The displacement current of 4.425 ,mu A is de | vedclass

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(C) The displacement current $I_{d}$ in a parallel plate capacitor is given by the formula: $I_{d} = \varepsilon_{0} A \frac{dE}{dt} = \varepsilon_{0}

Vedclass · Accepted Answer

$8$

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(C) The displacement current $I_{d}$ in a parallel plate capacitor is given by the formula: $I_{d} = \varepsilon_{0} A \frac{dE}{dt} = \varepsilon_{0} A \frac{d}{dt} (\frac{V}{d}) = \frac{\varepsilon_{0} A}{d} \frac{dV}{dt}$.Given values:$I_{d} = 4.425 	imes 10^{-6} \,A$$\frac{dV}{dt} = 10^{6} \,V s^{-1}$$A = 40 \,cm^{2} = 40 	imes 10^{-4} \,m^{2} = 4 	imes 10^{-3} \,m^{2}$$\varepsilon_{0} = 8.85 	imes 10^{-12} \,C^{2} N^{-1} m^{-2}$Rearranging the formula to solve for $d$:$d = \frac{\varepsilon_{0} A}{I_{d}} \frac{dV}{dt}$Substituting the values:$d = \frac{8.85 	imes 10^{-12} 	imes 4 	imes 10^{-3} 	imes 10^{6}}{4.425 	imes 10^{-6}}$$d = \frac{8.85 	imes 4 	imes 10^{-9}}{4.425 	imes 10^{-6}}$$d = 2 	imes 4 	imes 10^{-3} \,m = 8 	imes 10^{-3} \,m$Comparing this with $x 	imes 10^{-3} \,m$,we get $x = 8$.

List-$I$ (Relation)	List-$II$ (Law)
$A$. $\oint \overrightarrow{E} \cdot d\vec{l} = -\frac{d}{dt} \oint \overrightarrow{B} \cdot d\vec{a}$	$I$. Ampere's circuital law
$B$. $\oint \vec{B} \cdot d\vec{l} = \mu_0(I + \epsilon_0 \frac{d\phi_E}{dt})$	$II$. Faraday's laws of electromagnetic induction
$C$. $\oint \overrightarrow{E} \cdot d\vec{a} = \frac{1}{\epsilon_0} \int \rho dv$	$III$. Ampere-Maxwell law
$D$. $\oint \overrightarrow{B} \cdot d\vec{l} = \mu_0 I$	$IV$. Gauss's law of electrostatics

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