Two grounded infinite conducting planes are kept along the \(XZ\) and \(YZ\) planes, see Fig.2. A charge \(q\) is placed at (4,3). Find the force acting on the charge \(q\). |
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Solution
The given charge is located at \(B\). There will be three image charges \(-q,q,-q\) located at \(P,Q,R\) with
coordinates \((-4,3)\),\((-4,-3)\) and \((4,-3)\) respectively. See above figure.
The force due to the image charge at \(P\) is along the negative \(X\)-axis and
\[ \vec{F}_1= -\frac{q^2}{4\pi\epsilon_0} \frac{1}{64} \hat{i}\]
The force due to the image charge at \(Q\) is along \(OB\) and
\begin{eqnarray}
\vec{F}_2 = \frac{q^2}{4\pi\epsilon_0}\frac{1}{100}\Big(\frac{4}{5}\hat{i}
+ \frac{3}{5}\hat{j}\Big)
\end{eqnarray}
The force due to the image charge at \(R\) is along the negative \(Y\)-axis and
\[ \vec{F}_3= -\frac{q^2}{4\pi\epsilon_0} \frac{1}{36} \hat{j}\]
Adding the three forces we get
\begin{eqnarray}
\vec{F} &=& \vec{F}_1+ \vec{F}_2 + \vec{F}_3\\
&=& \frac{q^2}{4\pi\epsilon_0}\Big[\Big(-\frac{1}{64}+ \frac{1}{125}\Big)
\hat{i} +\Big(\frac{3}{500}-\frac{1}{36}\Big) \hat{j} \Big]\\
&=&\frac{q^2}{4\pi\epsilon_0}\Big[\frac{-125+64}{64\times125} \hat{i}
+\frac{108-500}{500\times36}\hat{j} \Big]\\
&=& -\frac{q^2}{4\pi\epsilon_0}\Big[\frac{61}{8000} \hat{i} +
\frac{49}{2250}\hat{j}\Big].
\end{eqnarray}