How do you solve the differential equation for a discharging capacitor?

[rad0786]

K this is more math than physics, but i thought the physics people could help on this.

A 10^-8 capacitor is charged to 50V and then disconnected. One can model the charge leakage of the capacitor with a RC circuit with no voltage source and the resistance of air between the capacitor plates. On a cold dry day, the resistance of the air gap is 5E13 (omega's); on a humid day, the resistance is 7E6. How long will it take the capacitor voltage to dissipate to half its original value on each day?

I'll work with the cold dry day - i assume the humid day is solved the same way.

Anyways, i used the diff equation:

R(dq/dt) + q/C = E

I sub in the values and i get

(5E13)(dq/dt) + (50/10^-8) = E

which reduces to
(dq/dt) + (1E-5) = E(2E-12)

now i have no idea how to solve this diff equation. I mean, it is seperable right? The only thing is i have no idea what to do with E?

 

[stunner5000pt]

E is the voltage provided by an external agent (is there any battery connected here?) You're only give nteh initial voltage which s 50 V

$$R \frac{dq}{dt} + \frac{q}{C} = 0$$
are you aware on how to solve first order differential equations?
because you can't simply substitute numbers into the above equation... to get an answer... at least with the info that you are given in this question
your text probably does give a ready formula for q in terms of time ,t for the discharging of a capacitor. Use that.