When a capacitor discharges, the potential difference between its plates and the resistor's terminals is the same. Hence,(adsbygoogle = window.adsbygoogle || []).push({});

[tex] \frac{Q}{C}=R \frac{dq}{dt}[/tex]

Solving this equation we get

[tex] q(t)= Q_{0} exp \frac{t}{RC} [/tex]

Obviously, this isn't the solution. It is actually [tex] q(t)= Q_0 exp \frac{-t}{RC} [/tex]. So, I'm missing a minus sign on the original equation.

But why should it be there?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# RC circuit - Capacitor discharge

**Physics Forums | Science Articles, Homework Help, Discussion**