Sea-water at a frequency $\mathrm{f}=9 \times 10^{2} \mathrm{~Hz}$, has permittivity $\varepsilon=80 \varepsilon_{0}$ and resistivity $\rho=0.25 \Omega \mathrm{~m}$. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source $\mathrm{V}(\mathrm{t})=\mathrm{V}_{0} \sin (2 \pi \mathrm{ft})$. Then the conduction current density become $10^{\mathrm{x}}$ times the displacement current density after time $t=\frac{1}{800} s$. The value of $x$ is ______ (Given: $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}$)

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · EM Waves

Sea-water at a frequency

f = 9 \times 1 0^{2} Hz

, has permittivity

ε = 80 ε_{0}

and resistivity

ρ = 0.25Ω m

. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source

V (t) = V_{0} sin (2 π ft)

. Then the conduction current density become

1 0^{x}

times the displacement current density after time

t = \frac{1}{800} s

. The value of

x

is ______
(Given:

\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} Nm^{2} C^{- 2}

)

Answer: 6

Between the plates the field is

E = \frac{V _{0} sin ( 2 π f t )}{d}

. The conduction current density is

J_{c} = \frac{E}{ρ}

and the displacement current density is

J_{d} = ε \frac{\partial E}{\partial t}

, so

\frac{J _{c}}{J _{d}} = \frac{sin ( 2 π f t )}{ρ ε ( 2 π f ) cos ( 2 π f t )} = \frac{tan ( 2 π f t )}{2 π f ρε}

With

f = 9 \times 1 0^{2}

Hz and

t = \frac{1}{800}

2 π f t = \frac{9 π}{4}

and

tan \frac{9 π}{4} = tan \frac{π}{4} = 1

. Using

ε = 80 ε_{0} = \frac{80}{36 π \times 1 0 ^{9}} F m^{- 1}

2 π f ρε = 2 π (900) (0.25) \frac{80}{36 π \times 1 0 ^{9}} = 1 0^{- 6}

\frac{J _{c}}{J _{d}} = \frac{1}{1 0 ^{- 6}} = 1 0^{6}

and

x = 6

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Integer answer