91Ó°ÊÓ

There is a point charge qata distance dabove a grounded conducted plane.

The field due to a point charge q at a distance r from it is

${\stackrel{\mathbf{→}}{\mathbf{E}}}_{\mathbf{q}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\frac{\mathbf{q}}{{\mathbf{r}}^{\mathbf{3}}}\hat{\mathbf{r}}$…… (1)

Here, ${\mathrm{Î\mu }}_0$is the permittivity of free space.

The field due to a uniform plane surface charge density $\mathrm{σ}$ is:

${\stackrel{\mathbf{→}}{\mathbf{E}}}_{\mathbf{σ}}\mathbf{=}\frac{\mathbf{σ}}{\mathbf{2}{\mathbf{Î\mu }}_{\mathbf{0}}}\hat{\mathbf{n}}$…… (2)

Here, $\hat{n}$is the unit vector perpendicular to the plane and pointing away from it.

From equation (2), the z component of electric field on any point on the conducting plane due to the point charge q at a height d is

$E_{qz}=-\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{qd}{{\left(x^2+y^2+d^2\right)}^{3/2}}$

Close to the surface, the induced surface charge can be considered constant. From equation (2) the field just below the surface is

$E_{\mathrm{σ}z}=-\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}$

The field inside a conductor should be zero. Hence, the density is solved as:

$$\begin{gathered} -\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{qd}{{\left(x^2+y^2+d^2\right)}^{3/2}}-\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}=0 \\ \mathrm{σ}\left(x,y\right)=\frac{-qd}{2\mathrm{Ï€}{\left(x^2+y^2+d^2\right)}^{\frac{3}{2}}} \end{gathered}$$

Thus, the induced charge density is $\frac{-qd}{2\mathrm{Ï€}{\left(x^2+y^2+d^2\right)}^{\frac{3}{2}}}$.