Consider two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of elec

Question

Consider two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is :NEET 2026 (Cancelled)

TestHub · Accepted Answer

Initial energy of the charged capacitor is given by $U_{i} = \frac{1}{2} C V^{2}$ .

When it is connected to an identical uncharged capacitor, the common potential becomes $V^{'} = \frac{C V}{C + C} = \frac{V}{2}$ .

Final energy of the system is $U_{f} = \frac{1}{2} (2 C) V^{'2} = \frac{1}{2} (2 C) (\frac{V}{2})^{2} = \frac{1}{4} C V^{2}$ .

Energy lost during the process is $Δ U = U_{i} - U_{f} = \frac{1}{2} C V^{2} - \frac{1}{4} C V^{2} = \frac{1}{4} C V^{2}$ .

Substituting the given values $C = 200 \times 1 0^{- 12}$ F and $V = 100$ V:

$Δ U = \frac{1}{4} \times 200 \times 1 0^{- 12} \times (100)^{2}$

$Δ U = 50 \times 1 0^{- 12} \times 1 0^{4} = 0.5 \times 1 0^{- 6}$ J.

Answer: $0.5 \times 1 0^{- 6}$ J

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