Question

An infinite sheet carrying a uniform surface charge density $\sigma$ lies on the xy-plane. The work done to carry a charge q from the point $\overrightarrow{A}=a(\hat{i}+2j+3\hat{k})$ to point $\overrightarrow{B}=a(\hat{i}-2\hat{j}+6\hat{k})$ (where a is constant with the dimension of length and ${\epsilon }_0$ is the permittivity of free space) is :

• A. $\frac{3\sigma aq}{2{\epsilon }_0}$
• B. $\frac{2\sigma aq}{{\epsilon }_0}$
• C. $\frac{5\sigma aq}{{2\epsilon }_0}$
• D. $\frac{3\sigma aq}{{\epsilon }_0}$

Answer 1

Answer: (A)

$\frac{3\sigma aq}{2{\epsilon }_0}$

The electric field due to a uniform surface charge $\sigma$ is $\frac{\sigma }{2{ϵ}_0}\hat{k}$.

Hence the work done in moving from A to B is $q\overrightarrow{E}.(\overrightarrow{B}-\overrightarrow{A})$

$=q\left(\frac{\sigma }{2{ϵ}_0}\hat{k}\right).(-4a\hat{j}+3a\hat{k})$

$=\frac{3\sigma aq}{2{ϵ}_0}$