Question

As per this diagram a point charge $+q$ is placed at the origin $O$. Work done in taking another point charge $-Q$ from the point $A$ [coordinates $(0,a)]$

• A. zero
• B. $\left(\frac{-qQ}{4\pi {\epsilon }_0}\frac{1}{a^2}\right)\sqrt{2}a$
• C. $\left(\frac{qQ}{4\pi {\epsilon }_0}\frac{1}{a^2}\right)$
• D. $\left(\frac{qQ}{4\pi {\epsilon }_0}\frac{1}{a^2}\right)\sqrt{2}a$

Answer 1

Answer: (A)

The work done in carrying a test charge consists in product of difference of potentials at points $A$ and $B$ and value of test charge.

Potential at $A$
$V_A=\frac{1}{4\pi {\epsilon }_0}\frac{q}{a}$
Potential at $B$
$V_B=\frac{1}{4\pi {\epsilon }_0}\frac{q}{a}$
Thus, work done in carrying a test charge $-Q$ from $A$ to $B$
$W=\left(V_A-V_B\right)(-Q)=0$