JEE MAIN - Physics (2021 - 26th August Morning Shift - No. 19)
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)
9.0 $$\mu$$A
9.0 mA
0.9 mA
0.9 $$\mu$$A
Explanation
$$\rho$$ = 200 $$\Omega$$m
C = 2 $$\times$$ 10$$-$$12 F
V = 40 V
K = 56
$$i = {q \over {\rho k{\varepsilon _0}}} = {{{q_0}} \over {\rho k{\varepsilon _0}}}{e^{ - {t \over {\rho k{\varepsilon _0}}}}}$$
$${i_{\max }} = {{2 \times {{10}^{ - 12}} \times 40} \over {200 \times 50 \times 8.85 \times {{10}^{ - 12}}}}$$
$$ = {{80} \over {{{10}^4} \times 8.85}}$$ = 903 $$\mu$$A = 0.9 mA
C = 2 $$\times$$ 10$$-$$12 F
V = 40 V
K = 56
$$i = {q \over {\rho k{\varepsilon _0}}} = {{{q_0}} \over {\rho k{\varepsilon _0}}}{e^{ - {t \over {\rho k{\varepsilon _0}}}}}$$
$${i_{\max }} = {{2 \times {{10}^{ - 12}} \times 40} \over {200 \times 50 \times 8.85 \times {{10}^{ - 12}}}}$$
$$ = {{80} \over {{{10}^4} \times 8.85}}$$ = 903 $$\mu$$A = 0.9 mA
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