# Physics 2

Show that during the charging of the capacitor in an RC circuit, the charge of the capacitor, as a function of time, is: q(t) = C(1 – e-t/RC)

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1. the character that doesn't show up is the emf (E)

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2. The current is
i = dq/dt = C*E*(1/RC)*e^(-t/RC)
= (E/R)*e^(-t/RC)

iR + q/C = E*e^(-t/RC)+ E*(1 – e^-t/RC)
= E

Therefore the sum of the voltage drop across the capacitor and the resistor equals E at all times, as required.

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3. thank you very much drwls, this is a question from a sample exam, getting ready for my finals tomorrow!

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4. Good luck with finals! I took a lazy approach to your question, showing that what you wrote is the solution, rather than solving the general differential equation.

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5. thanks. It's all good, I understand your approach and I really appreciate it!

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