# Homework Help: Help with integration

1. Feb 5, 2012

### barry-

hello all new to the forum i have a calculus problem wonder if anyone can help the charge (q) can be obtained by intergrating (i) with respect to time determine the the charge on capacitor as a result of current flowing for three hundredths of a second if (i) is a constant 5mA, my calculus is very rusty but have i got the equation right
5
∫ 5 di which i get the answer to 23.5 if im doing it right.
0.300

Any help would be great thanks.

2. Feb 5, 2012

### Staff: Mentor

Welcome to the PF.

IF the current i is constant, you don't need to use integration. You only need to use integration if i varies with time...

3. Feb 5, 2012

### barry-

ah ok so q = integral of (idt) is that right

4. Feb 5, 2012

### Staff: Mentor

Again, since i(t)=constant, you don't need the integral. You just multiply.

The fundamental equation for when the current is constant is

i(t) = ΔQ / Δt

So the change in charge with time is constant if I(t) is constant. What multiplication should you use to find the charge accumulated on the capacitor in a time Δt?

5. Feb 6, 2012