The material filled between the plates of a parallel plate capacitor has resistivity 200 \Omega 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 permittivity of material is 50)

Capacitor PYQ — JEE Main 2021 · 26 Aug · Shift 1 · Q61 with Solution

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Correct answer: C

For a parallel plate capacitor filled with dielectric:

$C = \varepsilon_0 K \frac{A}{d}$

Given:

So,

$\frac{A}{d} = \frac{C}{\varepsilon_0 K}$

The dielectric is slightly conducting, so leakage current flows through it.

Resistance of slab:

$R = \rho \frac{d}{A}$

Since $\dfrac{d}{A} = \left(\dfrac{A}{d}\right)^{-1}$ ,

$R = \rho \cdot \frac{\varepsilon_0 K}{C}$

Substitute values:

$R = 200 \times \frac{8.85 \times 10^{-12} \times 50}{2 \times 10^{-12}}$

$R = 200 \times \frac{8.85 \times 50}{2}$

$R = 200 \times 221.25$

$R = 44250\,\Omega$

Using Ohm's law:

$I = \frac{V}{R}$

Given $V = 40\text{ V}$ ,

$I = \frac{40}{44250}$

$I \approx 9.04 \times 10^{-4}\text{ A}$

$I \approx 0.9\text{ mA}$

$0.9\text{ mA}$ corresponds to Option C.