A point charge of -3 μC is placed at the origin. The work done to move a +2 μC charge from (4 m, 0) to (2 m, 0) is:
Correct answer: B. -6.75 × 10⁻³ J
- A. 6.75 × 10⁻³ J
- B. -6.75 × 10⁻³ J
- C. Zero
- D. 3.38 × 10⁻³ J
Explanation
Work done is W = qΔV, where ΔV is the potential difference. Potential at r = 4 m is V₁ = k(-3 × 10⁻⁶)/4, and at r = 2 m is V₂ = k(-3 × 10⁻⁶)/2. Thus, ΔV = V₂ - V₁ = (9 × 10⁹)(-3 × 10⁻⁶)(1/2 -1/4) = -3375 V. For q = 2 × 10⁻⁶ C, W = (2 × 10⁻⁶)(-3375) = -6.75 × 10⁻³ J.
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