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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.
Within a spherical charge distribution of charge density , N equipotential surfaces of potential , are drawn and have increasing radii , respectively. If the difference in the radii of the surfaces is constant for all values of and then :
A
constant
B
C
D
2

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

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
A
Electric field near A in the cavity = Electric field near B in the cavity
B
Charge density at A = charge density at B
C
Potential at A = potential at B
D
Total electric flux through the surface of the cavity is
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## Capacitance of cylindrical capacitors

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

But

Three long concentric conducting cylindrical shells have radii and . Inner and outer shells are connected to each other. The capacitance across middle and inner shells per unit length is:
A
B
C
D
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## Charging and Discharging of capacitors

Charging:
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.

Discharging:
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,
The instantaneous charge on capacitor in two discharging RC circuits is plotted with respect to time in figure. Choose the correct statement(s) (where and are emfs of two DC sources in two different charging circuits and capacitors are fully charged).
A
B
C
if
D
if
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## Capacitance of parallel plate capacitor with a combination of dielectrics

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 capacitor has plates of area 'A' separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as where 'x' is the distance measured from one of the plates. If , the total capacitance of the system is best by the expression:
A
B
C
D
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## 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.
The potential energy U of an electric dipole of moment in an electric field is given by  and electric field due to at a point of position vector is given by where Now, consider two dipoles and ; the dipole is aligned along x - axis at the origin O (0, 0) and the dipole is kept at A (a,a). The minimum energy in terms of , of the dipole system is:
A
B
C
D
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## Discuss and explain the relation between electric field intensity and potential

The relationship between electric field and scalar potential is given as:
Consider a thin spherical shell of radius with its center at the origin, carrying uniform positive surface charge density. The variation of the magnitude of the electric field and the electric potential with the distance from the centre, is best represented by which graph ?
A
B
C
D
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## Dielectrics in series

where , are the dielectric constants
A parallel plate capacitor having plates of area S and plate separation d, has capacitance in air. When two dielectrics of different relative permittivities and are introduced between the two plates as shown in the figure, the capacitance becomes . The ratio is :
A
6/5
B
5/3
C
7/5
D
7/3
The equivalent capacitance of the arrangement shown in figure, if A is the area of each plate is :
A
B
C
D
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## Solving capacitor circuits

Example:
In the given figure, find charge across each capacitor in steady state.
Solution:
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.

Also,

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

In the given circuit, a charge of is given to the upper plate of the capacitor. Then in the steady state, the charge on the upper plate of the capacitor is
A
B
C
D
In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance . The switch is pressed first to fully charge the capacitor and then released. The switch is then pressed to charge the capacitor . After some time, is released and then is pressed. After some time,
This question has multiple correct options
A
the charge on the upper plate of is .
B
the charge on the upper plate of is .
C
the charge on the upper plate of is .
D
the charge on the upper plate of is .