- #1

FranzDiCoccio

- 315

- 35

## 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: