A battery of 10 V is connected to a capacitor of 0.1 F. The battery is now removed and the capacitor is then connected to a second uncharged capacitor of same capacitance. Calculate the total energy stored in the system.

Solution

Given,
Capacitance of the capacitor, C= 0.1 F
Potential applied across the capactor, V = 10 V
Therefore,
Energy stored in the first capacitor is

U_{i} = \frac{1}{2} {CV}^{2} = \frac{1}{2} \times 0.1 \times (10)^{2} = 5 J .

Given, that the battery is removed and the charged capacitor is connected to an uncharged capacitor of the same capacitance.

Suppose the common potential is V.

Then, using the law of conservation of charge,
Charge on each capacitor, q' = CV'
and,

q' = \frac{q}{2}

∴ Total energy stored in the capacitor is

U_{f} = 2 \times \frac{1}{2} q' V' = q' \times \frac{q'}{C} = \frac{1}{4} \frac{q^{2}}{C}

[q' = q / 2, q = CV]

= \frac{1}{2} \times \frac{1}{2} {CV}^{2} = \frac{1}{2} \times U_{i} = \frac{1}{2} \times 5 = 2.5 J .

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Electrostatic Potential And Capacitance

A battery of 10 V is connected to a capacitor of 0.1 F. The battery is now removed and the capacitor is then connected to a second uncharged capacitor of same capacitance. Calculate the total energy stored in the system.

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