QJEEM21Ph3d83: JEE Main 2021 Physics - Class 12 JEEM21Ph3d83 Solution: Seawater at a frequency f = 9 10 2 Hz, has permittivity = 80 0…

Question

Verified Solution • 2025–26 Session moderate

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 10^x 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})

Accepted Answer

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

Step-by-Step Verified Solution