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,
$E=\frac{\sigma }{\epsilon _{o}}$ 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.