Seawater at a frequency f =9 × 102,Hz, has permittivity =80, 0 and resistivity =0.25, m. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source V ( t)= V 0 (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.......... ( 1 4 π 0 =9 × 109 Nm 2 C -2)

JEE Mains · Physics · STD 12 - Electromagnetic waves

Seawater at a frequency \(f =9 \times 10^{2} \,Hz ,\) 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 \(V ( t )= V _{0} \sin (2 \pi ft ) .\) Then the conduction current density becomes \(10^{x}\) times the displacement current density after time \(t =\frac{1}{800} \,s\). The value of \(x\) is ......... .
\(\left(\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} C ^{-2}\right)\)

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Verified Solution

Correct Answer

(D) \(6\)

Detailed explanation

\(J _{ c }=\frac{ E }{\rho}=\frac{ V }{\rho d }\)

\(J _{ d }=\frac{1}{ A } \frac{ dq }{ dt }\)

\(=\frac{ C V _{ c }}{ d } dt\)

\(\Rightarrow \frac{ V _{0} \sin 2 \pi ft }{\rho d }=10^{ x } \times \frac{80 \varepsilon_{0}}{ d } V _{0}(2 \pi f ) \cos 2 \pi ft\)

\(\tan \left(2 \pi \times \frac{900}{800}\right)=10^{ x } \times \frac{40}{9 \times 10^{9}} \times 900\)

\(= x =6\)

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