JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Capacitance

The material filled between the plates of a parallel plate capacitor has resistivity

200Ω m

. The value of capacitance of the capacitor is 2 pF . If a potential difference of 40 V is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is : (given the value of relative permitivity of material is 50 )

(A) $0.9 μ A$
(B)9.0 mA
(C) $9.0 μ A$
(D)0.9 mA✓ Correct

Answer: (D)

Solution

With the same plate area

A

and separation

d

, the capacitance is

C = \frac{ε _{0} ε _{r} A}{d}

while the leakage resistance of the filling is

R = \frac{ρ d}{A}

, so the geometry cancels in the product

R C = ε_{0} ε_{r} ρ

The leakage current is therefore

I = \frac{V}{R} = \frac{V C}{ε _{0} ε _{r} ρ} = \frac{40 \times 2 \times 1 0 ^{- 12}}{8.85 \times 1 0 ^{- 12} \times 50 \times 200}

I = \frac{8 \times 1 0 ^{- 11}}{8.85 \times 1 0 ^{- 8}} = 9.0 \times 1 0^{- 4} A = 0.9 mA

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Capacitance

Solution

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