Mathematics Electrostatic Potential and Capacitance

### Definition Based Problems

Q 3211578429

Why is electrostatic potential constant throughout the volume of the conductor and has the same value (as inside) on its surface?
CBSE 12 Delhi
Solution:

As we know that the electrostatic field inside the conductor is zero and on the surface, the field is normal to the surface at every point (Gauss's theorem).

No work is done in moving a small test charge, within the conductor and on its surface. We find there is no potential difference between the two points inside or on the surface, which implies the potential being constant throughout.
Q 3211678520

Explain the "For any charge configuration, equipotential surface through a point is normal to the electric field."
CBSE 12 Delhi 2014
Solution:

If the electric field is not normal, it will have non-zero component along the surface. In that case, the work done in moving a charge on the equipotential surface will not be zero.
Q 3221678521

Define the geometrical shape of equipotential surfaces due to a single isolated charge?
CBSE 12 Delhi 2013
Solution:

The geometrical shape is spherical.
Q 3241678523

There no work done in moving a charge from one point to another on an equipotential surface Why ?
CBSE 12 Foreign 2012
Solution:

Potential difference between any two points of an equipotential surface is zero. So no work is done in moving a charge from one point to another.
Q 3271678526

Can two equipotential surfaces intersect each other?

Solution:

Two equipotential surfaces cannot intersect. The direction of electric field is always perpendicular to the equipotential surface. If they intersect, there will be two directions of the electric field at the point of intersection which is not possible.
Q 3281678527

Why should electrostatic field be zero inside a conductor?
CBSE 12 Delhi 2012
Solution:

In the static situation, there is no current inside, or on the surface of the conductor. Hence, electric field is zero everywhere inside the conductor.
Q 3201678528

Why does the electric field inside a dielectric decrease when it is placed in an external electric field?

Solution:

An electric field is developed inside the dielectric due to induction in a direction opposite to the direction of external electric field. Thus, net electric field decreases.
Q 3251191024

Where does the energy of a capacitor reside?

Solution:

Electrical energy resides in the space, within the plates.
Q 3251778624

Derivation for the electric potential at any point along the axial line of an electric dipole ?

Solution:

Potential at P due to dipole is given by

V_p = V_(PA)+V_(pB) = q/(4pi epsi_0) [ 1/(r-l) -1/(r+l)]

 = 1/(4piepsi_0) (q. 2l)/((r^2-l^2))

 = 1/(4pi epsi_0) p/((r^2-l^2))
Q 3261778625

Draw a plot showing the variation of (i) electric field (E) and (ii) electric potential (V) with distance r due to a point charge Q.

Solution:

Q 3271878726

Define is an electrostatic shielding? What is its practical importance?

Solution:

Whatever be the charge and field configuration outside, any cavity in a conductor remains shielded from the outside electric influence. This is known as an electrostatic shielding.

The effect can be made use of in protecting the sensitive instruments from the outside electrical influence.
Q 3281878727

Derive the expression for the capacitance of a parallel plate capacitor having plate area A and plate separation d.

Solution:

Consider a capacitor with surface charge density cr on its plates. Suppose area of each plate is A and separation between the plates is d.

We know Q = CV => C = Q/V

Here Q = sigmaA .............(i)

V = E_0 d = sigma/epsi_0 d \ \ \ \ \ \ \ \ ( because E_0 = sigma/epsi_0) .............(ii)

From equations (i) and (ii), we get

C_0 = ( epsi_0 A)/d
Q 3211878729

Define the capacitor? Write its two uses.

Solution:

A capacitor is an arrangement of primarily two conductors which is used for storing charge.
The capacitor is used for storing: (i) electrical charge, and (ii) electrical energy.
Q 3251078824

Explain 'dielectric constant' of a medium. Briefly explain why the capacitance of a parallel plate capacitor increases, on introducing a dielectric medium between the plates.

Solution:

Dielectric constant of a medium is the ratio of the permittivity of the medium to that of vacuum.

An electric field is developed inside the dielectric due to induction in a direction opposite to the direction of external electric field. Thus, net electric field decreases.
Q 3221178921

Drive the expression for the energy stored in a parallel plate capacitor C having charges +Q and -Q on its plates.
CBSE 12 Delhi 2011
Solution:

Let a capacitor of capacity C be charged to a potential V by a charge Q. Work has to be done in charging a capacitor. This work is stored in the capacitor in the form of potential energy. It exists in the electric field between the plates of the capacitor.

Let q be the charge on the capacitor at any stage and V the potential, then

V = q/C

Then small amount of work done dw in giving a further small charge dq to the capacitor is

dw = V dq = q/C dq

therefore Work done in charging the capacitor to a charge Q is

W = int_0^Q q/C dq = 1/2 [q^2/C]_0^Q = 1/(2C) [ Q^2-0] = 1/2 Q^2/C

Also, Q = CV

therefore W = 1/2 ( C^2 V^2)/C = 1/2 CV^2
Q 3211180929

Explain the working principle of a parallel plate capacitor. On what factors, the capacitance of a parallel plate capacitor depends?

Solution:

Principle: When an uncharged, earthed conductor is brought near to a charged conductor, then the potential of later decreases and its charge holding capacity increases.

The capacitance depends on:
(i) Geometrical configuration (shape, size and separation) of the system of two conductors.
(ii) Nature of the dielectric separating two conductors.
Q 3251178924

Derive an expression for the potential energy of an electric dipole of dipole moment vecp in an electric field vecE

Solution:

Torque acting on the dipole, tau = p E sin theta

It tends to rotate the dipole in clockwise direction. To rotate the dipole anticlockwise
work has to be done on the dipole.

W = int_(theta_1)^(theta_2) tau d theta = int_(theta_1)^(theta_2) p E sin theta d theta = -p E ( costheta_2- costheta_2)
Q 3201178928

Why is the dielectric constant of conductors taken as oo?

Solution:

When the conductors are placed in the external field, then the induced electric field is equal and opposite to the external field E_0

therefore  Net field, E_("Net") = E_0 - E_("in") = E_0 - E_0 = 0

because Dielectric constant, K = E_0/E_("Net") = E_0/0 = oo
Q 3251180024

Drive the expression for the potential energy of a system of two point charges q_1 and q_2 brought from infinity to the points vecr_1 and vecr_2 respectively in the presence of external electric field vecE
CBSE 12 Delhi 2010
Solution:

Work done in bringing the charge q_1 from infinity to vecr_1

against the external electric field W_1 = q_1 V ( r_1)

Work done in bringing the charge q_2 from infinity to vecr_2 , W_2 = q_2 V ( r_2)

Work done on q_2 against the field due q_1 , W_3 =( q_1 q_2)/(4pi epsi_0 r_(12))

As work done is stored in the form of potential energy, therefore

Potential energy of the system  = W_1+W_2+W_3 = U_1 + U_2+U_3

Potential energy of the system = q_1 V (r_1) + q_2V (r_2) + ( q_1 q_2)/(4pi epsi_0 r_(12))
Q 3211280120

Define the polarization of charge? With the help of a diagram show why the electric field between the plates of capacitor reduces on introducing a dielectric slab. Define the dielectric constant on the basis of these fields.
CBSE 12 Delhi
Solution:

The induced dipole moment developed per unit volume in a dielectric slab on placing it inside the electric field is called polarisation.

Let vecE_0 be the uniform external electric field. When a dielectric slab is placed in uniform electric field, then the molecules get polarised, due to which - sigma_p (charge density due to polarisation) will appear near the positive plate and + sigma_p will appear in the dielectric near the negative plate.

Therefore, due to polarization of molecules, electric field will appear will appear in the opposite

direction, and the net electric field inside the dielectric will be

vecE= vecE_0 - vecE_p < vecE_0

to the electric field in medium.