Question

An infinite sheet carrying a uniform surface charge density o lies on the $xy$-plane. The work done to carry a charge $q$ from the point $A=a(\hat{i}+2\hat{j}+3\hat{k})$ to the point $B=a(\hat{i}-2\hat{j}+6\hat{k})$ (where $a$ is a 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)

The given
$A=a(\hat{i}+2\hat{j}+3\hat{k})$
$B=a(\hat{i}-2\hat{j}+6\hat{k})$
$AB=OB-OA$
$=a(\hat{i}-2\hat{j}+6\hat{k})-a(\hat{i}+2\hat{j}+3\hat{k})$
$AB=a(-4\hat{j}+3\hat{k})$
Work done $=q\left(\frac{\sigma }{2{\epsilon }_0}\right)\hat{k}\cdot AB$ (along to $Z$-axis)
$=q\left(\frac{\sigma }{2{\epsilon }_0}\right)\hat{k}\cdot a(-4\hat{j}+3\hat{k})=\frac{3q\sigma a}{2{\epsilon }_0}$
$(\because \hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1$ and $\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{k}=\hat{k}\cdot \hat{i}=0)$