Electric potential of charged concentric spheres

Click For Summary

Homework Help Overview

The problem involves two conducting concentric spheres with specified radii and charges, where the goal is to find the charge on the inner sphere that results in zero electric potential at its surface. The context is rooted in electrostatics and the application of Gauss's law.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply Gauss's law to derive the relationship between the charges on the spheres and the resulting potentials. They express concern about an alternative solution that suggests the inner sphere's charge should be zero.

Discussion Status

Some participants affirm the original poster's reasoning and question the validity of the alternative solution. There is an acknowledgment of the need for clarity regarding the influence of the inner sphere's charge on the potential at its surface.

Contextual Notes

Participants note that the problem assumes the electric potential is zero at infinity and that the inner sphere's charge must be considered to achieve the required potential condition.

FranzDiCoccio
Messages
350
Reaction score
43

Homework Statement



Two conducting concentric spheres of negligible thickness. The radii of the spheres are R_1 and R_2, respectively, with R_2>R_1. A charge q_2 is placed on the external sphere.
A charge q_1 is placed on the internal sphere.

Assume that the electric potential is zero infinitely far from the center of the spheres.

Find q_1 such that the potential on the inner sphere is zero.

Homework Equations



  • Gauss law: \Phi(\vec{E}) = \frac{Q}{\varepsilon_0}
  • Potential of a point charge (such that V=0 for d=\infty): \displaystyle V= k \frac{Q}{d}.

The Attempt at a Solution



Using Gauss' law I can say that, if only the outer sphere was present,
V_2 = k \frac{q_2}{R_2}
on its surface and inside it. Hence, at radius R_1,
V_1 = k \frac{q_2}{R_2}
If only the inner sphere was present
V_1 = k \frac{q_1}{R_1}
on its surface
When both spheres are present, on the surface of the inner sphere
V_1 = k \frac{q_2}{R_2}+k \frac{q_1}{R_1}
If we want V_1=0, it should be
\frac{q_2}{R_2}=- \frac{q_1}{R_1}
and hence
q_1=- \frac{R_1}{R_2} q_2

It seems to me that this makes sense. The potential would be
V_2 = \frac{k} q_2 \left\{<br /> \begin{array}{cc}<br /> \frac{1}{r}\left(1-\frac{R_1}{R_2}\right) &amp; r\geq R_2 \\<br /> \frac{1}{R_2}-\frac{1}{r}\frac{R_1}{R_2} &amp; R_1 \leq r \leq R_2<br /> \end{array}<br /> \right.<br />

Assuming that q_2&gt;0, the potential grows when the outer sphere is approached from outside, reaches its maximum on the surface of the spheres and drops to 0 when it reaches the inner radius...
Inside that, it stays zero.I'm asking here just to be on the safe side.
I came across a (kind of "official") solution for this exercise that in my opinion does not make any sense. It concludes q_1=0.

[edited]
 
Last edited:
Physics news on Phys.org
Your solution looks good to me. The "kind of official" solution does not: If the charge on the inner sphere were zero it would have no influence on the potential at its surface, and so the potential there would have to be the same as that of the surface of the outer sphere which is not zero as required by the problem.
 
Hi gneill,

thanks for your insight. I was pretty sure that the proposed solution could not be correct. Other than the final result, the entire derivation seems really shaky, but there's no point in repeating a wrong reasoning.
I needed some support because I'm a bit rusty on this stuff, and I was afraid I was missing a subtle point.
Thanks again
 
There are online video tutorials that cover this topic which may be of interest. Google: Potential in a System of Concentric Shells
 

Similar threads

Replies
4
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K
Replies
6
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
1K