- #1

TwinGemini14

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

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Gauss' Law = |(E . DA) = Qenclosed / (Epsilon-not)

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1) Compare the magnitude of the electrical flux through a cylindrical surface, 1 m long, (centered on the z-axis) with a radius of 1.5 cm to that of a cylindrical surface, 1 m long, (centered on the z-axis) with a radius of 3.5 cm.

A) Flux1.5 cm > Flux3.5 cm

B) Flux1.5 cm = Flux3.5 cm

C) Flux1.5 cm < Flux3.5 cm

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My logic is this. At 1.5cm the field is closer to the surface with a +2u C/m. Near 3.5, it is near a surface with a net charge of -5u C/m. So the magnitude of the e field is probably greater near the -5u C/m surface. According to Gauss' Law, since the magnitude of the e field is greater and the area is greater (3.5cm radius > 1.5cm radius), then the answer should be C.

So I said the answer is C.

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2) Compare the magnitude of the electric field at 2.5 cm from the z-axis and 4.5 cm from the z-axis.

A) E2.5 cm > E4.5 cm

B) E2.5 cm = E4.5 cm

C) E2.5 cm < E4.5 cm

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Here, E2.5 cm is zero because it is within the conducting shell. Since the E field is not zero at 4.5 cm, the magnitude must be greater.

So I said the answer is C.

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3) Compare the magnitude (i.e., the absolute value) of the electrical flux through a cylindrical surface, 1 m long, (centered on the z-axis) with a radius of 4.5 cm to that of a cylindrical surface, 1 m long, (centered on the z-axis) with a radius of 5.5cm.

A) Flux4.5 cm > Flux5.5 cm

B) Flux4.5 cm = Flux5.5 cm

C) Flux4.5 cm < Flux5.5 cm

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Since 5.5 cm is within a conducting shell, it's e field = 0. So then its flux must also be zero due to Gauss' Law. So the magnitude of the flux is greater in 4.5cm.

So I said the answer is A.

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Can somebody please help me with this problem and review my answers because I am not sure about it. How do I go about this problem in a more precise fashion? Thanks for the help in advance!