A conducting wire connects two charged metallic spheres A and B of ...

Question

A conducting wire connects two charged metallic spheres A and B of radii r$_{1}$ and r$_{2}$ respectively. The distance between the spheres is very large compared to their radii. The ratio of electric fields, (EA/EB) at the surfaces of spheres A and B will be 1 (A) r (B) r

Accepted Answer

When connected by a conducting wire, charge flows until both spheres reach the same potential:

$$V=\frac{kQ_1}{r_1}=\frac{kQ_2}{r_2}$$

Surface field on a sphere: $E=\dfrac{kQ}{r^2}=\dfrac{V}{r}$ (using $Q=Vr/k$).

Since $V$ is the same for both spheres:

$$\frac{E_A}{E_B}=\frac{V/r_1}{V/r_2}=\frac{r_2}{r_1}$$

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