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

To solve this problem, we need to determine the charge on capacitor \(Q\) when equilibrium is established after being connected to a charged capacitor \(P\). Initially, capacitor \(P\) is charged to a potential difference \(V = 6.0\,\text{V}\). The capacitance of \(P\) is \(C_P = 10 \times 10^{-6}\,\text{F}\). Thus, the initial charge on \(P\) is given by:  

Initial Charge on \(P\):
\[ q_P = C_P \times V = 10 \times 10^{-6}\,\text{F} \times 6\,\text{V} = 6 \times 10^{-5}\,\text{C} \] 
Upon connecting the capacitors \(P\) and \(Q\), they will reach a common potential difference \(V_f\) because they form a series circuit. The total charge is conserved:
\[ q_{\text{total}} = q_P = 6 \times 10^{-5}\,\text{C} \] The total capacitance \(C_{\text{total}}\) for capacitors in parallel is: 
\[ C_{\text{total}} = C_P + C_Q = 10 \times 10^{-6}\,\text{F} + 20 \times 10^{-6}\,\text{F} = 30 \times 10^{-6}\,\text{F} \] The final potential difference \(V_f\) across both capacitors is: 
\[ V_f = \frac{q_{\text{total}}}{C_{\text{total}}} = \frac{6 \times 10^{-5}\,\text{C}}{30 \times 10^{-6}\,\text{F}} = 2\,\text{V} \] Now, the charge \(q_Q\) on capacitor \(Q\) is given by: 
\[ q_Q = C_Q \times V_f = 20 \times 10^{-6}\,\text{F} \times 2\,\text{V} = 4 \times 10^{-5}\,\text{C} \] Therefore, in the form \( \alpha \times 10^{-5}\,\text{C} \), we have \( \alpha = 4 \). The computed value \( \alpha = 4 \) falls within the given range 4,4, confirming the solution's accuracy.