Question

A large sheet carries uniform surface charge density $\sigma$. A rod of length $l$ has linear charge density $\lambda$ on one half and $-\lambda$ on the second half. The rod is hinged at mid-point $O$ and makes an angle $\theta$ with the normal to the sheet. The torque experienced by the rod is

• A. Zero
• B. $\frac{\sigma \lambda l^2}{8{ϵ}_0}\sin \theta$
• C. $\frac{\sigma \lambda l^2}{{ϵ}_0}\sin \theta$
• D. $\frac{\sigma \lambda l^2}{2{ϵ}_0}$

Answer 1

The correct option is B $\frac{\sigma \lambda l^2}{8{ϵ}_0}\sin \theta$


We know that electric field due to infinite sheet is given as

$E=\frac{\sigma }{2{ϵ}_0}$

Charge on $\text{OB},q_1=\frac{\lambda l}{2}$

So, Force on $q_1$,

$F_1=q_{1}E=\frac{\lambda \sigma l}{4{ϵ}_0}$ (towards right)

Further torque due to $F_1$ about $O$,

${\tau }_1=\frac{\lambda \sigma l}{4{ϵ}_0}\times \frac{l}{4}\sin \theta$

$\Rightarrow {\tau }_1=\frac{\sigma \lambda l^2}{16{ϵ}_0}\sin \theta$ (clockwise)

Similarly, for $\text{OA}$,

${\tau }_2=\frac{\sigma \lambda l^2}{16{ϵ}_0}\sin \theta$ (clockwise)

Further,
${\tau }_{net}=\overrightarrow{{\tau }_1}+\overrightarrow{{\tau }_2}$

$\Rightarrow {\tau }_{net}=\frac{\sigma \lambda l^2}{8{ϵ}_0}\sin \theta$ (clockwise)

Hence, option $\left(b\right)$ is correct answer.