- #1
TwinGemini14
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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!
http://i662.photobucket.com/albums/uu347/TwinGemini14/elecshell.gif
-----------
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?
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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
----------
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.
---------------
Can somebody please help me with these problems? I'm not entirely sure and would appreciate the assistence. Thanks in advance!