# Another Gauss' Law Problem!

1. Sep 6, 2009

### TwinGemini14

An infinite line charge lies on the z-axis with l = 2 µC/m. Coxaial with that line charge are: an infinite conducting shell (with no net charge) with thickness 1 cm and with inner radius 2 cm and outer radius 3 cm, an infinite shell with a radius of 4 cm and with a net charge of -5 µC/m, and another infinite conducting shell (with no net charge) with a thickness of 1 cm and with an inner radius of 5 cm and outer radius of 6 cm. A cross sectional view of this setup is shown below:

http://i662.photobucket.com/albums/uu347/TwinGemini14/elecshell.gif

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4) Calculate the magnitude of the electric field at r = 10 cm from the z-axis.

A) 0 N/C
B) 1.15 x 10^5 N/C
C) 2.89 x 10^5 N/C
D) 4.22 x 10^5 N/C
E) 5.40 x 10^5 N/C

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Since our cylinders have infinite length, I simply added the charge densities to get the charge enclosed by our cylinder.
So Q = (2 uC)*(-5 uC) = -3 uC.
Using Gauss' Law, EA = Q/(epsilon-not)
E = Q/(Epsilon-not * A)
E = (3*10^-6) / ((8.85*10^-12)*(2pi*0.1))
E = |-5.4*10^5| = 5.4*10^5 N/C

I believe I did this one correct, but could somebody double check please?
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5) Calculate the induced surface charge per meter on the inner conductor's outer surface (at r = 3 cm).

A) -4 µC/m
B) -3 µC/m
C) 0 µC/m
D) +2 µC/m
E) +5 µC/m

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So I assumed this logic. Since the inner conductor's inner shell (2cm) must be -2 uC/m to balance the infinite line of charge. So then at the outer shell (3m), it must be -3 uC/m since at 4cm, the net charge is -5 uC. ANSWER = B.
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2. Sep 7, 2009

### Staff: Mentor

Looks good.

Good! Since you know the net charge on the conducting shell is zero, what must be the induced charge on the outer surface?
:yuck: Does a charge at r = 4 cm have any impact on anything going on at r = 3 cm?

3. Sep 7, 2009

### TwinGemini14

So then if the charge at 2cm is -2 uC/m, then at 3cm, it must be +2 uC/m since the charge between 2cm and 3cm must be zero.

Right?