# Solved Problem-6

Problem-6

(a) A point charge â€˜
(b) An infinitely long line charge of uniform line charge density â€˜

*Q*â€™ 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€˜-*Q*â€™.*Î»*â€™ is situated parallel to and at a distance â€˜*x*â€™ 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 -
For the charge

At
Thus, we see that the magnitude of the image charge is -
Hence, the potential at any point
The electric field at any point
The electric field component
The parallel field components
The induced surface charge density is given by

(b) The finite line charge
The field at point
Potential at
At the conducting plane,
The surface charge induced on the conducting plane is,
The induced charge per unit length on the conducting plane is,

*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.*Q*at (0, 0,*h*) and its image*Q*â€™ at (0, 0, -*h*), the potential at any point*P*(*x, y, z*) is,*z*= 0, i.e., on the conducting plane,*V*= 0.*Q*.*P*(*x, y, z*) is,*P*(*x, y, z*) can be found out by using the relation or may also be written as,*E*on the_{z}*xy*plane, i.e., on the conductor surface is,*E*_{x }and*E*vanish as they should._{y}*Ïƒ*=*ÎµE*. The total induced surface charge is given by,_{z}*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.*P*(*x, y, z*) is, Here,*P*is,*V*= 0 â‡’*Î»*â€™ = -*Î»*(with*z*= 0)