- #1
jpas
- 45
- 0
When a capacitor discharges, the potential difference between its plates and the resistor's terminals is the same. Hence,
[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?
[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?