Question

A capacitor \(P\) with capacitance \(10\times10^{-6}\,\text{F}\) is fully charged with a potential difference of \(6.0\,\text{V}\) and disconnected from the battery. The charged capacitor \(P\) is connected across another capacitor \(Q\) with capacitance \(20\times10^{-6}\,\text{F}\). The charge on capacitor \(Q\) when equilibrium is established will be \( \alpha \times 10^{-5}\,\text{C} \) (assume capacitor \(Q\) does not have any charge initially). The value of \( \alpha \) is __________.

Hint:

When charged capacitors are connected together, total charge remains conserved but potential redistributes.

Answer 1

Correct Answer: 4

Step 1: Find initial charge on capacitor \(P\).
Initial charge on \(P\) is given by
\[ Q_0 = C_P V \] \[ Q_0 = (10\times10^{-6})(6.0)=6\times10^{-5}\,\text{C} \]
Step 2: Apply charge conservation.
After connecting capacitors \(P\) and \(Q\), total charge is conserved.
Let the common final potential be \(V_f\).
\[ Q_0 = (C_P+C_Q)V_f \] \[ 6\times10^{-5} = (10\times10^{-6}+20\times10^{-6})V_f \] \[ 6\times10^{-5} = 30\times10^{-6}V_f \] \[ V_f = 2\,\text{V} \]
Step 3: Find charge on capacitor \(Q\).
\[ Q_Q = C_Q V_f \] \[ Q_Q = (20\times10^{-6})(2)=4\times10^{-5}\,\text{C} \]
Step 4: Compare with given form.
\[ Q_Q = \alpha\times10^{-5}\,\text{C} \Rightarrow \alpha=4 \]
Final Answer:
\[ \boxed{4} \]