Question

A large sheet carries uniform surface charge density $\sigma$. A rod of length $2l$ has a linear charge density $\lambda$ on one half and $-\lambda$ on the other half. The rod is hinged at mid point O and makes angle $\theta$ with the normal to the sheet. The torque experienced by the rod is :-

• A. $\frac{\sigma \lambda l^2}{2\epsilon_0} \cos \theta$
• B. $\frac{\sigma \lambda l}{\epsilon_0} \cos^2 \theta$
• C. $\frac{\sigma \lambda l^2 \sin \theta}{2\epsilon_0}$
• D. $\frac{\sigma \lambda l \sin^2 \theta}{\epsilon_0}$

Answer 1

Answer: (C)

$\frac{\sigma \lambda l^2 \sin \theta}{2\epsilon_0}$

Answer 2

To determine the torque experienced by the rod:

1. Electric Field due to a Large Sheet:
   $E = \frac{\sigma}{2\epsilon_0}$

2. Electric Dipole Moment of the Rod:
   The dipole moment $\vec{p}$ is calculated as:
   $\vec{p} = \int \vec{r} dq$
   Which results in:
   $p = \lambda l^2$

3. Torque on the Rod:
   The torque $\vec{\tau}$ is given by:
   $\vec{\tau} = \vec{p} \times \vec{E}$
   The magnitude of the torque is:
   $\tau = p E \sin\theta = \frac{\sigma \lambda l^2}{2\epsilon_0} \sin\theta$