How Does Charge Distribution Affect Electric Fields in Nested Spheres?

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Homework Help Overview

The discussion revolves around the electric field generated by a conducting sphere and a surrounding spherical shell, both carrying negative charges. Participants are tasked with determining the electric field at specific distances from the center of the spheres, considering the effects of charge distribution.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants attempt to apply Gauss's law to calculate the electric field in different regions based on the charge distribution. Questions arise regarding the charge enclosed by Gaussian surfaces and the accuracy of unit conversions.

Discussion Status

The discussion is ongoing, with participants questioning the charge enclosed at specific distances and the correctness of their calculations. Some guidance is provided regarding the need to consider both the sphere and the shell's charges when determining the total enclosed charge.

Contextual Notes

Participants are reminded to consider the charge on both the inner sphere and the outer shell when calculating the electric field, and there is a focus on ensuring correct unit conversions from nano-coulombs to coulombs.

roam
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Homework Statement


A conducting sphere of radius, R = 5.5 cm with an excess charge of Q = -35.5 nC is surrounded by a concentric, conducting, spherical shell of inner radius, Rin = 9.5 cm and outer radius, Rout = 11.5 cm that carries an excess charge of q = -13.0 nC.

[PLAIN]http://img571.imageshack.us/img571/7821/imagex.gif

Determine the electric field at the following radii for the aforementioned arrangement:

(a) r = 42.5 cm.

(b) r = 8.5 cm.

The Attempt at a Solution



(a) In the indicated region r>Rout. Therefore I model the charge distribution as a sphere with charge -Q and the expression for the field in this region would be

[tex]E=-k_e \frac{Q}{r^2}[/tex]

[tex]-(9 \times 10^9) \frac{35.5}{42.5^2}=176885813.1[/tex]

even if I convert r to meters I still get the wrong answer (correct answer: 2420)

(b) Again, I can apply Gauss's law to find the electric field. Since R<r<Rin I think I should use:

[tex]k_e \frac{Q}{r^2}=(9 \times 10^9) \frac{35.5}{8.5^2} = 4422145329[/tex]

The correct answer is 44200 N/C, did I forget to convert something?
Thanks.
 
Last edited by a moderator:
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What is the charge enclosed by a Gaussian surface at 42.5 cm? Don't forget that both the sphere and the shell have charge on them. Also, how many coulombs to a nano-coulomb are there?
 
kuruman said:
What is the charge enclosed by a Gaussian surface at 42.5 cm? Don't forget that both the sphere and the shell have charge on them. Also, how many coulombs to a nano-coulomb are there?

Yes, I got all the unit conversions correct but I'm still getting the wrong answer:

[tex]-(9 \times 10^{9})\frac{35.5 \times 10^{-9}}{(0.425)^2}=-1768.8[/tex]

Why is that?
 
Last edited:
roam said:
Yes, I got all the unit conversions correct but I'm still getting the wrong answer:

[tex]-(9 \times 10^{9})\frac{35.5 \times 10^{-9}}{(0.425)^2}=-1768.8[/tex]

Why is that?
It is because the charge enclosed by a spherical Gaussian surface at 42.5 cm is not -35.5 nC. What is it?
 

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