# Solved Problem-6

Problem-6
(a) A point charge â€˜Qâ€™ is located at a distance â€˜hâ€™ from a perfect conducting infinite plane. Obtain the image charge and show that the induced charge on the conducting plane is â€˜-Qâ€™.

(b) An infinitely long line charge of uniform line charge density â€˜Î»â€™ is situated parallel to and at a distance â€˜xâ€™ from the grounded infinite plane conductor. Obtain the image charge and show that the induced surface charge on the conductor per unit length is â€˜-Î»â€™.

Solution
(a) The image charge -Q is placed at a distance h in the conducting region and the conducting grounded plane is replaced by an equipotential surface of zero potential.

For the charge Q at (0, 0, h) and its image Qâ€™ at (0, 0, -h), the potential at any point
P(x, y, z) is,

At z = 0, i.e., on the conducting plane, V = 0.

Thus, we see that the magnitude of the image charge is -Q.

Hence, the potential at any point P(x, y, z) is,

The electric field at any point P(x, y, z) can be found out by using the relation  or may also be written as,

The electric field component Ez on the xy plane, i.e., on the conductor surface is,

The parallel field components Ex and Ey vanish as they should.

The induced surface charge density is given by Ïƒ = ÎµEz. The total induced surface charge is given by,

(b) The finite line charge Î» is assumed at x = 0, z = h and the image Î»â€™ is assumed at x = 0, z = -h, so that the two are parallel to y-axis.

The field at point P (x, y, z) is,
Here,

Potential at P is,

At the conducting plane, V = 0 â‡’ Î»â€™ = -Î» (with z = 0)

The surface charge induced on the conducting plane is,

The induced charge per unit length on the conducting plane is,