a) What is the flux of this field through a square of 10cm on a side whose plane is parallel to the yz plane?

 (b) What is the flux through the same  square if the normal to its plane makes a 60° angle with the x; axis?

Ans (a) Φ = = 3 × 10³ × 0.01 × cos0° = 30 N m² /C

 (b) Plane makes an angle of 60° with the x; axis. Hence, θ=60°   

Flux, Φ=3×10³ ×0.01×cos60°=15Nm²/C

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?

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.

Question 33: In a region of space the electric field is given by. Calculate the  

electric flux  through  a Surface of area 100 units in x-y plane. 

Solution:

A surface of area 100 units in the xy plane is represented by an area vector  =100

(Direction along the Normal to the area)

 The electric flux through the surface is given by

  =  = 300 units 

Question 34: Calculate the electric flux through a cube of side‘a’ as shown, where E_x = 

bx^1/2; E_y = E_z = 0, a = 10 cm and b = 800 N/C-m^1/2. (in Nm²/C)   

Solution:   E_x = bx^1/2, where b = 800 N/C^1/2.

For the left face perpendicular to the x-axis,

we have x = a = 10 cm, while for the right face x = 2a = 20cm.

Hence for the left face, the x-component of the field is

 E_x = 800 × (10 × 10^-2 m)^1/2 = 253 N/c

For the right face,

we have E_x  = 800 × (20 × 10^–2 )^1/2 = 358 N/C

The area of each face is

 S = 100 cm² = 10^–2m² 

Hence, the flux through the left face

 = –E_xS = (253) (10^-2) = – 2.53 N-m²/C

The flux through the right face

= E_x S = (358) (10^–2) = 3.58 N-m²/C

The net flux through the other faces is zero, because

E_y = E_z = 0

Hence, the net flux through the cube

 _E = 3.58 – 2.53 = 1  (approx)

32. Gaussian Surface:

Any hypothetical closed surface around a charge having same electric field at all points is called Gaussian Surface. There are three types of Gaussian surface.

1. Spherical Gaussian surface:-

A Gaussian surface around a point charge is called spherical Gaussian surface. As shown in fig (a)

2. Cylindrical Gaussian Surface:-

A Gaussian surface around a line charge is called cylindrical Gaussian Surface. As shown in fig (b)

3. Plane Gaussian surface:-

An infinite small part of spherical or cylindrical Gaussian surface is called plane Gaussians surface. As shown in fig (c) 

Given the electric field in the  , region; find the net electric flux through the cube and the charge enclosed by it.

Ans: Only  the  faces  perpendicular  to  the  direction  of  x-axis,  contribute  to  the  electric  flux.  The remaining faces of the cube give zero contribution. 

 =   

Q. 35: A point charge causes an electric flux of −1.0 × 10³ Nm² /C to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge. 

(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? 

(b) What is the value of the point charge? 

Answer: (a) Electric flux, Φ =  −1.0 × 10³ N m² /C and, r = 10.0 cm

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.

i.e., −1.0 × 10³ N m² /C

(b)   ,

Where, q = Net charge enclosed and ∈₀ = 8.854 × 10⁻¹² N⁻¹C ² m⁻²

∴       q =   −1.0 × 10³ × 8.854 × 10⁻¹² 

= −8.854 × 10⁻⁹ C   = −8.854 nC

33. Gauss Theorem:- ^M.Imp 

According to Gauss theorem, electric flux or surface integral of electrical field over a closed surface is equal to 1/ times the total charge enclosed by the surface.

I,e  =   =  

Proof:-  Suppose a point p having r distance from +q charge on the Gaussians Surface of small area ds. 

Then electric field at point p is 

  =  

& electric flux at point p is 

 = 

=

cos0°

 = E 

=   ×4π (Here  = 4π is surface area of sphere)

Or   =   =  

Hence Gauss theorem is proved 

Question 36: 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?

Ans:     = 

Question 37: What is the electric flux through a cube of side 1 cm which encloses an electric dipole? 

Ans:  Zero.

Question 38 : The electric flux through a closed Gaussian surface depends upon:

A Net charge enclosed and permittivity of the medium*.

B Net charge enclosed permittivity of the medium and the size of the Gaussian surface.

C Net charge enclosed only.       

D Permittivity of the medium only.

Question 39 : Electric flux through a spherical surface shown in the figure is ________.

Ans:  = 

 

Question 40 : If the net electric flux through a closed surface is zero, and then we can infer:

A No net charge is enclosed by the surface*. 

B Uniform electric field exists within the surface.

C Electric potential varies from point to point inside the surface. 

D Charge is present inside the surface.

34. Deduction of Coulomb’s law from Gauss theorem:- Or 

Electric field at a point due to point charge:- 

Suppose a point p on a Gaussian surface having charge q₀around a charge q. If r is the distance between q & q₀ then electric flux may be given by gauss theorem is

 = =  

 =   

=     = 

cos0°= 

 

= E   = 

E  = 

Here   = area of Gaussian sphere       

= 4π   

⇒ E × 4π = 

Or   E = 

Now force may be given as 

F  =  q₀E      

So   F =

This is coulomb’s law. Hence coulombs law can be obtained by Gauss theorem.

     Gauss theorem is used to calculate electric field at a point, but Coulomb’s cannot simplify problems related to electric field, Gauss’s law is more easily, suitable & more useful in situations involving symmetry.

35 APPLICATION OF GAUSS THEOREM ^M.Imp

(I). Electric field due to an infinitely long charged wire:-

Suppose a cylindrical Gaussian surface of radius r around a infinite long charged wire of length l & charge q having three small area elements, &  as shown in fig.

Now According to Gauss theorem 

∅_e= = 

= + + = 

 ∅_e   = =  

∅_e = E= 

Here    = area of the cylinder =   2rl

So   ∅_e  = E  2rl   = 

Or   E=  

Or  E=  

• Thus electric field of a line charge is inversely proportional to the distance from the line charge.

 (II). Electric Field due to uniformly charge infinite plane sheet:-

Suppose a uniformly charged infinite plane sheet of charge having charge density σ and two Gaussian surfaces around sheet. Now According to Gauss theorem 

  =  ES + ES   

= 2ES =  

2ES = 

E =  

Or E =   (∵ = σ)

Clearly electric field due to plane sheet of charge does not depend on distance on from the plane sheet. 

 (III). Electric Field due to two infinite plane sheet of charge:-

Suppose there are two infinite plane sheets of charge having charge densities σ₁and σ₂ and electric field of plane sheets of charge toward Right hand side is +ve & toward L.H.S. is –ve.

Now electric field in region (I) may be given as 

   =  +   =  —-1

Again electric field in region (II) may be given as 

 =     –     =  — –2

Similarly electric field in region (III) region me be given as 

 = +  =——-3

If σ₁ = σ₂ then eqⁿ1, 2 & 3 becomes.

 =     +    == 

&   =   –  = 0

 == 

 (IV). Electric Field due to a uniformly charged thin spherical shell.

Suppose a thin spherical charged shell having charge q & radius R. 

1.  When p point lies outside the spherical shell.

 Then electric field at point p is same as that of due to a point charge.

 I,e  E =   ( For   r >  R  ) 

2.   Now if point p lies on the spherical shell than according to Gauss theorem.

 E ×4πR²    =  

Or E =    (For r = R)

Or E  =       (∵)

3.  When p point lies inside the spherical shell.

As we know that charge enclosed by a Gaussian surface is zero 

 = E ∮ ds = o     

E = o

Hence electric field inside a charged spherical shell is zero.

Graphically variation of electric field with distance r is as shown fig electric field is zero when   , maximum at  & decreases when  

(V). Electric Field due to a charged insulating sphere:-

(A) When p point lies outside the sphere

 Then electric field at point p is same as that of due to a point charge 

I,e     ( For r >R) .

 (B) Now if point p lies on the sphere than according to Gauss theorem.            

So    

Or    (for)

Or E =        =)

1.  When p point lies inside the sphere

Now electric field at a point p lying inside the sphere at point p having r distance from centre of the sphere is given by Gauss theorem.

  = 

Cleary E =  = 

Or E  4 r² =   

E =   —–1 

Graphical variation of E with distance:

     If point p lies on the surface of the sphere then r = R. 

          So from eqⁿ 1    E   = (maximum) 

     If point p lies at centre of the sphere then r =0 

So eqⁿ 1 becomes E= 0

     If point p lies outside the sphere then the case becomes same as in case of point charge. 

 As   E=the variation of E with is as shown in graph.

Q41 Consider two hollow concentric spheres, and   , enclosing charges 2Q and 4Q respectively as shown in the figure.

(i)  Find out the ratio of the electric flux through them. 

(ii)  How will the electric flux through the sphere S  change if a medium of dielectric constant ‘ ’ is introduced in the space inside S  in place of air? Deduce the necessary expression.                                         

Ans: Electric Flux =  = 

For sphere   , =      

For sphere  , = 

So     

When  a  medium  of  dielectric  consistent  is  introduced  in  sphere   the  flux  through =

Directions: Read the following questions and choose

 (A)  If both the statements are true and statement-2 is the correct explanation of statement-1.

 (B) If both the statements are true but statement-2 is not the correct explanation of statement-1.

 (C) If statement-1 is true and statement-2 is False.

 (D)  If statement-1 is False and statement-2 is true.

1. Statement-1:  A positively charged particle always moves along an electric line of force.

 Statement-2:  Force on a charged particle is tangential to electric line of force.

 (a) (A)  (b) (B) (c) (C) (d) (D)

2. Statement-1:  If the electric field intensity  is zero at a point, then electric potential should be necessarily zero at that point (assuming potential is zero at infinity)

 Statement-2:  Electric field is zero at a point exactly midway between two equal and similar charges.

 (a) (A)  (b) (B) (c) (C) (d) (D)

3. Statement-1:  Electric field is discontinuous across the surface of a charged conductor.

 Statement-2:  Electric potential is constant on the surface of conductor.

 (a) (A)  (b) (B) (c) (C) (d) (D)

4. Statement-1:  Work done by the electrostatic field on charge moving on circular or elliptical path will be zero.

 Statement-2:  electrostatic field is a conservative field.

 (a) (A)  (b) (B) (c) (C) (d) (D)

5. Statement-1:  Electric lines of forces are perpendicular to equipotential surface.

 Statement-2:  Work done by electric field on moving a positive charge on equipotential surface is always zero.

 (a) (A)  (b) (B) (c) (C) (d) (D)