A parallel plate capacitor consists of two ci | vedclass

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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 $C$.

$A$ 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

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