ExamBro
ExamBro
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.
Verified Solution

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 \(K =5\)

\(C ^{\prime}= KC\)

\(C ^{\prime}=5\,C\)

\(E _{2}=\frac{1}{2}(5 C ) V ^{2}+\frac{( CV )^{2}}{2(5 C )}\)

\(E _{2}=\frac{5 CV ^{2}}{2}+\frac{ CV ^{2}}{10}\)

\(E _{2}=\frac{13 CV ^{2}}{5}\)

\(\frac{ E _{1}}{ E _{2}}=\frac{ CV ^{2}}{\frac{13 CV ^{2}}{5}}=\frac{5}{13}\)

\(\frac{ E _{1}}{ E _{2}}=\frac{5}{13}\)

Same subject
Explore more questions on app