Question

A parallel plate capacitor of capacitance 'C' is connected to a battery and charged to a potential 'V'. Another capacitor of capacitance '\(3C\)' is charged to a potential '\(3V\)'. The charging battery is then disconnected and both the capacitors are connected in parallel to each other such that positive terminal of one is connected to negative terminal of the other. The final energy of the configuration is

• A. \(1.5\,CV^2\)
• B. \(6.5\,CV^2\)
• C. \(8.0\,CV^2\)
• D. \(18.0\,CV^2\)

Hint:

Opposite polarity means the charges subtract; then find the common voltage and the stored energy.

Answer 1

The Correct Option is C

Step 1: Initial Charges:
First capacitor: \(q_1=CV\). Second capacitor: \(q_2=(3C)(3V)=9CV\).

Step 2: Connect Positive to Negative:
When the positive plate of one is joined to the negative plate of the other, the charges partly cancel. The net charge is
\[ q=9CV-CV=8CV \]
The equivalent capacitance in parallel is \(C+3C=4C\).

Step 3: Common Voltage:
\[ V'=\frac{q}{C_{eq}}=\frac{8CV}{4C}=2V \]

Step 4: Final Energy:
\[ U=\frac12C_{eq}V'^2=\frac12(4C)(2V)^2=8CV^2 \]
The values \(1.5CV^2\), \(6.5CV^2\) and \(18CV^2\) do not follow from a net charge of \(8CV\) on a total capacitance of \(4C\). So (C) is correct.

Final Answer:
The final energy is \(8.0\,CV^2\), option (C). \[ \boxed{\text{(C) } 8.0\,CV^2} \]