RC circuit - Capacitor discharge

[jpas]

When a capacitor discharges, the potential difference between its plates and the resistor's terminals is the same. Hence,

\frac{Q}{C}=R \frac{dq}{dt}
Solving this equation we get

q(t)= Q_{0} exp \frac{t}{RC}

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

But why should it be there?

 

[intwo]

According to Kirchoff's Second Law, the sum of the voltages is equal to zero. R(dQ/dt) + (Q/C) = 0, therefore you need -R(dQ/dt) = (Q/C). I apologize for the lack of proper script.

 

[jpas]

Thank you.
Didn't know such laws.

 

[willem2]

If the capacitor discharges, then dq/dt is negative, so according toyour originial equation, Q/C which is positive equals something negative