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: Understanding the Question:
A 800 pF capacitor is charged to 100 V and then cut off from the battery. It is then joined to an identical uncharged 800 pF capacitor. We need the total energy afterwards.

Step 2: Charge Stored at the Start:
\[ Q = CV = 800\times10^{-12}\times100 = 8\times10^{-8}\ \text{C} \] Since the battery is disconnected, this total charge stays the same.

Step 3: Energy Stored at the Start:
\[ U_i = \frac{1}{2}CV^2 = \frac{1}{2}\times800\times10^{-12}\times(100)^2 = 4\times10^{-6}\ \text{J} \]

Step 4: After Connecting:
The two capacitors are joined in parallel, so the common capacitance is \(C' = 1600\) pF. The charge is still \(Q\). The common voltage is \[ V' = \frac{Q}{C'} = \frac{8\times10^{-8}}{1600\times10^{-12}} = 50\ \text{V} \]

Step 5: Final Energy:
\[ U_f = \frac{1}{2}C'V'^2 = \frac{1}{2}\times1600\times10^{-12}\times(50)^2 = 2\times10^{-6}\ \text{J} \] Half of the energy is lost, mostly as heat and radiation while the charge flows.

Step 6: Checking the Options:
Option 1 gives \(2\times10^{-6}\) J and matches. Options 2 and 3 are not obtained by any correct step. Option 4 has a positive power, \(10^6\), which is far too large.

Final Answer:
The final energy is \(2\times10^{-6}\) J, so option 1 is correct. \[ \boxed{2\times10^{-6}\ \text{J}} \]