PrepRung
Sign inSign up free

JEE Advanced 2025 Paper 1 · Physics · Capacitors · hard

A parallel plate capacitor has plate area A, plate separation d and vacuum between the plates; its capacitance is C0 and it is connected to a battery…

One or more options correct

A parallel plate capacitor has plate area AA, plate separation dd and vacuum between the plates; its capacitance is C0C_0 and it is connected to a battery of emf VV, so the field in vacuum is E0=V/dE_0=V/d. A dielectric slab of constant K=3K=3, thickness d/2d/2 and the same area as the plates is slid in so that it lies flush against one plate, while the battery stays connected. Choose the correct statement(s).
  1. A. The capacitance becomes 1.5 C01.5\,C_0.
  2. B. The electric field in the air gap remains E0E_0.
  3. C. The electric field inside the slab is 0.75 E00.75\,E_0.
  4. D. The electrostatic energy stored in the air gap is 33 times that stored in the slab.
Show answer and solution

Answer: A, D

With the slab flush on a plate, the air gap (thickness d/2d/2) and slab (thickness d/2d/2) form two capacitors in series: 1C=d/2ε0A+d/23ε0A=2d3ε0A\dfrac1C=\dfrac{d/2}{\varepsilon_0A}+\dfrac{d/2}{3\varepsilon_0A}=\dfrac{2d}{3\varepsilon_0A}, so C=1.5 C0C=1.5\,C_0 (A true). The normal displacement DD is continuous, so Eslab=Eair/3E_{slab}=E_{air}/3 and V=Eaird2+Eair3d2=23EairdV=E_{air}\frac d2+\frac{E_{air}}{3}\frac d2=\frac23E_{air}d, giving Eair=1.5E0E_{air}=1.5E_0 and Eslab=0.5E0E_{slab}=0.5E_0. So (B) is false (the field in air rises to 1.5E01.5E_0) and (C) is false (0.5E00.5E_0, not 0.75E00.75E_0). (D) Energy densities: air 12ε0Eair2\tfrac12\varepsilon_0E_{air}^2, slab 12Kε0Eslab2=12ε0Eair2/3\tfrac12 K\varepsilon_0E_{slab}^2=\tfrac12\varepsilon_0E_{air}^2/3; the volumes are equal, so the ratio is 33 (true).
Report a mistake

Original question written in the JEE Advanced 2025 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake