Question 2014.S3.31: A perfectly conducting metal plate is placed in x-y plane in......
A perfectly conducting metal plate is placed in x-y plane in a right handed coordinate system. A charge of +32 \pi \varepsilon_0 \sqrt{2} Coulombs is placed at coordinate (0, 0, 2). {ε}_{0} is the permittivity of free space. Assume \hat{i}, \hat{j}, \hat{k} to be unit vectors along x, y and z axes, respectively. At the coordinate (\sqrt{2}, \sqrt{2},0), the electric field vector \vec{E} (Newtons/Coulombs) will be
(a) 2 \sqrt{2} \hat{k} (b)-2 \hat{k} (c)2 \hat{k} (d)-2 \sqrt{2} \hat{k}
Since there is a conducting plane, we can assume that an image of the charge (—Q) exists on the negative z-axis as shown in the figure below.
The electric field vector in the x-y plane is cancelled and hence the net is zero. The electric field vector acts along the negative z-direction. The field intensities E_1 \text { and } E_2 due to charges Q (0, 0, 2) and —Q (0, 0, —2) at point (\sqrt{2}, \sqrt{2},0), are given by
E=E_1+E_2=\frac{1}{4 \pi \varepsilon_0}\left[\frac{Q_1 r_1}{r_1^3}+\frac{Q_2 r_2}{r_2^3}\right]=\frac{1}{4 \pi \varepsilon_0}\left[\frac{Q_1}{r_1^2}+\frac{Q_2}{r_2^2}\right]
where Q_{1}=Q{\mathrm{~and~}}Q_{2}=-Q
\begin{aligned}& r_1=(\sqrt{2}, \sqrt{2}, 0)-(0,0,2)=\sqrt{2} \hat{i}+\sqrt{2} \hat{j}-2 \hat{k} \\& r_2=(\sqrt{2}, \sqrt{2}, 0)-(0,0,-2)=\sqrt{2} \hat{i}+\sqrt{2} \hat{j}+2 \hat{k}\end{aligned}
Substituting in the above expression, we have
\begin{aligned}E= & \frac{1}{4 \pi \varepsilon_0}\left[\left\lgroup \frac{Q}{16 \sqrt{2}}(\sqrt{2} \hat{i}+\sqrt{2} \hat{j}-2 \hat{k} \right\rgroup -\left\lgroup \frac{Q}{16 \sqrt{2}}(\sqrt{2} \hat{i}+\sqrt{2} \hat{j}+2 \hat{k} \right\rgroup\right] \\& =\frac{Q}{16 \sqrt{2} 4 \pi \varepsilon_0}[-4 \hat{k}]\end{aligned}
Substituting for Q, we have
E=\frac{32 \sqrt{2} \pi \varepsilon_0}{16 \sqrt{2} \pi \varepsilon_0}(-\hat{k})=-2 \hat{k}
