ElectrostaticsHard
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 2C is similarly charged to a potential difference 2V. The charging battery is now disconnect and the capcitors 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
Options
A.zero
B.
C.
D.
Solution
The digramatic representation of given problem is shown in the figure
The net charge shared between the two capacitors is
Q′ = Q2 - Q1 = 4CV - CV = 3CV
he two capacitors will have the same potential, say V′
The net capacitance of the parallel combination of two capacitance will be
C′ = C1 + C2 = C + 2C = 3C
The potential difference across the capacitance will be
The electrostatic energy of the capacitors will be
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