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[QUIZ/EM-04001] A charge and two infinite planes

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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 \(q\). 

 Point charge and two perpendicular planes


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}

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