Question

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

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

Answer 1

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

Torque on the rod is given by,
$\tau =Flsin\theta$ $\because F=qEandq=\lambda l$
Putting all the values will get,
$\tau =\left(\lambda lE\right)lsin\theta$
$=\lambda l^2\frac{\sigma }{2{\epsilon }_0}sin\theta =\frac{\sigma \lambda l^2}{2{\epsilon }_0}sin\theta$