Electric Fields, Insulators, Conductors

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Homework Help Overview

The discussion revolves around understanding electric fields in the context of insulators and conductors, specifically focusing on the charges associated with a hollow conducting sphere and its inner and outer surfaces.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the application of Gauss's law to determine the charges on the inner and outer surfaces of a conducting sphere. Questions arise regarding the relationship between the total charge on the conductor and the charges on its surfaces.

Discussion Status

Some participants have provided guidance on using Gaussian surfaces to analyze the charge distribution. There is an ongoing exploration of the differences between the total charge on the conductor and the charges on its inner and outer surfaces, with various interpretations being discussed.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information available for discussion. The original poster expresses uncertainty about the differences in charge values and seeks clarification on the setup of the problem.

Aerospace
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I am going to attach the question along with this post so that you guys can see the questions and the diagram and be able to help me in answering my question better.
I found the ccharge on the inner sphere and it is -3.928e-9 and the charge on the outer sphere is 8.15698e-9.
Now I have to find the total charge on the inner and outer surface of the hollow conducting sphere.
I'm not sure why it is any different from the charge on the outer sphere written above. If someone could help me by explaining and helping me set up the problem, it would be DEFINITELY appreciated. Thanks.

I love this forum, people always come through :)
 

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How did you find the charges on the spheres?

Consider a guassian surface at three different radii:
- just outside the insulator (I assume this is how you found the charge on the inner surface)
- outside the conducting sphere (What charge would that give?)
- and one inside the conducting sphere (What's the field there?)
 
I still don't understand...
I found the charge on the insulating sphere using the formula (E)(Eo)(SA of sphere)=-q
E is the electric field, Eo is Epsilon naught, SA is Surface Area, and the negative in front of q (charge) is because the electric field is radially inward.

For the conducting sphere, I used (Ec)(Eo)(SAc of sphere)=Q-q
Where Ec is Electric field for the conductor, Eo is epsilon naught, SAc is Surface Area of conducting sphere, Q is charge on the conductor and q is the inside charge of the conducting sphere...


So my initial question was, what's the difference between the charge on the conductor and the charge on the inner and outer surface of the conducting sphere? because I have to find the total charge on both of those.
 
Originally posted by Aerospace
For the conducting sphere, I used (Ec)(Eo)(SAc of sphere)=Q-q
Where Ec is Electric field for the conductor, Eo is epsilon naught, SAc is Surface Area of conducting sphere, Q is charge on the conductor and q is the inside charge of the conducting sphere...
Gauss's law gives you the total charge within the surface. Setting this equal to Q - q, means that Q is the total charge on the conducting sphere: including both inner and outer surfaces.
So my initial question was, what's the difference between the charge on the conductor and the charge on the inner and outer surface of the conducting sphere? because I have to find the total charge on both of those.
I don't understand your question. The total charge on anything is the sum of all the charges. For a conductor, this is the sum of the charge on its inner and outer surfaces.

What's the charge on the inner surface? Use a Gaussian surface in the interior of the conductor and see what the net charge must be. (It's trivial.)
 
Figured it out :) Thanks

Thank you so much. I used your idea of using the Gaussian surface and therefore, the total charge on the inner surface is therefore q, which is 3.928e-9 and the total charge on the outer surface is Q-q which is 8.15698e-9 - 3.928e-9 = 4.22898e-9

Yay...haha...
 

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