A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. Electrostatic energy lost in the process is __________ [2023]

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Solution

Q.
A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. Electrostatic energy lost in the process is __________ $μ J$ [2023]

Ans.
(6)

$Q = C V = 600 \times 10^{- 12} \times 200 = 12 \times 10^{- 8} C$

$Initial energy = \frac{1}{2} C V^{2}$

$= \frac{1}{2} \times 600 \times 10^{- 12} \times {(200)}^{2} = 12 μ J$

When connected to another uncharged capacitor

charge will be equally distributed on identical capacitors

$Q' = \frac{Q}{2} = 6 \times 10^{- 8}$

$Final energy = 2 \times \frac{Q'^{2}}{2 C} = \frac{Q'^{2}}{C}$

$= \frac{{(6 \times 10^{- 8})}^{2}}{600 \times 10^{- 12}} = 6 μ J$

$Energy lost = Initial energy - Final energy$

$= (12 - 6) μ J = 6 μ J$

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