Electrostatics — Easy JEE physics MCQ
Work done in moving a charge of $2$ μC between two points with potential difference $10$ V is:
- A. $20$ μJ
- B. $5$ μJ
- C. $0.2$ J
- D. $2$ mJ
Solution
$W = q\Delta V = 2 \times 10^{-6} \times 10 = 20 \times 10^{-6}$ J $= 20$ μJ
