JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A capacitor \(C _{1}\) of capacitance \(5\,\mu F\) is charged to a potential of \(30\,V\) using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor \(C _{2}\) of capacitance \(10\,\mu F\) as shown in figure. When the switch is closed charge flows between the capacitors. At equilibrium, the charge on the capacitor \(C _{2}\) is________ \(\mu C\)

- A \(100\)
- B \(101\)
- C \(105\)
- D \(111\)
Answer & Solution
Correct Answer
(A) \(100\)
Step-by-step Solution
Detailed explanation
Before closing the switch \(Q = C _{1} V _{0}=5 \times 30=150\,\mu C\) After closing the switch \(V =\frac{ Q }{ C _{1}+ C _{2}}=\frac{150}{10+5}=10\,V\) \(Q _{2}= C _{2} V =10 \times 10=100\,\mu C\)
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