Question

A large sheet carries uniform surface charge density $\sigma$. A rod of length $2\ell$ has a 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 (The rod is contained in a plane containing the normal to sheet).

• A. 0
• B. $\frac{\sigma \lambda {\ell }^2}{2{\epsilon }_0}\sin \theta$
• C. $\frac{\sigma \lambda {\ell }^2}{{\epsilon }_0}\sin \theta$
• D. $\frac{\sigma \lambda \ell }{2{\epsilon }_0}$

Answer 1

Answer: (B)

The situation is shown clearly in figure, the two elements of same width dx are as shown in figure.

The torque experienced by these elements is
$d\tau =2\lambda Edx\times x\sin \theta$
$\tau =\int d\tau =\underset{0}{\overset{1}{\int }}2\lambda \times \frac{\sigma }{2{ϵ}_0}\times \sin \theta \times xdx$
$=\frac{\sigma \lambda }{{ϵ}_0}\sin \theta \times \frac{1^2}{2}$