A parallel plate capacitor with circular plat | vedclass

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Question

(A) The time constant of the $CR$ circuit is $	au = CR = (10^{-9} \; F) 	imes (10^6 \; \Omega) = 10^{-3} \; s$.The charge on the capacitor at time $

Vedclass · Accepted Answer

$B=0.74 	imes 10^{-13}\; T$

Vedclass · Answer

(A) The time constant of the $CR$ circuit is $	au = CR = (10^{-9} \; F) 	imes (10^6 \; \Omega) = 10^{-3} \; s$.The charge on the capacitor at time $t$ is $q(t) = CV[1 - \exp(-t/	au)] = (10^{-9} \; F)(2 \; V)[1 - \exp(-t/10^{-3})] = 2 	imes 10^{-9} [1 - \exp(-t/10^{-3})] \; C$.The electric field $E$ between the plates is $E = \frac{q(t)}{\varepsilon_0 A}$,where $A = \pi r_0^2 = \pi(1)^2 = \pi \; m^2$.Consider a circular loop of radius $r = 0.5 \; m$ passing through point $P$. The electric flux through this loop is $\Phi_E = E 	imes (\pi r^2) = \frac{q(t)}{\varepsilon_0 A} 	imes (\pi r^2) = \frac{q(t)}{\varepsilon_0 \pi} 	imes \pi (0.5)^2 = \frac{q(t)}{4 \varepsilon_0}$.The displacement current $i_d$ is $i_d = \varepsilon_0 \frac{d\Phi_E}{dt} = \varepsilon_0 \frac{d}{dt} \left( \frac{q(t)}{4 \varepsilon_0} ight) = \frac{1}{4} \frac{dq}{dt}$.Since $\frac{dq}{dt} = \frac{d}{dt} [CV(1 - e^{-t/	au})] = \frac{CV}{	au} e^{-t/	au} = \frac{V}{R} e^{-t/	au}$,at $t = 10^{-3} \; s$ (which is $t = 	au$),$\frac{dq}{dt} = \frac{2 \; V}{10^6 \; \Omega} e^{-1} = 2 	imes 10^{-6} e^{-1} \; A$.Thus,$i_d = \frac{1}{4} (2 	imes 10^{-6} e^{-1}) = 0.5 	imes 10^{-6} e^{-1} \; A$.Using the Ampere-Maxwell law,$\oint B \cdot dl = \mu_0 i_d \implies B(2 \pi r) = \mu_0 i_d$.$B = \frac{\mu_0 i_d}{2 \pi (0.5)} = \frac{\mu_0 i_d}{\pi} = \frac{4 \pi 	imes 10^{-7} 	imes 0.5 	imes 10^{-6} 	imes e^{-1}}{\pi} = 2 	imes 10^{-13} 	imes e^{-1} \; T$.Using $e^{-1} \approx 0.3678$,$B \approx 2 	imes 10^{-13} 	imes 0.3678 \approx 0.7356 	imes 10^{-13} \; T \approx 0.74 	imes 10^{-13} \; T$.

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