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Electrostatic Potential and Capacitance
Sharing of Charge between Capacitors
3 questions, each with its answer.
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A capacitor of capacitance $C=900\ pF$ is charged fully by 100 V battery B as shown in figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance $C=900\ pF$ as shown in figure (b). The electrostatic energy stored by the system (b) is
Common potential $V_c = \frac{C_1V_1 + C_2V_2}{C_1 + C_2} = \frac{C \times 100 + C \times 0}{C + C} = 50$ V Energy stored in system (b) $= 2 \times \frac{1}{2}CV_c^2 = CV_c^2$ $= 900 \times 10^{-12} \times 50 \times 50 = 2.25 \times 10^{-6}$ J
That is every 2022 question in Sharing of Charge between Capacitors.
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