Question

Three infinite long plane sheets carrying uniform charge densities
${\sigma }_1=-\sigma ,{\sigma }_2=+2\sigma$ and ${\sigma }_3=+3\sigma$ are placed parallel to the x-z plane at y =a, y=3a and y=4a as shown in Fig. 20.18. The electric field at point P is

• A. zero
• B. $-\frac{2\sigma }{{ϵ}_0}\hat{j}$
• C. $-\frac{3\sigma }{{ϵ}_0}\hat{j}$
• D. $\frac{3\sigma }{{ϵ}_0}\hat{j}$

Answer 1

The correct option is C

$-\frac{3\sigma }{{ϵ}_0}\hat{j}$

The electric field at a point P due to an infinite long plane sheet carrying a uniform charge density \sigma is given by
$E=\frac{\sigma }{2{ϵ}_0}$
It is independent of the distance of point P from the sheet and is, therefore, uniform. The direction of the electric field is away from the sheet and perpendicular to it if $\sigma$ is positive
and is towards the sheet and perpendicular to it if $\sigma$ is negative. Hence
$E_1=\frac{\sigma }{2{ϵ}_0}(-\hat{j})$ along -ve y-direction
$E_2=\frac{2\sigma }{2{ϵ}_0}(-\hat{j})$ along -ve y-direction
and $E_3=\frac{3\sigma }{2{ϵ}_0}(-\hat{j})$ along -ve y-direction
From the superposition principle, the net electric field at point P is
$E=E_1+E_2+E_3=\frac{\sigma }{2{ϵ}_0}(-\hat{j})+\frac{2\sigma }{2{ϵ}_0}(-\hat{j})+\frac{3\sigma }{2{ϵ}_0}(-\hat{j})=-\frac{3\sigma }{{ϵ}_0}\hat{j}$,which is choice (c).