JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2022
- Hard
- MCQ
A source of potential difference \(V\) is connected to the combination of two identical capacitors as shown in the figure. When key ' \(K\) ' is closed, the total energy stored across the combination is \(E _{1}\). Now key ' \(K\) ' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now \(E _{2}\). The ratio \(E _{1} / E _{2}\) will be :

Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(\frac{5}{13}\)
Step-by-step Solution
Detailed explanation
\((1)\) Switch is closed
\(C _{ eq }=2 C\)
Energy \(E_{1}=\frac{1}{2} C_{\text {oq }} V^{2}\)
\(=\frac{1}{2} 2 C ^{2} \times V ^{2}\)
\(E _{1}= CV ^{2}\)
\((ii)\) When switch is opened charge on right capacitor remain CV while potential on left capacitor remain same
Dielectric \(K =5\)
\(C ^{\prime}= KC\)
\(C ^{\prime}=5\,C\)
\(E _{2}=\frac{1}{2}(5 C ) V ^{2}+\frac{( CV )^{2}}{2(5 C )}\)
\(E _{2}=\frac{5 CV ^{2}}{2}+\frac{ CV ^{2}}{10}\)
\(E _{2}=\frac{13 CV ^{2}}{5}\)
\(\frac{ E _{1}}{ E _{2}}=\frac{ CV ^{2}}{\frac{13 CV ^{2}}{5}}=\frac{5}{13}\)
\(\frac{ E _{1}}{ E _{2}}=\frac{5}{13}\)
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