How Does Grounding Affect Charge Distribution on a Metal Sphere?

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Homework Help Overview

The discussion revolves around the effects of grounding on charge distribution in a metal sphere when a positive point charge is placed at its center. Participants are exploring the implications of grounding and the resulting charge on the inner and outer surfaces of the sphere.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using Gauss' Law to analyze the charge distribution and the implications of grounding on electric potential. Some express uncertainty about the application of Gauss' Law and seek alternative approaches. Questions arise regarding the potential difference and its relationship to charge distribution.

Discussion Status

The discussion is active, with participants providing insights and exploring different interpretations of the problem. Some guidance has been offered regarding the use of Gauss' Law and the concept of electric potential, but no consensus has been reached on the correct answer to the original question.

Contextual Notes

Participants mention a lack of familiarity with Gauss' Law and potential difference, indicating varying levels of understanding of the concepts involved. The grounding of the sphere is a key aspect under consideration, influencing the charge distribution and electric potential.

songoku
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Homework Statement


A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sphere is then grounded as shown. A is the inner surface and B is the outer surface of the sphere. Which of the following statements is correct?

a. the charge on A is -q ; that on B is +q
b. the charge on B is -q ; that on A is +q
c. there is no charge on either A or B
d. the charge on A is -q ; there is no charge on B

charge.jpg



Homework Equations





The Attempt at a Solution


The charge on A will be negative because the electrons on the sphere will be attracted by the positive charge q. A will be more positive than B because B is grounded (not sure about this). So the answer is d ??

Thanks
 
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To see if your answer is correct, use Gauss' Law with two kinds of Gaussian surfaces concentric with the shell.

One inside the conductor between A and B.
One outside B.

What is the flux through each and what is the enclosed charge by each?
 
Hi kuruman

I haven't studied Gauss Law yet. Maybe there are other approaches?

Thanks
 
Have you studied "potential difference"? You must have otherwise grounding the sphere is meaningless to you. So what is the potential difference between zero and infinity?
 
Hi kuruman

Yes I know a little biit about potential difference. When grounded, the potential becomes 0.

Potential difference between zero and infinity = infinity ??
 
songoku said:
Hi kuruman

Yes I know a little biit about potential difference. When grounded, the potential becomes 0.

Potential difference between zero and infinity = infinity ??

What is the electrical potential at infinity? It is not infinity.
 
Hi RoyalCat

V=k\frac{q}{r}

Electrical potential when r = infinity is zero
 
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songoku said:
Hi RoyalCat

V=k\frac{q}{r}

Electrical potential when r = infinity is zero

Very good, then we've discovered that the potential at infinity, is 0 (That's how we've defined it, mind you)

The potential difference between infinity and a grounded object, is therefore 0 as well (0-0=0, as best I can recall. ;p)

Looking at the formula you posted, what does that tell us about the charge on the outer rim of the conductor?

Heh, it's really hard to avoid Gauss' Law reasoning with this problem. ;)
 
Last edited:
Hi RoyalCat

The charge on the outer of the rim of the conductor will be zero. But I don't see the connection of the potential difference (between infinity and ground) and the charge on the outer rim. I think we can determine the charge by only looking on that formula ?

Thanks
 
  • #10
1. If the potential difference between the outer rim and infinity is zero, there is no electric field between the outer rim and infinity.
2. Electric field lines start at positive charges and end at negative charges.

Therefore

If the there is no electric field between the outer rim and infinity there can be no charges at the boundaries of that region.

By grounding the conducting shell you essentially bring "infinity" (where the potential is zero) closer to charge q.
 
  • #11
Oh so the answer is really d.

Thanks a lot for your explanation, kuruman and RoyalCat. I learn a lot. Thanks again
 

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