Seawater 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 $V(t)=V_{0} \sin (2 \pi f t)$. Then the conduction current density becomes $10^{x}$ times the displacement current density after time
$\mathrm{t}=\frac{1}{800} \mathrm{~s}$. The value of $\mathrm{x}$ is
(Given : $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}$ )
(6)
$J_{c}=\frac{E}{\rho}=\frac{V}{\rho d}$
$J_{\mathrm{d}}=\frac{1}{\mathrm{~A}} \frac{\mathrm{dq}}{\mathrm{dt}}$
$=\frac{\mathrm{C}}{\mathrm{A}} \frac{\mathrm{dV}_{\mathrm{c}}}{\mathrm{dt}}$
$=\frac{\epsilon}{\mathrm{d}} \frac{\mathrm{d} \mathrm{V}_{\mathrm{c}}}{\mathrm{dt}}$
$\Rightarrow \frac{V_{0} \sin 2 \pi f t}{\rho d}=10^{x} \times \frac{80 \varepsilon_{0}}{d} V_{0}(2 \pi f) \cos 2 \pi f t$
$\tan \left(2 \pi \times \frac{900}{800}\right)=10^{\mathrm{x}} \times \frac{40}{9 \times 10^{9}} \times 900$
$=x=6$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.