JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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
- A \(\frac{V}{{K + n}}\)
- B \(V\)
- C \(\frac{{\left( {n + 1} \right)\,V}}{{\left( {K + n} \right)}}\)
- D \(\frac{{nV}}{{K + n}}\)
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)…
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