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JEE · NEET Previous Year Question

$Two positive point charges of 10 μC and 12 μC are placed 10 cm apart in air. The work done to bring them 6 cm closer is$

Topic: JEE PYQ

Answer: D. Full step-by-step solution below.

A $8.1 J$
B $3.2 J$
C $9 J$
D $None$

Correct answer: D

Solution

(d) We have r_{i} = 10 cm and r_{f} = 4 cm Work done = Δ U = U_{f} - U_{i} = 9 \times 10^{9} \times 10 \times 10^{- 6} \times 12 \times 10^{- 6} (\frac{1}{0.04} - \frac{1}{0.1}) = 16.2

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