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Two charges 10 $μ$ C and 5 $μ$ C are 9 cm apart. If they are shifted by 3 cm closer, then the work done by the field is

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2.5 J

- 2.5 J

12.5 J

-5 J

answer is B.

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Detailed Solution

$U_{i} = \frac{1}{4 π ε_{o}} \frac{10 \times 10^{- 6} \times 5 \times 10^{- 6}}{9 \times 10^{- 2}}$

$U_{f} = \frac{1}{4 π ε_{o}} \frac{10 \times 10^{- 6} \times 5 \times 10^{- 6}}{6 \times 10^{- 2}}$

$W = U_{f} - U_{i}$

$= \frac{50 \times 10^{- 12} \times 9 \times 10^{9}}{10^{- 2}} [\frac{1}{9} - \frac{1}{6}]$

$= 45 \times \frac{- 1}{18} = - 2.5 J$

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