TwoG What G epslic t ) C | Not Secure ! com/ibiscms/modfibis/view.php?id-6.OD rn
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TwoG What G epslic t ) C | Not Secure ! com/ibiscms/modfibis/view.php?id-6.OD rnia, Merced- PHYS 009-20-Fall18-CALLAGHAN > ric Field and Gauss Law 9/13/2018 1 1:59 PM 9 87/109/10/2018 12:55 AM 9 Gra Map 3 Sapling Learning A long pipe of outer radius 3.00 cm and inner radius 2.55 cm carries a uniform charge density of 7.22 mC/m. Using Gauss's law and assuming the pipe is sufficiently long to consider it infinitely long calculate the electric ield r-6.30 cm from the centerine of the pipe Side view Top view Number 143x 10- NIC There is additional feedback View this feedback bottom dirider bar on the divider bar again to hide the addstional leedback incorrect Close OPrevous ®Give Up & View Solden Try Again ONext EExplanation / Answer
Draw a "hypothetical" pipe of radius r= 0.063m (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.
Solving, we get:
*E = 1.62*10^6 N/C*
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