Question

A 800 pF capacitor is charged by a 100 V battery. After some time, the battery is disconnected. The capacitor is then connected across another 800 pF capacitor. The electrostatic energy stored in the combination will be

• A. \(2 \times 10^{-6}\) J
• B. \(3.8 \times 10^{-6}\) J
• C. \(4.2 \times 10^{-6}\) J
• D. \(6 \times 10^{6}\) J

Hint:

Charge stays constant and capacitance doubles, so the energy \(Q^2/2C\) becomes half of \(\frac{1}{2}CV^2\).

Answer 1

The Correct Option is A

Step 1: Method: Use the Energy Formula in Terms of Charge:
Write energy as \(U = Q^2/2C\). The charge is fixed at \(Q\) after the battery is removed, so this form is handy.

Step 2: Initial Energy:
\[ U_i = \frac{Q^2}{2C} \] with \(Q = CV = 8\times10^{-8}\) C, this gives \[ U_i = \frac{(8\times10^{-8})^2}{2\times800\times10^{-12}} = \frac{6.4\times10^{-15}}{1.6\times10^{-9}} = 4\times10^{-6}\ \text{J} \]

Step 3: Final Energy:
After joining in parallel the capacitance doubles to \(2C\), and Q is unchanged. \[ U_f = \frac{Q^2}{2(2C)} = \frac{U_i}{2} = 2\times10^{-6}\ \text{J} \]

Step 4: Conclusion:
The energy is halved, giving \(2\times10^{-6}\) J. This is the first printed option.

Final Answer:
\[\boxed{U_f = 2\times10^{-6}\ \text{J}}\]