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

Options

A. `0`
B. `(sigma lambda l^(2))/(2 epsilon_(0)) sin theta`
C. `(sigma lambda l^(2))/(epsilon_(0)) sin theta`
D. `(sigma lambda l)/(2 epsilon_(0))`

Answer

B

Explanation

`tau=(lambda l E)l sin theta`
`=lambda l^(2) (sigma)/(2 epsilon_(0))=(sigma lambda l^(2))/(2 epsilon) sin theta`