A parallel-plate capacitor with circular plat | vedclass

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(A) The displacement current $I_d$ is given by the formula $I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$,where $\Phi_E$ is the electric flux through the su

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(A) The displacement current $I_d$ is given by the formula $I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$,where $\Phi_E$ is the electric flux through the surface bounded by the loop.Since the electric field $E$ exists only between the plates (within the radius $R = 10 \ cm$),the flux through the loop of radius $r = 20 \ cm$ is $\Phi_E = E \cdot A_{plate} = E \cdot \pi R^2$.Thus,$I_d = \varepsilon_0 \frac{d}{dt}(E \cdot \pi R^2) = \varepsilon_0 \pi R^2 \frac{dE}{dt}$.Given: $R = 10 \ cm = 0.1 \ m$,$\frac{dE}{dt} = 3.6 	imes 10^{12} \ V/(m \cdot s)$,and $\frac{1}{4 \pi \varepsilon_0} = 9 	imes 10^9 \ N \cdot m^2/C^2 \Rightarrow \varepsilon_0 = \frac{1}{36 \pi 	imes 10^9} \ F/m$.Substituting these values:$I_d = \left( \frac{1}{36 \pi 	imes 10^9} ight) \cdot \pi \cdot (0.1)^2 \cdot (3.6 	imes 10^{12})$$I_d = \frac{1}{36 	imes 10^9} \cdot 0.01 \cdot 3.6 	imes 10^{12}$$I_d = \frac{3.6 	imes 10^{10}}{36 	imes 10^9} = \frac{36 	imes 10^9}{36 	imes 10^9} = 1 \ A$.

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