Question

A \(1\,\mu F\) capacitor is charged to \(12\,V\) and then connected to an identical uncharged capacitor. What will be the common potential?

• A. \(12\,V\)
• B. \(6\,V\)
• C. \(3\,V\)
• D. \(4\,V\)

Hint:

For two identical capacitors, one charged and the other uncharged, \[ \boxed{V_f=\frac{V}{2}} \] because the total charge remains constant while the total capacitance doubles. Always apply the principle of conservation of charge in capacitor-sharing problems.

Answer 1

The Correct Option is B

Concept: When a charged capacitor is connected to an identical uncharged capacitor, charge redistributes between the two capacitors until both attain the same potential. The total charge remains conserved throughout the process. The charge stored in a capacitor is given by \[ Q=CV, \] where \[ Q=\text{Charge}, \] \[ C=\text{Capacitance}, \] \[ V=\text{Potential difference}. \] If two identical capacitors are connected together, the equivalent capacitance becomes \[ C_{\text{eq}}=C+C=2C. \] The common potential after redistribution is \[ V_f=\frac{Q_{\text{total}}}{C_{\text{eq}}}. \]

Step 1: Write the given data.
Given, \[ C=1\,\mu F, \] Initial potential of the charged capacitor, \[ V=12\,V. \] The second capacitor is initially uncharged.

Step 2: Calculate the initial charge stored on the charged capacitor.
Using, \[ Q=CV, \] we get \[ Q=(1\,\mu F)(12\,V). \] Hence, \[ Q=12\,\mu C. \] Thus, \[ \boxed{Q_{\text{total}}=12\,\mu C.} \]

Step 3: Find the equivalent capacitance after connection.
Since both capacitors are identical, \[ C_{\text{eq}} = 1+1 = 2\,\mu F. \] Therefore, \[ \boxed{C_{\text{eq}}=2\,\mu F.} \]

Step 4: Calculate the common potential.
Using \[ V_f=\frac{Q_{\text{total}}}{C_{\text{eq}}}, \] we obtain \[ V_f = \frac{12\,\mu C}{2\,\mu F} = 6\,V. \] Therefore, \[ \boxed{V_f=6\,V.} \] Hence, the correct answer is \[ \boxed{\textbf{Option (B)}}. \]