Problem:

What is the electrostatic energy of a point charge placed near an grounded infinite conducting plane. Show that this is not equal to the energy of the point charge and it image.

Solution

A point charge \(q\) is placed at position \((0,0,d)\) in front of an grounded infinite conducting plane taken to be the \(X-Y\) plane. The electric field at a point, \(z>0\) is given by the field of the point charge and an image charge \(-q\) placed at \((0,0,-d)\). However care is required when using this result to compute the electrostatic energy of the system. The electrostatic energy of \(n\) point charges \(q_k, k=1,...,n\), located at \(\vec{r}_k\), is

\begin{equation}\label{EQ01} W_n = \frac{1}{4\pi\epsilon_0} \frac{1}{2} \frac{q_kq_j}{|\vec{r}_j -\vec{r}_k|} \end{equation} For two charges, \(q, -q\) separated by a distance \(d\) this expression takes the form \begin{equation}\label{EQ02} W_2 = -\frac{1}{4 \pi\epsilon_0}\frac{q^2}{2d} \end{equation} We will now show that the above expression does not give correct answer for the electrostatic energy of a point charge and an infinite conducting plane held to zero potential. We will compute the energy as work done in bringing the point charge \(q\) to a position \((0,0,d)\) in front of the conducting plane. When the charge \(q\) is at position \(z\), it experiences an attractive force \( F= - \frac{1}{4\pi\epsilon_0}\frac{q^2}{4z^2}\) coming from the image charge at \((0,0,-z)\). Therefore the electrostatic energy required to bring the charge from point \((0,0,a)\) to the position \((0,0,b)\) is

\begin{eqnarray}\label{EQ03}

W_{a\to b} &=& - \int_{a}^z \vec{F}dz\\\label{EQ04}

&=&\frac{1}{4\pi\epsilon_0} \int_a^b \frac{q^2}{4z^2} \, dz\\\label{EQ05}

&=& \frac{1}{4\pi\epsilon_0}\frac{q^2}{4z}\Big|_a^b\\\label{EQ06}

&=& - \frac{1}{4\pi\epsilon_0}\left(\frac{q^2}{4b} - \frac{q^2}{4a}\right).

\end{eqnarray}

The required electrostatic energy is obtained by taking limit \(a\to\infty\) and setting \(b=d\). This gives \begin{equation} W = -\frac{1}{4\pi\epsilon_0} \frac{q^2}{4d}. \end{equation} The above expression is half the value \eqref{EQ02} for two point charges separated by a distance \(2d\).