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← JEE Main 2022 · 28 Jul · Shift 2
PhysicsSingle correct+4 / -1Question 9 of 90

JEE Main 2022 · 28 Jul · Shift 2, Physics question 9

A slab of dielectric constant K has the same cross-sectional area as the plates of a parallel plate capacitor and thickness 34d\dfrac34d, where d is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be :

(Given C0C_0 = capacitance of capacitor with air as medium between plates.)

Choose one option
Answer and worked solution

Official answer: A

Slab. C=ε0Ad−t+t/K=C014+34K=4KC03+KC=\frac{\varepsilon_0A}{d-t+t/K}=\frac{C_0}{\frac14+\frac3{4K}}=\frac{4KC_0}{3+K}.

(A)