Seawater at a frequency f = 9 × 102 Hz, has permittivity ε = 80 ε 0 and resistivity ρ = 0.25 Ω m. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source V(t) = V0 sin(2 π ft). Then the conduction current density becomes 10x times the displacement current density after time t = 1/800 s. The value of x is _____________. (Given: (1)/(4πε₀) = 9 × 10⁹ Nm2C ⁻²) Numerical answer type. Work it out, then open the answer.

Answer: 6. E = V₀/dsin ω t: Jc = E/ρ = (V₀)/(ρ d)sin ω t, Jd = /dt = (ε ω V₀)/(d)cos ω t. (Jc)/(Jd) = (tan ω t)/(ρ ε ω). ω t = 2π × 900 × 1/800 = 2.25π, so tan ω t = tan π/4 = 1. ρεω = 0.25 × 8036π × 10⁹ × 1800π = 10⁻⁶, so (Jc)/(Jd) = 10⁶. Answer: x = 6.

Seawater at a frequency f = 9 × 102 Hz, has permittivity ε = 80 ε 0… | JEE Main 2021 (17 Mar, Shift 2)

Q12JEE Main 2021 · 17 MAR · SHIFT 2Difficult

Displacement Current, Maxwell's Equations & Production

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 = $\frac{1}{800}$ s. The value of x is _____________. (Given: $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}$ Nm2C ${}^{-2}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 6

$E = \dfrac{V_0}{d}\sin \omega t$ : $J_c = \dfrac{E}{\rho} = \dfrac{V_0}{\rho d}\sin \omega t$ , $J_d = \varepsilon\dfrac{dE}{dt} = \dfrac{\varepsilon \omega V_0}{d}\cos \omega t$ .
$\dfrac{J_c}{J_d} = \dfrac{\tan \omega t}{\rho \varepsilon \omega}$ .
$\omega t = 2\pi \times 900 \times \dfrac{1}{800} = 2.25\pi$ , so $\tan \omega t = \tan\dfrac{\pi}{4} = 1$ .
$\rho\varepsilon\omega = 0.25 \times \dfrac{80}{36\pi \times 10^{9}} \times 1800\pi = 10^{-6}$ , so $\dfrac{J_c}{J_d} = 10^{6}$ .
Answer: $x = 6$ .

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