Please consider the following RC circuit as context:
Assume that the circuit has been connected for a long time. If switch S has been opened at $t=0,$ the differential equation used to solve for the charge on the capacitor $Q$ would be, by using Kirchhoff's loop rule:
$$\frac{Q}{C}-R\frac{dQ}{dt}=0, \quad Q(0)=C\mathcal{E}$$
since, discarding the portion to the left of the capacitor, the voltage drop through the capacitor would oppose that of the resistor following the clockwise current flow. However, a professor told me that the right differential equation in this case would be:
$$\frac{Q}{C}+R\frac{dQ}{dt}=0, \quad Q(0)=C\mathcal{E}$$
I simply do not understand how the voltage drop of the capacitor is negative from its bottom to top plate. Wouldn't it act the same as a battery? Namely, wouldn't it add the the voltage to the circuit?
Best Answer
I believe that your Professor is correct. The equation by bobD is correct but what you are missing is that the charge on capacitor on that time is Q and therefore current flowing through that instant is $d(Q(0)-Q)/dt$ and that gives your correct equation. Even if you consider your equation correct then when you integrate it the charge will increase exponentially which is not possible. HOPE THIS HELPS