Question

Here's an alternative derivation of Eq. 3.10 (the surface charge density

induced on a grounded conducted plane by a point charge qa distance dabove

the plane). This approach (which generalizes to many other problems) does not

rely on the method of images. The total field is due in part to q,and in part to the

induced surface charge. Write down the zcomponents of these fields-in terms of

qand the as-yet-unknown $\mathit{\sigma }\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$-just below the surface. The sum must be zero,

of course, because this is inside a conductor. Use that to determine $\mathit{\sigma }$.

Short answer

The surface charge density induced on a grounded conducted plane by a point charge qa distance dabove the plane is $\frac{-qd}{2\pi {\left(x^2+y^2+d^2\right)}^{\frac{3}{2}}}$.

Step-by-step solution

Given data

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

 Step 2: Define electric field due to point charge and uniform surface charge density

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

${\overrightarrow{\mathbf{E}}}_{\mathbf{q}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{{\mathbf{r}}^{\mathbf{3}}}\hat{\mathbf{r}}$…… (1)

Here, ${\epsilon }_0$is the permittivity of free space.

The field due to a uniform plane surface charge density $\sigma$ is:

${\overrightarrow{\mathbf{E}}}_{\mathbf{\sigma }}\mathbf{=}\frac{\mathbf{\sigma }}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}}\hat{\mathbf{n}}$…… (2)

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

Charge density induced on the plane

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

$E_{qz}=-\frac{1}{4\pi {\epsilon }_0}\frac{qd}{{\left(x^2+y^2+d^2\right)}^{3/2}}$

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

$E_{\sigma z}=-\frac{\sigma }{2{\epsilon }_0}$

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

$$\begin{gathered} -\frac{1}{4\pi {\epsilon }_0}\frac{qd}{{\left(x^2+y^2+d^2\right)}^{3/2}}-\frac{\sigma }{2{\epsilon }_0}=0 \\ \sigma \left(x,y\right)=\frac{-qd}{2\pi {\left(x^2+y^2+d^2\right)}^{\frac{3}{2}}} \end{gathered}$$

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