It\'s part b of question 2. http://www.flickr.com/photos/53399890@N05/6215386991

Question

It's part b of question 2. http://www.flickr.com/photos/53399890@N05/6215386991/in/photostream

There is a semiconducting cylindrical shell of inner radius a and outer radius b. The electrons (charge q = -e) flow parallel to the axis of the shell in one direction and the holes (charge q = +e) flow parallel to the axis of the shell in the opposite direction. A material that is positively charged is inside the shell (r < a). The magnitude of the electron current density is given as J_= J( b-r)^2/(b-a)^2 and the magnitude of the hole current density is J+ = J(r-a)^2/(b-a)^2. Prove that the current in the semiconductor is i=(2/3)p J(b^2 - a^2).

Explanation / Answer

i=(integral from a to b) (J++J-)2r.dr

The top equation is drived based on the fact that the shel is considered as the integral of co-centered circles of radius r whith athickness of dr. So if ya integrate the current densities othese areas, it gives the total current. 2r.dr is the total area of co-centered circles of radius r whith athickness of dr.

i=(integral from a to b) {[(b-r)^2+(r-a)^2]/(b-a)^2}.2rdr

If ya continue calculating this, it should give the answer

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