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(C) Given:Radius of plates, $r = 2 \,cm = 2 	imes 10^{-2} \,m$Distance between plates, $d = 0.1 \,mm = 10^{-4} \,m$Rate of change of potential differ

Vedclass · Accepted Answer

$0.556 \,mA$

Vedclass · Answer

(C) Given:Radius of plates, $r = 2 \,cm = 2 	imes 10^{-2} \,m$Distance between plates, $d = 0.1 \,mm = 10^{-4} \,m$Rate of change of potential difference, $\frac{dV}{dt} = 5 	imes 10^6 \,Vs^{-1}$Permittivity of free space, $\varepsilon_0 = 8.85 	imes 10^{-12} \,F/m$Area of the plates, $A = \pi r^2 = \pi 	imes (2 	imes 10^{-2})^2 = 4\pi 	imes 10^{-4} \,m^2$The displacement current $I_d$ is given by:$I_d = \varepsilon_0 \frac{d\phi_E}{dt} = \varepsilon_0 A \frac{dE}{dt}$Since $E = \frac{V}{d}$, we have $\frac{dE}{dt} = \frac{1}{d} \frac{dV}{dt}$Therefore, $I_d = \varepsilon_0 \frac{A}{d} \frac{dV}{dt}$Substituting the values:$I_d = (8.85 	imes 10^{-12}) 	imes \frac{4\pi 	imes 10^{-4}}{10^{-4}} 	imes (5 	imes 10^6)$$I_d = 8.85 	imes 10^{-12} 	imes 4\pi 	imes 5 	imes 10^6$$I_d = 8.85 	imes 20\pi 	imes 10^{-6} \,A$$I_d \approx 8.85 	imes 62.83 	imes 10^{-6} \,A$$I_d \approx 556 	imes 10^{-6} \,A = 0.556 \,mA$Thus, the correct option is $C$.

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

$A$ parallel plate capacitor consists of two circular plates each of radius $2 \,cm$, separated by a distance of $0.1 \,mm$. If the potential difference across the plates is varying at the rate of $5 \times 10^6 \,Vs^{-1}$, then the value of displacement current is

Similar Questions

What did Hertz observe in his experiment to produce electromagnetic waves?

If the rate of change of electric field across the plates of a parallel plate capacitor is $E$ and the displacement current is $I$,then the area of one plate of the capacitor is ($\varepsilon_{0}$ is permittivity of free space).

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