An emf self-induced in a solenoid of inductance L changes . . .

1. Oct 31, 2006

An emf self-induced in a solenoid of inductance L changes in time as
E=E_0e^(-kt)
Find the total charge that passes through the solenoid, if the charge is finite. (Use E_0 for 0, k, and L as necessary.)

Now I know that the induced emf is

E_L=-L(dI/dt)
I'm having a hard time on this problem since I'm looking for the charge, and not for the current which I believe I have to determine in the first place. I'm not sure how to exactly find di/dt and then relate it to the charge. Could anyone assist with this please?

2. Oct 31, 2006

I know that I have to take the integral of an equation twice, though I do not know how to incorporate L and the given equation into one really.

3. Oct 31, 2006

Mindscrape

I don't see any integration, but rather differentiation. They tell you that $$E = E_0 e^{-kt}$$ - this is a given correct? So, what about using chain rule to find $$\frac{di}{dt}$$, which will give you charge?

4. Oct 31, 2006

I know how to differentiate the equation, but I don't know how that gives me dI/dt, could you elaborate?

5. Oct 31, 2006

Mindscrape

Oh, whoops, I think I misinterpreted the problem. It is not asking you to find the rate of change of voltage with respect to charge, that's what I thought at least. This is really just some algebra to find Q(t).

I can put you on the right track if you want. Your equation for the voltage (EMF) in an inductor could be wrong, though I'm not really sure what sign convention was used for the polarity through the inductor in the circuit you are analyzing (so it could be either positive or negative, but it is usually positive). Anyway, you have already said that you know charge as being proportional to voltage and inversely proportional to inductance. You also said that you know voltage is proportional to the exponential function. Use equalities and you can get charge in terms of the exponential.