JEE MAIN - Physics (2021 - 17th March Evening Shift - No. 28)
Seawater at a frequency f = 9 $$\times$$ 102 Hz, has permittivity $$\varepsilon $$ = 80$$\varepsilon $$0 and resistivity $$\rho$$ = 0.25 $$\Omega$$m. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source V(t) = V0 sin(2$$\pi$$ft). Then the conduction current density becomes 10x times the displacement current density after time t = $${1 \over {800}}$$s. The value of x is _____________. (Given : $${1 \over {4\pi {\varepsilon _0}}} = 9 \times {10^9}$$ Nm2C$$-$$2)
Answer
6
Explanation
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Given f = 9 $$\times$$ 102 Hz
$$ \in $$ = $$ \in $$0$$ \in $$r
$$ \in $$ = 80 $$ \in $$0
So $$ \in $$r = 80
$$\rho $$ = 0.25 $$\Omega$$m
V(t) = V0 sin (2$$\pi$$ft)
$${I_d} = {{dq} \over {dt}} = {{cdv} \over {dt}}$$
$${I_d} = {{{ \in _0}{ \in _r}A} \over d}{d \over {dt}}({v_0}\sin (2\pi ft))$$
$${I_d} = {{{ \in _0}{ \in _r}A} \over d}{V_0}(2\pi f)\cos (2\pi ft)$$ .......... (1)
& $${I_c} = {V \over R}$$
$${I_c} = {{{V_0}\sin (2\pi ft)} \over {\rho {d \over A}}} = {{A{v_0}\sin (2\pi ft)} \over {\rho d}}$$ ....... (2)
divide equation (1) and (2)
$${{{I_d}} \over {{I_c}}} = { \in _0}{ \in _r}2\pi f(\rho )\cot (2\pi ft)$$
$${{{I_d}} \over {{I_c}}} = {1 \over {4\pi \times 9 \times {{10}^9}}} \times 80 \times 2\pi \times 9 \times {10^2} \times (0.25) \times \cot (2\pi \times 9 \times {10^2} \times {1 \over {800}})$$
$$ = {{{{10}^3}} \over {{{10}^9}}}\left( {\cot \left( {{{9\pi } \over 4}} \right)} \right)$$
$$ = {{{{10}^3}} \over {{{10}^9}}}$$
$${{{I_d}} \over {{I_c}}} = {1 \over {{{10}^6}}}$$
$${I_c} = {10^6}{I_d}$$
So x = 6
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