Two spherical conductors of radii r1 and r 2 are separated by a distance much greater than the radius of either sphere. The spheres are connected by a conducting wire, as shown in Figure 25.25. The charges on the spheres in equilibrium are q1 and q2, respectively, and they are uniformly charged.

Question

Two spherical conductors of radii [latex] r_{1} ext { and } r_{2} [/latex] are separated by a distance much greater than the radius of either sphere. The spheres are connected by a conducting wire, as shown in Figure 25.25. The charges on the spheres in equilibrium are [latex] q_{1} ext { and } q_{2} [/latex], respectively, and they are uniformly charged. Find the ratio of the magnitudes of the electric fields at the surfaces of the spheres.

Accepted Answer

Because the spheres are connected by a conducting wire, they must both be at the same electric potential:

[latex] V=k_{e} \frac{q_{1}}{r_{1}}=k_{e} \frac{q_{2}}{r_{2}} [/latex].

Therefore, the ratio of charges is

(1) [latex] \frac{q_{1}}{q_{2}}=\frac{r_{1}}{r_{2}} [/latex].

Because the spheres are very far apart and their surfaces uniformly charged, we can express the magnitude of the electric fields at their surfaces as

[latex] E_{1}=k_{e} \frac{q_{1}}{r_{1}^{2}} \quad ext { and } \quad E_{2}=k_{e} \frac{q_{2}}{r_{2}^{2}} [/latex]

Taking the ratio of these two fields and making use of Equation (1), we find that

(2) [latex] \frac{E_{1}}{E_{2}}=\frac{r_{2}}{r_{1}} [/latex].

Hence, the field is more intense in the vicinity of the smaller sphere even though the electric potentials of both spheres are the same.

Question 25.9: Two spherical conductors of radii r1 and r 2 are separated b...

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