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.

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  1. 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

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  2. Ah I see, thank you very much.

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