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. $\frac{\sigma \lambda \ell^{2}}{2 \varepsilon_{0}} \sin \theta$
• C. $\frac{\sigma \lambda \ell^{2}}{\varepsilon_{0}} \sin \theta$
• D. $\frac{\sigma \lambda \ell}{2 \varepsilon_{0}}$

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=\int\limits_{0}^{1} 2 \lambda \times \frac{\sigma}{2 \epsilon_{0}} \times \sin \theta \times xdx $
$=\frac{\sigma \lambda}{\epsilon_{0}} \sin \theta \times \frac{1^{2}}{2} $