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.

Question

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. Help Please.

Answer 1

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.

Wrong, 2.5mm^2 doesn't equal .0025m^2... It equals 2.55*10^-6 m^2... the rest is perfect...