# Deriving current from decharging capacitor and homogenous DE

1. Nov 30, 2014

### Jonsson

Hello there,

I want learn to do homogenous differential equations and I cannot work out what is wrong in this example. Hope you'll be able to tell me why my DE is unsatisfactory.

Assume that a capacitor is has charge Q and that the terminal with the high voltage is connected to a switch in line with a resistor which in turn is connected to the low terminal on the cap.

When the switch is closed, I write ohms law:
$i = \frac{v}{R}$
The current in the circuit will be proportional to the voltage across the capacitor plates.

The from there I rewrite ohms law using the definition of capacitance and differentiate to get the DE:

$$i=\frac{v}{R} = \frac{q}{RC} \implies \frac{d i}{dt} = \frac{i}{RC}$$

I solve the DE and find:

$$i = C e^{t/(RC)},$$
Where $C$ is some constant. The problem is that the current diverges as a function of time.

I wanted expected something along the lines of:

$i = Ce^{-t/(RC)},$

Where the current converge to zero of as time pass.

What is unsatisfactory with the DE I started with? I think it looks correct, and cannot find anything wrong with it. Perhaps you can help. :)

Thank you for your time.

Kind regards,
Marius

Last edited: Nov 30, 2014
2. Dec 1, 2014

### vanhees71

You simply made a sign mistake in your derivation of the DE. Here you integrate the electric along the circuit from the positively charged plate to the negatively charged plate, this gives you $-U=Q/C$. Further, since $\vec{j} = \sigma \vec{E}$ along the wire, you have $i$ positive in this direction. Thus the right equation is
$$i=-\frac{Q}{R C},$$
and the rest of the calculation is analogous to the one you've presented.

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