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Electrostatic Potential And Capacitance

JEE Mains

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. 

Electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface

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Since E = 0 inside the conductor and has no tangential component on the surface, no work is done in moving a small test charge within the conductor and on its surface. That is, there is no potential difference between any two points inside or on the surface of the conductor. Hence, the result. If the conductor is charged,electric field normal to the surface exists; this means potential will be different for the surface and a point just outside the surface.In a system of conductors of arbitrary size, shape and charge configuration, each conductor is characterised by a constant value of potential, but this constant may differ from one conductor to the other.
An ellipsoidal cavity is carved within a perfect conductor. A positive charge q is placed at the centre of the cavity. The points A and B are on the cavity surfaces as shown in the figure. Then
This question has multiple correct options
Electric field near A in the cavity = Electric field near B in the cavity
Charge density at A = charge density at B
Potential at A = potential at B
Total electric flux through the surface of the cavity is

Capacitance of cylindrical capacitors

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A cylindrical capacitor consists of a hollow or a solid cylindrical conductor surrounded by another concentric hollow spherical cylinder. The capacitance of a cylindrical capacitor can be derived as:
By Gauss law charge enclosed by gaussian cylinder of radius r and length L is



Charging and Discharging of capacitors

When a capacitor is connected to a battery, positive charge appears on one plate and negative charge on the other. The potential difference between the plates ultimately becomes equal to the emf of the battery. The whole process takes some time and during this time there is an electric current through the connecting wires and the battery.
 where is the charge on the capacitor at time is called the time constant, is the emf of the battery.

If the plates of a charged capacitor are connected through a conducting wire, the capacitor gets discharged. Again there is a flow of charge through the wires and hence there is a current.

where is the charge on the capacitor at time is the charge on the capacitor at time and is called the time constant,

Capacitance of parallel plate capacitor with a combination of dielectrics

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In the figure shown, find the effective capacitance across P and Q. (Area of each plate is )
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

JEE Advanced

Potential due to a dipole at a general point

Potential due to a dipole is given by:

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
Approximation is made that the length of dipole is negligible as compared to the distance of the point from the dipole.

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.

Dielectrics in series

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where , are the dielectric constants 
The equivalent capacitance of the arrangement shown in figure, if A is the area of each plate is :

Solving capacitor circuits

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In the given figure, find charge across each capacitor in steady state.
Replace capacitors of and by a single capacitor of as equivalent capacitance of capacitors in parallel is the sum of each capacitance.
Since capacitors are in series, charge on each capacitor is same by law of conservation of charge.


Hence, charge on each capacitor is .
Now is distributed between and .