# [YMP/EM-04003] Electrostatic energy of a point charge and a grounded infinite conducting plane.

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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
$$\label{EQ01} W_n = \frac{1}{4\pi\epsilon_0} \frac{1}{2} \frac{q_kq_j}{|\vec{r}_j -\vec{r}_k|}$$ For two charges, $$q, -q$$ separated by a distance $$d$$ this expression takes the form $$\label{EQ02} W_2 = -\frac{1}{4 \pi\epsilon_0}\frac{q^2}{2d}$$ 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 $$W = -\frac{1}{4\pi\epsilon_0} \frac{q^2}{4d}.$$ The above expression is half the value \eqref{EQ02} for two point charges separated by a distance $$2d$$.

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