JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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 :

- A \(\frac{1}{10}\)
- B \(\frac{2}{5}\)
- C \(\frac{5}{13}\)
- D \(\frac{5}{26}\)
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…
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