Verified Solution • 2025–26 Session moderate

QJEEM21Ph3321: JEE Main 2021 Physics - Class 12 JEEM21Ph3321 Solution: A parallel plate capacitor of capacitance 200 F is connected to …

A parallel plate capacitor of capacitance 200 μ\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.

Step-by-Step Verified Solution

ΔU=12(ΔC)V2\Delta U = {1 \over 2}(\Delta C){V^2}

ΔU=12(KC−C)V2\Delta U = {1 \over 2}(KC - C){V^2}

ΔU=12(2−1)CV2\Delta U = {1 \over 2}(2 - 1)C{V^2}

ΔU=12×200×10−6×200×200\Delta U = {1 \over 2} \times 200 \times {10^{ - 6}} \times 200 \times 200

ΔU=4\Delta U = 4 J

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring