What Is the Capacitance for a Discharge to 27% in 2.8ms?

[cwesto]

Homework Statement

A capacitor is discharged through a 100$$\Omega$$ resistor. The discharge current decreases to 27.0% of its initial value in 2.80ms.

What is the value of the capacitor?

Homework Equations

Q_f=Q_i*e($$\frac{-t}{CR}$$)

The Attempt at a Solution

ln(.27)=$$\frac{.0028s}{C100}$$

$$\frac{ln(.27)}{.0028}$$=4.68×10⁴

Answer is wrong, which means I need help. :D Thanks!

 

[Thaakisfox]

So you need to express C.

Your equation is ok, just when you put in the numbers... try that part again ;)

 

[astrorob]

Firstly, be careful of the equation you use. The equation you state is for charge - but you're interested in current. In this case you're safe since the equations are exactly similar, but you might faulter in the future if you're not careful..

It may be useful to rearrange the equation before you insert the values, then you will see where you've made the mistake. In your answer the value you've calculated is the recipricol of capacitance rather than capacitance itself.

 

[Redbelly98]

You have

4.68x10⁴ = 1/C​

So what is C ?

EDIT:
Ah, I really should refresh my screen more often. Two people got their responses in before I got around to it