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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2021
  • Hard
  • Numerical

A parallel plate capacitor of capacitance \(200 \,\mu {F}\) is connected to a battery of \(200 \, {V} .\) A dielectric slab of dielectric constant \(2\) is now inserted into the space between plates of capacitor while the battery remain connected. The change in the electrostatic energy in the capacitor will be ......\( J.\)

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Answer & Solution

Correct Answer

(D) \(4\)

Step-by-step Solution

Detailed explanation

\(\Delta {U}=\frac{1}{2}(\Delta {C}) {V}^{2}\)

\(\Delta {U}=\frac{1}{2}({KC}-{C}) {V}^{2}\)

\(\Delta {U}=\frac{1}{2}(2-1) {CV}^{2}\)

\(\Delta {U}=\frac{1}{2} \times 200 \times 10^{-6} \times 200 \times 200\)

\(\Delta {U}=4 {J}\)

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