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FranzDiCoccio
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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]
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