Physics Circle home Student Login

Home›Chapters›Capacitors PYQs›Dielectrics›Question 77

JEE Main 2021 (31 Aug, Shift 2) · Dielectrics

Dielectrics question from JEE Main 2021 (31 Aug, Shift 2)

Q77JEE Main 2021 · 31 AUG · SHIFT 2Medium

Dielectrics

A parallel plate capacitor of capacitance 200 μ\muF 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.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 4

  1. Battery connected, VV constant: ΔU=12(K−1)CV2\Delta U=\frac{1}{2}(K-1)CV^2.
  2. ΔU=12(2−1)(200×10−6)(200)2\Delta U=\frac{1}{2}(2-1)(200\times10^{-6})(200)^2
  3. ΔU=4 J\Delta U=4\text{ J}
Report a mistake

More questions on Dielectrics