Question

A long pipe of outer radius 3.50 cm and inner radius 2.98 cm carries a uniform charge density of 5.22 mC/m3. Using Gauss\'s law and assuming the pipe is sufficiently long to consider it infinitely long, calculate the electric field r = 7.35 cm from the centerline of the pipe.

Answer 1

Draw a "hypothetical" pipe of radius r=.0735m (same radius you want to know the field) and concentric with the actual pipe. Let it have arbitrarly length "L".
From symmetry you assume the E-field is along "r" , hence perpendicular to your hypothetical pipe and also constant on the surface "A" of that pipe, since r is constant.
You then evaluate the left side of Gauss Law. Because E is constant on the surface and crosses the curved portion perpendicularly (its parallel to the area vector) the integral is just "EA" . There is no contribution from the caps of the pipe because E does not penetrate the caps, just slides over them.
So now Gauss' law reduces to;
eoEA = Q

Where A = 2pirL
and where "Q" is the net charge enclosed by your hypothetical pipe.
Since it encloses a section of actual pipe ,of length L, you find Q by multiplying the charge density "D" times the volume of the pipe thickness;
Q = D[piRo^2L - piRi^2L] , Where Ro & Ri are the outer and inner radii of the pipe, in meters.

So the E-field is then; after cancelling "L" and "pi";
E = (D/2eor)[Ro^2 - Ri^2]

Just plug in your given values and you're done. Be sure everything is in SI units.

E = sigma r/ 2e0

E = 5.22e-3 * 0.0735/(2*8.85e-12)

E = 2.167 *10^7 N/C