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The capacitance of a parallel plate capacitor becomes 4 3 times its original value. If a dielectric slab of thickness t = d 2 is inserted between the plates (where d is the distance of separation between the plates), what is the value of the dielectric constant of the slab?

Correct answer

2

Step-by-step solution

Let C be the original capacitance. ∴ New, capacitance, C ′ = 4 3 C Thickness, t = d 2 ⇒ ε 0 A d - t 1 - 1 K = 4 3 ε 0 A d ⇒ 2 1 + 1 K = 4 3 ⇒ K = 2

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