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 ......\( J.\)
- A \(400\)
- B \(0.4\)
- C \(40\)
- D \(4\)
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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