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 electric force experienced by the rod is,

• A. $0$
• B. $\frac{\sigma \lambda {\ell }^2}{2{ϵ}_0}sin\theta$
• C. $\frac{\sigma \lambda {\ell }^2}{{ϵ}_0}sin\theta$
• D. none of these.

Answer 1

Answer: (A)

Given,
surface charge density of sheet $=\sigma$ ,
length of rod $=2\ell$ ,
linear charge density of half rod $=\lambda$ ,
linear charge density of other half rod $=-\lambda$ ,
angle made by rod with the normal to the sheet $=\theta$ .

Electric field due to the sheet is given by,
$E=\frac{\sigma }{{ϵ}_o}$ normal to the sheet and uniform.

Total force on upper half rod,
$F_1=qE=\left(\lambda l\right)E$ .
Total force on lower half rod,
$F_2=qE=-\left(\lambda l\right)E$ .
Total force on rod,
$F=F_1+F_2=\left(\lambda l\right)E-\left(\lambda l\right)E=0$ .
So, near the plate, the field is uniform. Equal and opposite forces are experienced by upper and lower halves.