JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2019
- Hard
- MCQ
The parallel combination of two air filled parallel plate capacitors of capacitance \(C\) and \(nC\) is connected to a battery of voltage, \(V\). When the capacitor are fully charged, the battery is removed and after that a dielectric material of dielectric constant \(K\) is placed between the two plates of the first capacitor. The new potential difference of the combined system is
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Answer & Solution
Correct Answer
(C) \(\frac{{\left( {n + 1} \right)\,V}}{{\left( {K + n} \right)}}\)
Step-by-step Solution
Detailed explanation
After fully charging, battery is disconnected.
Total chare of the system \(=\mathrm{CV}+\mathrm{nCV}\)
\(=(n+1) C V\)
After the insertion of dielectric of constant \(\mathrm{K}\) New potential (common) \(\mathrm{V}_{\mathrm{c}}=\frac{\text { Total charge }}{\text { Total capacitance }}\)
\(=\frac{(n+1) C V}{K C+n C}=\frac{(n+1) V}{K+n}\)
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