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Electrostatic Potential And Capacitance

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1

Discuss and explain the relation between electric field intensity and potential

The relationship between electric field and scalar potential is given as:
, where gradient operator.

The potential at the origin is zero due to electric field . The potential at point  is:

A
 
B
C
D
2

Potential due to a dipole at a general point

Potential due to a dipole is given by:

where 
electric dipole moment
angle made by the line joining center of dipole and the point with the dipole moment vector
distance between the center of dipole and the point
Note:
Approximation is made that the length of dipole is negligible as compared to the distance of the point from the dipole.
Four points a, b, c and d are set at equal distance from the centre of a dipole as shown the figure. The electrostatic potential , and would satisfy the following relation :
A
B
C
D
3

Equipotential surfaces

Surfaces having same potential are termed as equipotential surfaces
The properties of equipotential surfaces can be summarized as follows:
  • The electric field lines are normal to the equipotentials and are directed from higher to lower potentials. 
  • By symmetry, the equipotential surfaces produced by a point charge form a family of concentric spheres, and for a constant electric field, a family of planes normal to the field lines. 
  • The tangential component of the electric field along the equipotential surface is zero, otherwise non-vanishing work would be done to move a charge from one point on the surface to the other.
  • Work done in moving a particle along an equipotential surface is zero. 

Equipotentials surfaces are shown in figure a and b. The field is uniform in 

A
a only
B
b only
C
a and b
D
none
4

Capacitance of parallel plate capacitor

We know that
But and
putting these values in capacitance formula we get 
  
Where, A = Area of plate, d = Distance between capacitor plates

If we increase the distance between two plates of the capacitor, the capacitance will

A
decrease
B
remain same
C
increase
D
first decrease then increase
5

Capacitance of parallel plate capacitor with a combination of dielectrics

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Example:
In the figure shown, find the effective capacitance across P and Q. (Area of each plate is )
Solution:
The above arrangement acts as the capacitor in parallel with and which are in series.

and are in series.


is in parallel with
Total capacitance

A parallel plate condenser is filled with two dielectrics as shown in figure. Area of each pate is and the separation is metre. The dielectric constants are and respectively. Its capacitance (in farad) will be :
A
B
C
D