Electrostatic Potential and Capacitance - Result Question 42
44. A capacitor is charged to store an energy $U$. The charging battery is disconnected. An identical capacitor is now connected to the first capacitor in parallel. The energy in each of the capacitor is
(a) $U / 2$
(b) $3 U / 2$
(c) $U$
(d) $U / 4$
[2000]
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Answer:
Correct Answer: 44. (d)
Solution:
- (d) In first case, when capacitor $C$ connected with battery.
Energy stored, $u_1=\frac{1}{2} CV^{2}=\frac{1}{2} \frac{Q^{2}}{C}$
When battery is disconnected and another capacitor of same capacity is connected in parallel to first capacitor, then
$C _{\text{eq }}=C^{\prime}=2 C$
Voltage across each capacitor, $V^{\prime}=\frac{Q}{2 C}$
$\therefore \quad$ Energy stored, $V_2=\frac{1}{2} CV^{2}$
$=\frac{1}{2} C(\frac{Q}{2 C})^{2}=\frac{1}{4} \cdot \frac{1}{2} \frac{Q^{2}}{C}=\frac{U_1}{4}$
