Electric potential of charged concentric spheres

  • #1
280
29

Homework Statement



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

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

Find [itex]q_1[/itex] such that the potential on the inner sphere is zero.

Homework Equations



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

The Attempt at a Solution



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

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

Assuming that [itex]q_2>0[/itex], 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 [itex]q_1=0[/itex].

[edited]
 
Last edited:

Answers and Replies

  • #2
gneill
Mentor
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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.
 
  • #3
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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
 
  • #4
gneill
Mentor
20,925
2,867
There are online video tutorials that cover this topic which may be of interest. Google: Potential in a System of Concentric Shells
 

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