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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2021
  • Hard
  • Numerical

A \(2\, \mu F\) capacitor \(C _{1}\) is first charged to a potential difference of \(10\, V\) using a battery.Then the battery is removed and the capacitor is connected to an uncharged capacitor \(C _{2}\) of \(8\, \mu F\). The charge in \(C _{2}\) on equilibrium condition is \(\ldots\,\mu C\). (Round off to the Nearest Integer)

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Answer & Solution

Correct Answer

(D) \(16\)

Step-by-step Solution

Detailed explanation

\(20=\left( C _{1}+ C _{2}\right) V \Rightarrow V =2\, volt\)

\(Q _{2}= C _{2} V =16\, \mu C\)

\(=16\)

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