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 $A = a(\hat i + 2\hat j + 3\hat k)$ to the point $B = a(\hat i - 2\hat j + 6\hat k)$ (where $a$ is a constant with dimension of length and ${ \in _o}$ is the permittivity of free space) is:
(A) $\dfrac{{3\sigma aq}}{{2{ \in _o}}}$
(B) $\dfrac{{2\sigma aq}}{{{ \in _o}}}$
(C) $\dfrac{{5\sigma aq}}{{2{ \in _o}}}$
(D) $\dfrac{{3\sigma aq}}{{{ \in _o}}}$

Answer 1

Hint
In these kinds of situations the first thing important is to notice the given problem statement. Specially this is a question regarding little manipulation. First we just have to sort what we got and then we just have to interpret the missing quantities for the equation. In this question we will get the answer simply by putting the values in the equation forwork done through the vector analysis.

Complete step by step answer
Here, first we have to find sort out the given quantities:
Point $A$ is given as: $A = a(\hat i + 2\hat j + 3\hat k)$
Point $B$ is given as: $B = a(\hat i - 2\hat j + 6\hat k)$
Uniform surface charge density: $\sigma$
Now, the solution:
Firstly the Electric field due to infinite plane: $\vec E = \dfrac{\sigma }{{2{ \in _o}}}\hat k$
And, the work done in the movement of the charge :
$W = qV W = q(\vec E.\vec AB)$
Now, the value of $AB$:
$\vec A\vec B = \vec B - \vec A = a(\hat i - 2\hat j + 6\hat k) - a(\hat i + 2\hat j + 3\hat k) \vec A\vec B = a( - 4\hat j + 3\hat k)$
From this we get:
$W = q(\dfrac{\sigma }{{2{ \in _o}}}k.a( - 4\hat j + 3\hat k)) \\\ W = q\dfrac{\sigma }{{2{ \in _o}}}a(0 + 3) \\\$\therefore$W = \dfrac{{3\sigma aq}}{{2{ \in _o}}}$
Option (A) is correct.

Note
Most important thing here is to notice the given values, then we can easily derive the missing quantities . Apart from that it is necessary to observe what process is straight and easy to find the rest values and then we can execute the formulas. In these types of questions interpreting the correct formula to be used is necessary unless it will get twisted.