A capacitor of capacitance C = 10 , mu F is c | vedclass

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(B) The charge on a capacitor during charging is given by $q(t) = q_0(1 - e^{-t/	au})$, where $q_0 = C\varepsilon$ is the steady-state charge and $

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(B) The charge on a capacitor during charging is given by $q(t) = q_0(1 - e^{-t/	au})$, where $q_0 = C\varepsilon$ is the steady-state charge and $	au = RC$ is the time constant.Given: $C = 10 \, \mu F = 10 	imes 10^{-6} \, F$, $\varepsilon = 2 \, V$, $t = 50 \, ms = 50 	imes 10^{-3} \, s$, and $q = 12.6 \, \mu C = 12.6 	imes 10^{-6} \, C$.The steady-state charge is $q_0 = C\varepsilon = (10 	imes 10^{-6} \, F)(2 \, V) = 20 	imes 10^{-6} \, C = 20 \, \mu C$.Substituting the values into the charging equation:$12.6 = 20(1 - e^{-t/	au})$$12.6/20 = 1 - e^{-t/	au}$$0.63 = 1 - e^{-t/	au}$$e^{-t/	au} = 1 - 0.63 = 0.37$Since $1/e \approx 0.37$, we have $e^{-1} = 0.37$. Therefore, $t/	au = 1$, which means $	au = t$.$	au = RC = 50 	imes 10^{-3} \, s$.$R = 	au / C = (50 	imes 10^{-3} \, s) / (10 	imes 10^{-6} \, F) = 5 	imes 10^3 \, \Omega = 5 \, k \Omega$.

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