How Does Gauss' Law Apply to Complex Cylindrical Charge Distributions?

[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

ANSWER = E
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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Can somebody please help me with these problems? I'm not entirely sure and would appreciate the assistence. Thanks in advance!

 

[Doc Al]

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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

-------------
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

ANSWER = E
I believe I did this one correct, but could somebody double check please?

Looks good.

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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.

Good! Since you know the net charge on the conducting shell is zero, what must be the induced charge on the outer surface?

So then at the outer shell (3m), it must be -3 uC/m since at 4cm, the net charge is -5 uC. ANSWER = B.

Does a charge at r = 4 cm have any impact on anything going on at r = 3 cm?

 

[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?

So the answer is D.

 

[Doc Al]

Right!