Positive and negative point charges of equal magnitude are kept at $\left(0,0, \frac{a}{2}\right)$ and $\left(0,0, \frac{-a}{2}\right)$, respectively. The work done by the electric field when another positive point charge is moved from $(-a, 0,0)$ to $(0, a, 0)$ is

(D) depends on the path connecting the initial and final positions

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Electrostatics

Positive and negative point charges of equal magnitude are kept at

(0, 0, \frac{a}{2})

and

(0, 0, \frac{- a}{2})

, respectively. The work done by the electric field when another positive point charge is moved from

(- a, 0, 0)

(0, a, 0)

Answer: (C)

This is an electric dipole configuration with charges

+ q

(0, 0, a /2)

and

- q

(0, 0, - a /2)

.

Initial position:

(- a, 0, 0)

Distance from

+ q

r_{1}^{i} = a^{2} + (a /2)^{2} = \frac{a 5}{2}

Distance from

- q

r_{2}^{i} = a^{2} + (a /2)^{2} = \frac{a 5}{2}

Final position:

(0, a, 0)

Distance from

+ q

r_{1}^{f} = a^{2} + (a /2)^{2} = \frac{a 5}{2}

Distance from

- q

r_{2}^{f} = a^{2} + (a /2)^{2} = \frac{a 5}{2}

Potential at initial point:

V_{i} = \frac{k q}{r _{1}^{i}} - \frac{k q}{r _{2}^{i}} = 0

Potential at final point:

V_{f} = \frac{k q}{r _{1}^{f}} - \frac{k q}{r _{2}^{f}} = 0

Work done by electric field:

W = q_{0} (V_{i} - V_{f}) = 0

Both points are equidistant from both charges, hence at the same potential. Work done is zero.

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Single correct option