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A capacitor is charged with a battery and energy stored is U. After disconnecting battery another capacitor of same capacity is connected in parallel with it. Then energy stored in each capacitor is :
A
$\dfrac{U}{2}$
B
$\dfrac{U}{4}$
C
4U
D
2U
Detailed Solution
Initial energy: $U = \dfrac{q^2}{2C}$, where q is the charge on the capacitor.
After the battery is disconnected and an identical uncharged capacitor is connected in parallel, the charge q is shared. Common potential:
$V_{common} = \dfrac{q_1 + q_2}{C_1 + C_2} = \dfrac{q + 0}{C + C} = \dfrac{q}{2C}$
Energy on each capacitor $= \dfrac{1}{2}CV_{common}^2 = \dfrac{1}{2}C\left(\dfrac{q}{2C}\right)^2 = \dfrac{q^2}{8C} = \dfrac{1}{4} \cdot \dfrac{q^2}{2C}$
$= \dfrac{U}{4}$
(The total energy is $\dfrac{U}{2}$; the other half is lost as heat in the connecting wires.)
After the battery is disconnected and an identical uncharged capacitor is connected in parallel, the charge q is shared. Common potential:
$V_{common} = \dfrac{q_1 + q_2}{C_1 + C_2} = \dfrac{q + 0}{C + C} = \dfrac{q}{2C}$
Energy on each capacitor $= \dfrac{1}{2}CV_{common}^2 = \dfrac{1}{2}C\left(\dfrac{q}{2C}\right)^2 = \dfrac{q^2}{8C} = \dfrac{1}{4} \cdot \dfrac{q^2}{2C}$
$= \dfrac{U}{4}$
(The total energy is $\dfrac{U}{2}$; the other half is lost as heat in the connecting wires.)