First slide

Capacitance

Question

A capacitor is charged until its stored energy is 3 J and the charging battery is removed. Another uncharged capacitor is then connected across it and we find that the charge distributes equally. Final value of total energy stored in the electric fields is :

Easy

A

3 J
B

2.5 J
C

2 J
D

1.5 J

Solution

Initial total energy = $\frac{Q^{2}}{2 C} = 3 J$
Final stored energy = $\frac{{(\frac{Q}{2})}^{2}}{2 C} + \frac{{(\frac{Q}{2})}^{2}}{2 C} = \frac{Q^{2}}{4 C} = 1.5 J$

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