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JEE Advanced 2025 Paper 2 · Physics · Capacitors · hard

A parallel-plate capacitor has plate area A and separation d; its capacitance with air between the plates is C0. A dielectric slab of constant K=3…

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A parallel-plate capacitor has plate area AA and separation dd; its capacitance with air between the plates is C0C_0. A dielectric slab of constant K=3K=3, with the same area as the plates and thickness d/2d/2, is placed flush against one plate, leaving an air gap of d/2d/2. The capacitor is connected to a battery of emf VV. Which of the following statements is/are correct?
  1. A. The electric field in the air gap is 3V2d\dfrac{3V}{2d}.
  2. B. The potential difference across the dielectric slab is V4\dfrac{V}{4}.
  3. C. The battery remains connected and the slab is slowly withdrawn completely; the external agent does work C0V24\dfrac{C_0V^2}{4}.
  4. D. If the battery is first disconnected and the slab then withdrawn slowly, the external agent again does work C0V24\dfrac{C_0V^2}{4}.
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Answer: A, B, C

The two layers are 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=1.5C0C=1.5C_0 and Q=1.5C0VQ=1.5C_0V. Field in air =σ/ε0=Qε0A=1.5Vd⋅=\sigma/\varepsilon_0=\dfrac{Q}{\varepsilon_0A}=\dfrac{1.5V}{d}\cdot using C0=ε0A/dC_0=\varepsilon_0A/d, i.e. 3V2d\dfrac{3V}{2d}: (A) correct. Field in dielectric is one third of this, V2d\dfrac{V}{2d}, over d/2d/2 gives a drop of V/4V/4: (B) correct. With battery connected, Ui=12(1.5C0)V2U_i=\tfrac12(1.5C_0)V^2, Uf=12C0V2U_f=\tfrac12C_0V^2, battery work =(C0−1.5C0)V2=−0.5C0V2=(C_0-1.5C_0)V^2=-0.5C_0V^2 (battery absorbs charge). Energy balance: Wext+Wbatt=ΔUW_{ext}+W_{batt}=\Delta U, so Wext=−0.25C0V2+0.5C0V2=C0V24W_{ext}=-0.25C_0V^2+0.5C_0V^2=\dfrac{C_0V^2}{4}: (C) correct. With the battery disconnected, QQ is fixed, Uf=Q2/2C0=1.125C0V2U_f=Q^2/2C_0=1.125C_0V^2 and Wext=1.125−0.75=3C0V28≠C0V24W_{ext}=1.125-0.75=\dfrac{3C_0V^2}{8}\neq\dfrac{C_0V^2}{4}: (D) is false.
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