The plates of a parallel plate capacitor are charged to a potential of 200V . Now, a dielectric slab of thickness 4mm is inserted between its plates and to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by 3.2mm . The dielectric constant of the slab is

Question

The plates of a parallel plate capacitor are charged to a potential of 200V . Now, a dielectric slab of thickness 4mm is inserted between its plates and to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by 3.2mm . The dielectric constant of the slab is (A) 1 (B) 4 (C) 5 (D) 6

Tardigrade Admin · Accepted Answer

(ϵ 0 A/d) = (ϵ 0 A/dâ² - t + (t)K) ⇒ d=dâ²-t+(t/K) ⇒ dâ²-d=t(1 - (1/K)) Here, dâ²-d=3.2 mm, t=4 mm ∴ 3.2=4(1 - (1/K)) ⇒ (3 .2/4)=1-(1/K) ⇒ 1-(1/K)=(4/5) ∴ K=5

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dϵ0​A​=d′−t+Kt​ϵ0​A​ ⇒d=d′−t+Kt​ ⇒d′−d=t(1−K1​) Here, d′−d=3.2mm,t=4mm ∴3.2=4(1−K1​) ⇒43.2​=1−K1​ ⇒1−K1​=54​ ∴K=5

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