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.

---

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.

---

Solution:

The geometrical shape is spherical.

---

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.

---

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.

---

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.

---

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.

---

Solution:

Electrical energy resides in the space, within the plates.

---

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))`

---

Solution:

---

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.

---

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`

---

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.

---

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.

---

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`

---

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.

---

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)`

---

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`

---

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))`

---

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.

---