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 [co-ordinates (0, a)] to another point B [co-ordinates (a, 0)] along the straight path AB is :

• A. zero
• B. $\frac{-qQ}{4\pi\epsilon_0}\frac{1}{a^2}\sqrt{2}a$
• C. $\frac{qQ}{4\pi\epsilon_0}\frac{1}{a^2}\frac{a}{\sqrt{2}}$
• D. $(\frac{qQ}{4\pi\epsilon_0}\frac{1}{a^2})\sqrt{2}a$

Answer 1

Answer: (A)

zero

Answer 2

The work done in moving a charge in an electrostatic field is path-independent and depends only on the initial and final points. The work done by an external agent is given by $W_{ext} = q_{test} (V_B - V_A)$.

The distance of point A (0, a) from the origin (0, 0) is $r_A = \sqrt{(0-0)^2 + (a-0)^2} = a$. The distance of point B (a, 0) from the origin (0, 0) is $r_B = \sqrt{(a-0)^2 + (0-0)^2} = a$.

The electric potential at a distance $r$ from a point charge $q'$ is $V = \frac{1}{4\pi\epsilon_0} \frac{q'}{r}$. The potential at A due to +q at the origin is $V_A = \frac{1}{4\pi\epsilon_0} \frac{q}{a}$. The potential at B due to +q at the origin is $V_B = \frac{1}{4\pi\epsilon_0} \frac{q}{a}$.

Since $V_A = V_B$, the potential difference $V_B - V_A = 0$. The work done in moving charge -Q from A to B is $W_{ext} = (-Q)(V_B - V_A) = (-Q)(0) = 0$.