JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2026
- Hard
- Numerical
A capacitor P with capacitance \( 10\times10^{-6} F \) is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor P is connected across another capacitor Q with capacitance \( 20\times10^{-6} F \). The charge on capacitor Q when equilibrium is established will be \( \alpha\times10^{-5} C \).(assume capacitor \(Q\) does not have any charge initially),The value of \( \alpha \) is __________ .
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) 4
Step-by-step Solution
Detailed explanation

\(V=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}=\frac{10^{-5} \times 6+0}{3 \times 10^{-5}}\)
V=2 volt
\(Q _2= C _2 V=2 \times 10^{-5} \times 2=4 \times 10^{-5} C\)
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