37. Electric Potential Due to a point charge:- ^imp

Electrostatic potential at a point is defined as the amount of work done to bring a charge from ∞ to that point against electrostatic force without acceleration.

Suppose a test charge q₀ is placed at point A & a charge + q is at point O. Now according to Coulomb’s law the force between the charges is 

F = 

if we move charge from A to B through small distance d𝑥 then small amount of work  may be given   

As   =   

= F d𝑥 cos180°  

= – F d𝑥

total work done to move charge from ∞ to p is   

W  =  –      

= –  d𝑥    

=. d𝑥    

= .d𝑥

=       

=

= 

=    

W =  []

Or    

Cleary    V ∝   

i.e. electric potential varies inversely with distance.

At   , V_B   = 0    

So    = V_A

Unit of electrostatic potential is also volt.    

1 volt =  

= 1 NmC^-1   

= 1 JC^-1

i,e  electrostatic potential at a point is said to be one volt if one joule is the amount of work done to move one coulomb charge from ∞ to that point against electrostatic force without acceleration.

Electrostatic potential difference:-

Suppose a point charge q₀ is placed at point B in the field of any other charge +q then work done to move a charge from B to A is given by 

 = V_A -V_B 

= V_BA

Hence Electrostatic Potential difference may be defined as the amount of work done to move a unit +ve charge from one point to other point again the electrostatic forces without acceleration.

• It is assumed that test charge q₀ is so small that it does not disturb the source charge q.
•  The external force is so small that it just balances the repulsive force between the charges & does not produce acceleration in source charge.

SI unit of potential difference is volt:- ^imp

i,e 1 volt =  

= 1 NmC^-1  

 = 1 JC^-1

Potential difference between two points is said to be one volt if one joule is the amount of work done to move one coulomb charge from one point to another point against the electrostatic forces without acceleration.              

Example 42: Two charges  + 10 µC and  + 20 µC are placed at a separation of 2 cm. Find the electric potential due to the pair at the middle point of the line joining the two charges.

Solution: Using the equation

The potential due to + 10 µC is

The potential due to + 20 µC is

The net potential at the given point is

9 MV + 18 MV = 27 MV.

If the charge distribution is continuous, we may use the technique of integration to find the electric potential.

Example 43(a) Calculate the potential at a point P due to a charge of 4 × 10^–7C located 9 cm away.

(b) Hence obtain the work done in bringing a charge of 2 × 10^–9 C from infinity to the point P. Does the answer depend on the path along which the charge is brought?

Solution (a) = 4 × 10⁴ V   

(b) = 8 × 10^–5 J

No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along r and another perpendicular to r. The work done corresponding to the later will be zero.

Q 44.Two point charges,   and C are separated by a distance of 60 cm in air.

1.Find at what distance from the 1 charge, q would the electric potential be zero.

Ans: 

Example 45 Two charges 3 × 10^–8 C and –2 × 10^–8 C are located 15 cm apart. 

At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Ans electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the formula for potential used in the calculation required choosing potential to be zero at infinity.

38. Electric potential at Axial Point of an electric dipole:- ^imp 

Suppose a test charge q₀ is placed at point p having r distance from centre of the dipole. 

Now potential at point p from charge –q at A is

 ….1

& potential at P due to q charge at B is

…….2

So the net potential at P will be 

   =   + 

=     

=    [ –] 

= 

Or V =

V =         (∵ q.2a=P)

Special cases

     If dipole is small then . Here a² can be neglected. 

In this case    =

     Potential varies as  in case of electric dipole.

     If point p lies near the dipole then we can take.           

   =  

39_.  potential at an equatorial point of a dipole:- ^M.Imp

Suppose a test charge q₀ is placed at point p equatorial to electric dipole ± q.2a. The distance of p from centre of the dipole o is r. As shown in fig. Now potential V_A at p due to –q charge at A & potential at p due to +q charge at B

Such that  

… 1

&

….2

Cleary from equation (1) & (2)  

Net electric field is  

V = + 

Or V = 

40. Electric Potential at any point due to a dipole:- ^imp

Suppose a test charge q₀ is placed at point P having r distance with the center of a dipole ± q. 2a placed at A & B. As shown in fig. 

Let AP =        &   BP =

Now net potential at p is 

=  +

=   [

V =  [ —–1

If p point lies far away from the dipole 

Than   r₁ –r₂ ⋍ AB cos  = 2a cos &     ⋍ r²

So 1 becomes      

V   = . —–2

=  

=     (p= q.2a)

V   = ——3

Special Cases:-

•  If the point P lies on the axial of electric dipole than = 0° or 180°       

So from eqⁿ 2  V =      

 i.e.   Potential is maximum.

• When the p point lies on the equatorial line of electric dipole 

Than Q= 90°⇒ Cos 90 = 0⇒ V= 0 

i.e.  Potential at equatorial point of electric dipole is zero.

41. Difference between electric potential of a dipole & A Single Charge.

1. Potential due to dipole depends upon the distance & angle between dipole moment p & distance r where as potential due to single charge depend only on distance.
2. Potential due to dipole is cylindrical symmetric while potential due to point charge is spherical Symmetric.
3. Potential due to dipole varies as   while potential due to single charge varies as.

42. Electric Potential Due to a System of charges:-

 Suppose q₁, q₂, q₃ …..q_n charges having distances r₁, r₂, r₃ …… r_n from a point p. then total potential at p may be calculated as given below.

Potential at p due to q₁ charge is 

—-1

Again potential at p due to q₂ charge 

 =  —-2

Similarly potential at p due to q₃ charge

 =  —-3

  =  —

Adding all the eqⁿs we get 

V = V₁ + V₂ + V₃ ……….V_n 

=   + +  ——  

 =  ( + +  ……)

 = 

43. Electric Potential due to Continuous charge distributions:-

(i) Suppose a point P having r distance from a continuous line distribution of charge.

Then potential at p due to small charge dq is 

  =   

Here    

So    = ∫

(ii) Suppose a point P having r distance from a continuous surface distribution of charge.

Then potential at p due to small charge dq is 

Here              

So   V  = ∫

(iii) Suppose a point p having r distance from a continuous volume distribution of charge.

Now potential at p due to charge  is    

Here           

So total potential at P becomes            

V  ∫

44. Equipotential Surface & their properties:- ^M.Imp
Any surface which has same electric potential at every point on it is called equipotential surface.

There are three types of equipotential surfaces.

(i). Spherical Equipotential surface:- 

An Equipotential surface around a point charge is of spherical in shape so called spherical equipotential surface.

(ii) Cylindrical Equipotential Surface:-

An Equipotential surface around a line charge is of cylindrical in shape so called cylindrical equipotential surface.

(iii) Plane Equipotential Surface:- 

An infinite small part of spherical or cylindrical equipotential surface becomes like a plane so called plane equipotential surface.

Properties:-

(i) The work done to move a test charge on a Equipotential surface is zero:-

Suppose a test charge is moved on an equipotential surface. Then amount of work done to move charge from A to B is          

 = – 

But  =  on an equipotential surface, because potential at every point in Equipotential surface is same   

⇒  = 0       

Here ≠ 0       

 ⇒ = 0

Hence the work done to move a test charge on an equipotential surface is zero

(ii) No two Equipotential surfaces can intersect at each other:-

If two equipotential surfaces will intersect each other, then at that point there will be two values of electric potential which is not possible. Hence No two equipotential surface will intersect each other.

(iii) Electric field is always normal to the equipotential surface:- 

As On Equipotential surface      

W  =  0    

Here q0 ≠ 0     

Here   

So   cosθ = 0 

 ⇒  = 90

Hence electric field is normal to the equipotential surface.

45. Relation between Electric Field and Potential Difference:- ^M.Imp

Suppose two equipotential surfaces having potential   . 

Now again suppose that a test charge q₀ is moved from point P to Q through small distance dr. Then amount of work done to move charge is    

   =——— (1)

Again we know that potential difference to move charge from P to Q is

Or   =   —- (2)

Comparing eq^{n (}1) & (2)

  = 

Or      

⇒      (∵ here –ve sign show that  are in opposite direction).

 Hence electric field intensity is equal to negative gradient of potential difference. 

Q46 .The electric field and electric potential at any point due to a point charge kept in air is  and   respectively. Compute the magnitude of this charge. 

Ans: 

Also

Q 47: The magnitude of electric field varies with the distance r as, E = 10r + 5. By how much does the electric potential increase in moving from point at r = 1m to a point at r = 10m.

As  

We know that

46. Electric potential due to a uniformly charged spherical shell:-

(i) Suppose a point p having r distance from a shell of radius R & charge +q. Such that  Then electric potential at P will be

I.e. electrostatic potential is inversely proportional to distance

(ii) If point P lies on the surface of the shell then r=R

V = (electrostatic potential is maximum)

(iii)  If point P lies inside the shell then electric field E=0.

As we know  

⇒          

• So electric potential is constant, it remains same as that on the surface of the shell. 
• The relation between electric potential & distance of the charge is as shown in graph.

Q48.A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. What is the potential at the centre of the sphere?

Ans: 10 V.

47. Electric potential Energy:-

 The electric potential energy of a system of charge may be defined as the amount of work done in assembling charge at their location by bringing them from  to a required point. 

Suppose a test charge is placed at point  & a another charge q₂ is bring toward q₁ from .Then electric potential between the charge is 

& the amount of work done to bring charge q₂ from to P₂ is  

So   

This work is stored in the charge in from of potential energy. So potential energy of the system of charge is   

U = 

• Potential energy of a system of charge may be given as 

U  =  

Q49. 1. Two point charges and   are placed r distance apart. Obtain the expression for the amount of work done to place a third charge  at the midpoint of the line joining the two charges.

2. At what distance from charge   on the line joining the two charges (in terms of  ,   and r) will this work done be zero.

Ans: According to the problem,

1. Work done: 

=

Q50. Three points chargesand    are initially infinite distance apart. Calculate the work done in assembling these charges at the vertices of an equilateral triangle of side 10cm.

Ans:  W = Energy of system,

Q.51: (a) Determine the electrostatic potential energy of a system consisting of two charges 7 µC and –2 µC (and with no external field) placed at (–9 cm, 0, 0) and (9 cm, 0, 0) respectively.

 (b) How much work is required to separate the two charges infinitely away from each other? 

(c) Suppose that the same system of charges is now placed in an external electric field ; . 

What would the electrostatic energy of the configuration be?

 Solution (a) U =   

(B) W = U₂ – U₁ = 0 – U = 0 – (–0.7) = 0.7 J     

(c) = 70 -20 -0.7 =49.3 J                                            

48. Capacitance:- ^imp

The ability of a body to store charge is called capacity or capacitance. 

If  we give charge to a conductor, its potential keep on rising. Thus 

Or 

Here C is constant of proportionality called capacity or capacitance of the conductor.

Or C = 

The value of C depends upon:- 

1. The size & shape of the conductor.
2. The Nature of medium surrounding the conductor.
3. It does not depend upon material of the conductor by which it is formed and the value of charge and potential.

The SI unit of capacitance is farad (f)    

I,e  = 

Thus capacitance of a conductor is said to be one farad if one coulomb charge given to the capacitor raises its potential through one volt.

•   The c, g, s unit of capacitance is one stat farad.     

i,e 1 stat farad  =   

The capacitance of a conductor is said to be one stat farad if one stat coulomb charge raises the potential of 1 stat volt of conductor.

1 Farad = 

= 

= 9  10¹¹ stat Farad 

Farad is a large unit of capacitance so smaller units are used is as 

1 micro farad = 1 f = 10^-6F

Or 1 micro micro farad = 1 Pico farad = 1 μ f = 1 pF = 10^-12F

Dimensional formula: – 

C =  =  = [  ]

49. Capacitance of an isolated spherical capacitor :-

Suppose a isolated spherical capacitor of capacitance C. If q is the amount of charge given to the sphere, then potential at point p at the surface of the sphere is 

v = 

So capacitance   

C =     =  

=     

= R 

So capacitance of isolated spherical capacitor is 

C  = R 

• Here we can see that capacitance of the conductor depends only upon the radius of the conductor.

50. Capacity of earth:-

As we know radius of earth = 6.4× 10⁶ m

So its capacity C= R   

=      

= 

     Here it should be noted that capacity of earth is less than 1 F. So any body lying on the surface of earth does not have capacity of 1F.

Q52: Calculate the radius of a conductor having capacitance 1F, and compare it with the radius of earth?

Ans. Here  C = 1F        

So from relation        

C = R 

Radius of the planet may be calculated as      

  =   =

 m 

= km 

And we know that radius of earth is 

Comparing both we get   

Thus a conductor having radius 1500 time more than radius of earth will have one farad capacitance.

51. Capacitor & its principle: ^imp

Capacitor: – A capacitor is a device consists of two conductor separated by a small medium & is capable of store large amount of charge.

Principle of capacitor:-

Suppose an uncharged plat B is placed near to a + vely charged plate A. Then due to induction of charges, the face of plate B toward plate A acquire –ve charge & face away from the plate A acquire + ve charge.

The charge on plate B increases when charge on plate A is increased & becomes constant at a certain limit.

In this condition if we earth the +ve side of plate B, then +ve charge transfer into the earth. Due to this charge density of plate B decreases, now a time same more charge can be stored in plate B.

Thus a large amount of charge can be stored in plate B.  Thus a large amount of charge can be stored between arrangements of two conductors.

 Symbol of capacitor:-   

A capacitor of fix capacitance may be represented as & a capacitor of variable capacitance may be represented as

Q 53 Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. 

Ans: when a parallel plate capacitor is connected across a battery. As soon as the charges from the battery reach one plate, due to insulating gap charge is not able to move further to the other plate.  

Thus,  positive  charge  is  developed  at  one  plate  and  negative  charge  is developed on the other. As the amount of charge increases on the plates a voltage is developed across the capacitor that is opposite to the applied voltage.  

Hence, the current flowing in the circuit decreases and gradually becomes zero. Thus, the charge is developed on the capacitor.

Q 54 When an ideal capacitor is charged by a dc battery, no current flows. 

However, when an ac source is used, the current flows continuously. How does one explain this, based on the concept of displacement current?

Ans: When an ideal capacitor is charged by dc battery, charge flows till the capacitor gets fully charged. When an ac source is connected then conduction current keeps on flowing in the connecting wires. 

Due to changing current, charge deposited on the plates of the capacitor changes with time. This Causes change in electric field between the plates of the capacitor which causes the electric flux to change  and  gives  rise to a displacement  current  in the  region between  the  plates of  the capacitor. 

52. Capacitance of a parallel plate capacitor:- ^imp

Suppose an arrangement of a parallel plate capacitor consist of two plates A & B separated by distance and plate A is given +ve charge due to which –ve charge induces in plate B toward A & +ve charge away from A which is earthed. 

 Now potential between the plates may be given as 

Or       (

Here σ is uniform surface charge density.

So          ()

So capacitance of parallel plate capacitor becomes

=  

Thus capacitance of a parallel plate capacitor depends upon.

1. Area of the plates, ()
2. Distance between the plates ()
3.  Permittivity of the mediums of the plates ()

Q55.A parallel plate capacitor is charged to V volt by a battery. The battery is disconnected and the separation between the plates is halved. The new potential difference across the capacitor will be:

Q56.A charged particle is placed between the two plates of a charged parallel plate capacitor. It experiences a force F. If one plate is removed, then the force on the particle will be:

A. 2F       B. F     C          D. Zero

Q57: The plates of a parallel plate capacitor are 5 mm apart and 2m² in area. The plates are in vacuum. A potential difference of 10,000 V is applied across a capacitor.

 Calculate: – (a) the capacitance: (in fm)    

(b) the charge on each plate; (in nC)

Solution: (a)  = 3540 

 (b) Q = CV = (0.00354 × 10^–6) × (10,000) = 3540

The plate at higher potential has a positive charge of +3.54 C and the plate at lower potential has a negative charge of – 3.54 C. 

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53 .Parallel plate capacitor with dielectric slab:- ^M.Imp

Suppose a parallel plate capacitor consist of two parallel Plates  separated by d distance apart and a dielectric slap of thickness t is inserted between the plates. 

If  is electric field between the Plates & E is electric field in dielectric slap then potential between the plates is given by

 …1

Also we know that   = K = dielectric constant     

⇒   E = 

Using in eqⁿ 1 we get    

Or    

So capacitance becomes      

C =      (  C_o =  

     Clearly C > C_o hence capacitance of capacitor increases when a dielectric slab is placed between the capacitors. 

Q 59.A parallel plate capacitor, each with plate area A and separation d, is charged to a potential difference V. 

The battery used to charge it is then disconnected. A dielectric slab of thickness d and dielectric constant K is now placed between the plates. 

What change, if any, will take place in. 

1.  Charge on the plates. 

2. Electric field intensity between the plates. 

3. Capacitance of the capacitor.

Ans: 1. No change. As the battery is disconnected

2. Decreases OR becomes   times, Due to polarization of the dielectric.

3. Increases OR becomes k times. As electric field and therefore, the p.d.  between the plate’s decreases

Q60. A capacitor of 4μF is charged by a battery of 12V. The battery is disconnected and a dielectric slab of dielectric constant 8 is inserted in between the plates of the capacitor to fill the space completely. 

Find the change in the: 

1. Charge stored in the capacitor. 

2. Potential difference between the plates of the capacitor.

 3. Energy stored in the capacitor.

Ans: 1. 

Charge remains same or no change.

2. Initial potential difference 

After battery is removed

Decrease in potential difference

 = 12 – 1.5 = 10.5V

3. Energy stored initially,

Final 

Energy reduces by 

Q61.The space between the plates of a parallel plate capacitor is completely filled in two ways. 

In the first case, it is filled with a slab of dielectric constant K. In the second case, it is filled with two slabs of equal thickness and dielectric constants  and  respectively as shown in the figure. The capacitance of the capacitor is same in the two cases. Obtain the relationship between K,  and.

Ans: In the first case, capacitor is completely filled with a slab of dielectric constant K So 

And parallel combination of two capacitors

∵ 

Q57.A parallel plate capacitor with air between the plates has a capacitance of 8 pF. The separation between the plates is now reduced by half and the space between them is filled with a medium of dielectric constant 5. Calculate the value of capacitance of the capacitor in the second case.

Ans:  

And

Q62.A parallel plate capacitor is charged by a battery. After some time the battery is disconnected and a dielectric slab of dielectric constant K is inserted between the plates. 

How would (i) the capacitance, 

(ii) The electric field between the plates and

 (iii) The energy stored in the capacitor, be affected? Justify your answer. 

Ans: Original capacitance  

When a dielectric is inserted:  

1. Capacitance 

2. Electric Field. Decreases             

3. Energy storeddecrease.

54. Parallel Plate Capacitor with Conduction Slab:- ^imp

Suppose a parallel plate capacitor consists of two parallel plates A & B separated by d distance apart and a conduction slab of thickness t is placed between the plates. 

As electric field between the plates of capacitor is E_o  & in conduction slab E is zero, So potential difference between the plates may be given as.

Or       (∵ E_o =)

Or 

So capacitance of parallel plate capacitor becomes   

C  =  (C_o = 

Or       Clearly    C > C_o 

Hence capacitance of parallel plate capacitor increases when a conducting slab is placed between the plates.

00. Capacitance of a spherical capacitor (^{not directly in syllabus)}

Suppose a spherical capacitor consist of two concentric rings of radius a & b such that a<b. 

Now if +q charge is given to inner sphere then –q charge induces on outer sphere then potential difference between the sphere is

V  =   – 

=     

So capacitance of capacitors becomes   

C =     

=  

=   

=  

0. Cylindrical Capacitor:- ^{(not directly in syllabus)}

Suppose a cylindrical capacitor consist of two Cylinders of radius a & b. Now electric Field at any point p having r distance from the axis of cylinder is  

E = 

Now potential difference between two cylinders is 

V = – 

=  –  

=  

V =   

=  

=  [log r    

=    [log b – log a]

V =  log (λ=)

So capacitance 

 C=  

=      

= 

55. Combinations of Capacitors:- ^M.Imp

(i) Capacitors in series:-

When the negative plate of one capacitor is connected to the positive plate of the Second & negative of the second to the positive of third & so on, then the Capacitors are said to be connected in series. 

Suppose three capacitors C_1, C₂, & C₃ are connected in series & q is the amount of charge stored in each capacitor. 

Then total rise in potential is 

Where  

⇒    =  +  + 

⇒    =( +  + )

Or =  +  + 

Here C_s is the equivalent capacitance due to series combination.

 Hence equivalent capacitance in series combination is equal to the sum of reciprocal of the all the capacitor connected in series.

     Equivalent capacitance is smaller than the smallest capacitor.

 (ii) Capacitor’s in parallel:-

When +ve plates of all the capacitors are connected to one common point &- ve plates of all the capacitors are connected  to other common point then the combinations of capacitor is called parallel combination.

Suppose  are three capacitor Connected in parallel. If V is the potential across each capacitor, 

Then total charge is 

But  ,  

Then     

Or      

Here  is the equivalent capacitance due to parallel combination.

     Hence equivalent capacitance is equal to the sum of individual capacitance.

     Equivalent capacitance is larger than the largest individual capacitance.

Q63.Two capacitors of capacitances  and  are connected in parallel. If a charge Q is given to the combination, the ratio of the charge on the capacitor  to the charge on  will be:

A     BC    D

Q 64: Find the equivalent capacitance in circuit

B.       

                                                           C.                                                                  D.

Q 65: Find the equivalent capacitance in circuit and total charge 

     C = 16 μF = 16 x 10^-6 Farad and Q = 160 μC

Q 66: Find the equivalent capacitance in circuit if each capacitor is of 2f

Ans 

Q 67: Find the equivalent capacitance in circuit if each capacitor is of 2f

Q 68: Find the equivalent capacitance in circuit if each capacitor is of 2f

56. Energy Stored in a capacitor:- ^M.Imp

 Suppose a parallel plate capacitor consist of two Parallel plates A & B. Initially both plates are neutral. Now suppose +q charge is transferred from plate B to plate A. Now to transfer some more charge dq from B to A, the work has to be done.

As   

⇒   

Now total amount of work done to move charge +Q from B to A is  

Or   [     

=  []

  []

This work will store in the capacitor in from of energy so, 

U =  ———- 1

As Q = CV      

⇒  U =  

 CV­­² ——2

Also  C = ⇒                         

U =    QV ——3

So we may write                                

 U =   CV²   

=    

 =  qV

Q69. Two identical capacitors of 12 pF each are connected in series across a battery of 

50 V. How much electrostatic energy is stored in the combination? If these were 

connected in parallel across the same battery, how much energy 

will be stored in the combination now? Also find the charge 

drawn from the battery in each case.

In series combination

U

And  

In parallel combination: 

And Energy 

  Charge  

Q70. 1. In the following arrangement of capacitors, the energy stored in the 6 μF capacitor is E. Find the Value of the following: 

2. Energy stored in 12 μF capacitor. 

3. Energy stored in 3 μF capacitor. 

4. Total energy drawn from the battery.

Ans: 1. 

Energy stored in 12 μF

. Charge on 6 μF   C

Charge on 12 μF   Coulomb

Charge on 3μF 

Energy stored in 3μF capacitor   18E

Question71: A parallel plate capacitor has plates of area 4 m² separated by a distance of 0.5 mm. The capacitor is Connected across a cell of emf 100 V. Find the energy store in the capacitor (in mJ) if a dielectric slab of Dielectric  Strength 3 thicknesses 0.5 mm is inserted inside this capacitor after it has been disconnected from the Cell.    

Solution: 

C    =      =    KC₀ =     0.2124 F             

V =  =

U   = = 118 

57. Behavior of a conductor in electric field:-

1. The net electric field inside a conductor is zero.
2. Just outside surface of a charged conductor, electric field is normal to the surface.
3. The net charged inside a conductor is zero & charge given to the conductor spread on its surface.
4.  Potential is constant inside & on the surface of a conductor.
5. Electric field at the surface of a charged conductor is proportional to the surface charge density.
6. Electric field is zero in the cavity of a hollow charged conductor.

58. Electrostatic shielding:-

The phenomenon of making a hollow region of conductor having no any electric field inside it is called electrostatic shielding. It is based on the fact that electric field vanishes inside the cavity of a hollow conductor.

Uses of electrostatic shielding:-

1. In thunderstorm, during lighting it is safe to sit in car rather than near a tree or in open ground because metallic body of car act as electrostatic shielding from lightning.
2. Sensitive components of electronic devices are protected from external electric disturbance by placing them in metal shields.
3. In coaxial cables electrostatic shielding is used.

59. Electric susceptibility:- 

The polarization  is directly proportional to the applied electric field across a dielectric slab.   

 i, e       ∝    

Or    

Here χ is called electric susceptibility.

It has no dimensions χ =   

Ration between K &  χ

The net electric field in a polarized dielectric is   

But  = 

  – 

Dividing both side by  we get      

1 =  – χ         

Or  1 = K – χ      

Or    K =1+ χ

60. Dielectric Strength:-

 The maximum electric field that can exist in a dielectric without causing the breakdown of its insulating property is called dielectric strength of the material. 

   SOME IMPORTANT MCQ FROM PREVIOUS EXAM ^imp

1. Si unit of charge is   Coulombs.
2. Si unit of permittivity is   C²N^-1m^-2
3. Si unit of k is   N¹m²c^-2.
4. Ratio of magnitude of electric force in air and water between an electron and proton is  K .
5. Charge on a neutron is  0.
6. Charge on proton is    1.6×10^-19C.
7. Charge on electron is   -1.6×10^-19C.
8. When the distance between the two charge particles is doubled then the force between them becomes   one fourth.
9. Nature of electric force between the two protons is   repulsive.
10. When the distance between the two charge particles is halved then the force becomes   four times.
11. Charge on an atom is   0C.
12. Torque acting on an electric dipole of dipole moment P placed at an angle 90⁰ to the electric field E will be   PE.
13. Torque acting on an electric dipole of dipole moment P placed parallel to the electric field E will be 0.
14. Si unit of electric potential is   volt.
15. Si unit of capacitance is   farad.
16. The energy density of electric field E is. 
17. Dielectric constant of metal is    ∞.