Induced charge!

The problem can be evaluated using the method of electric images . Consider a completely different charge distribution than our given system : A system of two point chargee $q$ and $-q$ placed at $( 0,0, d)$ and $( 0,0, -d )$ . Notice that this system satisfies the conditions for the potential in the previous system , that is ,

• Potential at all points in the x-y plane is zero

• Potential $V \rightarrow 0$ when $x^2 + y^2 + z^2 \ggg d^2$

• The only charge in the region $z \geq 0$ is point charge q (Note that here we are interested only in the region $z \geq 0$ and not the entire space because there is a point charge -q in the region $z \leq 0$ which was not there in our given system )

This system has the same boundary conditions on potential as that of the given system, But does this mean that the given system is exactly the same ( that is , has the same solution for $V(x ,y,z )$ ) as this system ?? Yes! This is guaranteed by the uniqueness theorem .

Now, evaluating the surface charge density is pretty straight forward:

Electric field near the surface of the conductor is $E = \dfrac { \sigma }{ \epsilon } \hat{z}$

Thus the charge density would be $\sigma = - \epsilon \dfrac{ dV }{ dz} at~ z= 0$ . Evaluate potential $V(x, y, z)$ using the " trick " and hence evaluate $\sigma$ . We get

$\sigma (x ,y ) = \dfrac{ - qd }{ 2 \pi ( x^2 + y^2 + z^2 )^{ \frac{3}{2} } }$

Thus $\color{#3D99F6}{ \sigma ( 3,4 ) = - 2.25079 × 10^{-3} ~C m^{-2}}$