Question

Using Gauss's law and assuming the pipe is sufficiently long to consider it infinitely long, calculate the electric field r = 6.51 cm from the centerline of the pipe.

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Answer 1

Draw a "hypothetical" pipe of radius r=0.0651m (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 = 2pi*r*L
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[pi*Ro^2*L - pi*Ri^2*L] , 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] = [(0.00122)/(2*8.85*10^-12*0.0651)]*[0.0310^2 - 0.0264^2]

E = 279560.17 N/C