Discharging Capacitor answer check

[latitude]

Homework Statement

I have finished this and gotten an answer, but I have no solutions on hand and would like to see if anyone else gets the same answer :]

A 2-nF capacitor with an initial charge of 5.1 microCoulombs is discharged through a 1.3- kiloOhm resistor.

(a) Calculatue the current in the resistor 9 microseconds after the resistor is connected acorss the terminals of the capacitor.
(b) What charge remains on the capacitor after 8 microseconds?

Homework Equations

I(t) = - Q/RC(e^(-t/RC))
q(t) = Qe^-t/RC

The Attempt at a Solution

(A) Find the current through the resistor. So I differentiated equation one to get

I = Q (e^(-t/RC)
= 0.16 microAmperes...

(B) Differentiate so
q = (-Q)(-RC) (e^(-t/RC))
= 6.11 E -11 C

My answers seem way too small to me. So my question is, was I supposed to differentiate?? Forgive my ignorance, but what does I(t) mean? Is it the same as dI/dt? Also, if I do not differentiate the question seems exceedingly simple (it's worth 10 marks.) Is there something in my process that I am over-simplifying?

Thanks so much to everyone in advance.

 

[Dadface]

I(t) stands for the current I at t seconds. and Q/RC is the initial current I(o).If you ln your equation you get
lnI=lnI(o)-t/RC

 

[ideasrule]

I don't get why it's necessary to differentiate, ln, use the logarithm rules, or do anything else. You have a formula for I:

I = Q (e^(-t/RC)

and the question asks for I. Just plug in the numbers. Similarly, you have a formula for q:

q(t) = Qe^-t/RC

and the question asks for q. Plug in the numbers.

 

[Dadface]

I don't get why it's necessary to differentiate, ln, use the logarithm rules, or do anything else. You have a formula for I:

I = Q (e^(-t/RC)

and the question asks for I. Just plug in the numbers. Similarly, you have a formula for q:

q(t) = Qe^-t/RC

and the question asks for q. Plug in the numbers.

True, just plug in the numbers.