Explanation

Distance of point A from origin:

\[ r_A = \sqrt{4^2 + 4^2 + 2^2} \]

\[ = \sqrt{16 + 16 + 4} \]

\[ = \sqrt{36} = 6 \]

Distance of point B from origin:

\[ r_B = \sqrt{2^2 + 2^2 + 1^2} \]

\[ = \sqrt{4 + 4 + 1} \]

\[ = \sqrt{9} = 3 \]

Potential due to point charge:

\[ V = \frac{kQ}{r} \]

Work done:

\[ W = q (V_B - V_A) \]

\[ = qkQ \left( \frac{1}{r_B} - \frac{1}{r_A} \right) \]

\[ = (2\times10^{-6})(9\times10^9)(10^{-8}) \left( \frac{1}{3} - \frac{1}{6} \right) \]

\[ = (2\times10^{-6})(90) \left( \frac{1}{6} \right) \]

\[ = 30\times10^{-6} \text{ J} \]

Correct Answer: 30 × 10⁻⁶ J

Related Theory

This problem belongs to Electrostatics, specifically work done in conservative electric fields.

1. Electric Potential

Potential due to point charge:

\[ V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r} \]

It depends only on distance from charge.

2. Conservative Nature of Electric Field

Work done in electrostatic field is path independent.

It depends only on initial and final positions.

3. Work-Energy Relation

\[ W = q(V_B - V_A) \]

If charge moves closer to positive source charge, potential increases.

4. Electric Potential Energy

\[ U = \frac{kQq}{r} \]

Change in potential energy equals negative work done by field.

5. Important Concept

Distance must be calculated using 3D distance formula.

6. JEE Trap

• Students forget to compute distances correctly.
• Forget factor 1/4πε₀.
• Confuse work done by field vs external agent.

7. Equipotential Surfaces

All points at same distance have same potential.

If rA = rB, work would be zero.

8. Exam Relevance

Electrostatics always has at least one question in JEE Main.

Work, potential and field relation is core concept.

Related JEE Main Questions