Question

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. \(\frac U2\)
• B. \(\frac U4\)
• C. \(4U\)
• D. \(2U\)

Answer 1

The Correct Option is B

To solve this problem, we need to analyze the situation where a charged capacitor, initially holding energy \( U \), is connected with another capacitor of the same capacity after the battery is disconnected. Here’s how the system behaves:

1. U = \frac{1}{2}CV^2: Initially, the energy stored in the first capacitor is \( U \). This energy is due to the charge \( Q = CV \) on the capacitor.
2. When this charged capacitor is connected to another uncharged capacitor of the same capacitance \( C \), the total system's capacitance becomes \( 2C \).
3. At this point, the charge \( Q \) originally on the first capacitor gets redistributed equally between the two capacitors as they are identical.
4. The new potential \( V' \) across both capacitors is given by dividing the original charge \( Q \) by the new total capacitance \( 2C \): V' = \frac{Q}{2C} = \frac{CV}{2C} = \frac{V}{2}.
5. The energy stored in each of the capacitors when they are connected can be calculated using: U' = \frac{1}{2}C(V')^2 = \frac{1}{2}C \left( \frac{V}{2} \right)^2 = \frac{1}{2}C \frac{V^2}{4} = \frac{1}{8}CV^2.
6. Since the initial energy \( U \) is \frac{1}{2}CV^2, the energy stored in each capacitor is: U' = \frac{1}{4}U.

Therefore, the energy stored in each capacitor after the connection is \(\frac{U}{4}\), which makes the correct answer:

\(\frac{U}{4}\)