A 5 μF capacitor is charged by a 100 V supply. The supply is then disconnected and the charged capacitor is connected to another uncharged 3 μF capacitor. How much electrostatic energy of the first capacitor is lost in the process of attaining the steady situation?

Given,
capacitance of the capacitor-C= 5 μF

potential applied across the capacitance-V= 100V

∴

Initial energy stored in capacitor,

U_{i} = \frac{1}{2} {CV}^{2} = \frac{1}{2} \times 5 \times 10^{- 6} \times (100)^{2} = 2.5 \times 10^{- 2} J .

Charge on

5 μF

capacitor-q=CV

= 5 \times 10^{- 6} \times 100

= 5 \times 10^{- 4} C .

When, supply is disconnected and, charged capacitor is connected to another uncharged capacitor of 3 μF then, the two capacitors attain a common potential V.

V_{f} = \frac{total charge}{total capacitor} = \frac{5 \times 10^{- 4}}{(5 + 3) \times 10^{- 6}} = \frac{125}{2} V .

Energy stored in the capacitor, after combination

U_{f} = \frac{1}{2} C_{f} V_{f}^{2} = \frac{1}{2} \times (5 + 3) \times 10^{- 6} \times {(\frac{125}{2})}^{2} = 1.56 \times 10^{- 2} J

Therefore,
Electrostatic energy lost in the process of attaining the steady state = U_i – U_f = (2.5 – 1.56) x 10^–2 = 0.94 x 10^–2 J.

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