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

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 ......\( J.\)

  1. A \(400\)
  2. B \(0.4\)
  3. C \(40\)
  4. D \(4\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(4\)

Step-by-step Solution

Detailed explanation

\(\Delta {U}=\frac{1}{2}(\Delta {C}) {V}^{2}\) \(\Delta {U}=\frac{1}{2}({KC}-{C}) {V}^{2}\) \(\Delta {U}=\frac{1}{2}(2-1) {CV}^{2}\) \(\Delta {U}=\frac{1}{2} \times 200 \times 10^{-6} \times 200 \times 200\) \(\Delta {U}=4 {J}\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app