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JEE Main 2021 (27 Jul, Shift 2) · Parallel Plate Capacitor

Parallel Plate Capacitor question from JEE Main 2021 (27 Jul, Shift 2)

Q13JEE Main 2021 · 27 JUL · SHIFT 2Medium

Parallel Plate Capacitor

A simple pendulum of mass 'm', length 'l' and charge '+ q' suspended in the electric field produced by two conducting parallel plates as shown. The value of deflection of pendulum in equilibrium position will be

Figure

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Answer: (c)

  1. Slab region and air region are in series. Take C2=Kε0AtC_2 = \frac{K\varepsilon_0 A}{t} (slab, beside the +V2+V_2 plate) and C1=ε0Ad−tC_1 = \frac{\varepsilon_0 A}{d-t} (air).
  2. Total p.d. =V2−(−V1)=V1+V2= V_2 - (-V_1) = V_1 + V_2, so Q=C1C2C1+C2(V1+V2)Q = \frac{C_1C_2}{C_1 + C_2}(V_1 + V_2).
  3. Field in air: E=Q/C1d−t=C2(V1+V2)(C1+C2)(d−t)E = \frac{Q/C_1}{d - t} = \frac{C_2(V_1 + V_2)}{(C_1 + C_2)(d - t)}
  4. Equilibrium: tan⁡θ=qEmg\tan\theta = \frac{qE}{mg}
  5. So the answer is (c) tan⁡−1[qmg×C2(V1+V2)(C1+C2)(d−t)]\tan^{-1}\left[\frac{q}{mg}\times\frac{C_2(V_1+V_2)}{(C_1+C_2)(d-t)}\right].
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