First slide

Capacitance

Question

A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system

Moderate

A

decreases by a factor of 2
B

remains the same
C

increases by a factor of 2
D

increases by a factor of 4

Solution

When the capacitor is charged by a battery of potential V, then energy stored in the capacitor,
$U_{i} = \frac{1}{2} C V^{2} . . . . (i)$

When the battery is removed and another identical uncharged capacitor is connected in parallel Common potential,
$V^{'} = \frac{C V}{C + C} = \frac{V}{2}$
$∴ Then the energy stored in the capacitor,$
$U_{f} = \frac{1}{2} (2 C) {(\frac{V}{2})}^{2} = \frac{1}{4} C V^{2} . . . . (i i)$
$∴ From eqns. (i) and (ii)$
$U_{f} = \frac{U_{i}}{2}$
that means the total electrostatic energy of resulting system will decreases by a factor of 2.

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