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A variable condenser is permanently connected to a 100 V battery. If the capacity is changed from 2 $μ$ F to 10 $μ$ F, then change in energy is equal to

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$2 \times 10^{- 2} J$

$2.5 \times 10^{- 2} J$

$3.5 \times 10^{- 2} J$

$4 \times 10^{- 2} J$

answer is D.

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Detailed Solution

Change in energy of condenser,

$ΔU = U_{2} - U_{1} = \frac{1}{2} C_{2} V^{2} - \frac{1}{2} C_{1} V^{2}$

$= \frac{V^{2}}{2} (C_{2} - C_{1}) = \frac{(100)^{2}}{2} (10 \times 10^{- 6} - 2 \times 10^{- 6})$

$= 4 \times 10^{- 2} J$

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