An ion beam has a cross-sectional of 2.5 mm^2 and the charge delivered by it to a target can be expressed as q(t)=(.12 C/s^3)t^3 + (.24 C/s^2)t^2 - .45 C. Calculate the current density in the beam at t = 3.0 seconds.

I used an integral from 0 to 3s. and got 3.24C, then I inserted it to current density formula j= I/A, and I got an answer of 1296 A/m^2 and I am not sure if I'm right.

your procedure was correct.

Integrate? Why, all you are looking for is at time 3 .

q=.12*27+.24*9-.45 Coulombs at time=3

current density= q/area

Ah I see, thank you very much.

To find the current density in the beam at t = 3.0 seconds, you need to correctly calculate the current (I) and the cross-sectional area (A) of the beam at that time.

First, let's calculate the current by evaluating the charge delivered by the ion beam at t = 3.0 seconds using the given equation:
q(t) = (0.12 C/s^3)t^3 + (0.24 C/s^2)t^2 - 0.45 C

Substituting t = 3.0 seconds:
q(3.0) = (0.12 C/s^3)(3.0 s)^3 + (0.24 C/s^2)(3.0 s)^2 - 0.45 C
q(3.0) = 0.12 C/s^3 * 27 s^3 + 0.24 C/s^2 * 9 s^2 - 0.45 C
q(3.0) = 3.24 C

So, the charge delivered is q(3.0) = 3.24 C.

Next, we need to calculate the cross-sectional area of the beam. Given that the cross-sectional area is 2.5 mm^2, we should convert it to square meters by dividing by 1,000,000 (since there are 1,000,000 square millimeters in a square meter):
A = 2.5 mm^2 / 1,000,000
A = 2.5 * 10^(-6) m^2

Now we can calculate the current density (j) using the formula j = I / A:
j = q(3.0) / A
j = 3.24 C / 2.5 * 10^(-6) m^2
j ≈ 1,296,000 A/m^2

So, the current density in the ion beam at t = 3.0 seconds is approximately 1,296,000 A/m^2.