A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V . Another capacitor of capacitance 2 C is similarly charged to a potential difference 2

Question

A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V . Another capacitor of capacitance 2 C is similarly charged to a potential difference 2 V . The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

TestHub · Accepted Answer

The diagramatic representation of given problem is shown in the figure.

The net charge shared between the two capacitors is,

         $Q^{'} = Q_{2} - Q_{1} = 4 CV - CV = 3 CV$

The two capacitors will have the same potential, say $V^{'}$ .
The net capacitance of the parallel combination of the two capacitors will be,

           $C^{'} = C_{1} + C_{2} = C + 2 C = 3 C$

The potential difference across the capacitors will be,

           $V^{'} = \frac{Q^{'}}{C^{'}} = \frac{3 CV}{3 C} = V$

The electrostatic energy of the capacitors will be,

         $U^{'} = \frac{1}{2} C^{'} V^{'^{2}} = \frac{1}{2} (3 C) V^{2} = \frac{3}{2} {CV}^{2}$

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