Two spherical conductors $$A$$ and $$B$$ of radii $$1\,mm$$ and $$2\,mm$$ are separated by a distance of $$5\,cm$$ and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surface of spheres $$A$$ and $$B$$ is

When the two conducting spheres are connected by a wire, charge will flow between them until they reach electrostatic equilibrium. At equilibrium, both spheres will be at the same electrical potential.

Let this common potential be $$V$$.

The potential at the surface of a spherical conductor is given by the formula:

$$V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$$

Since the potentials are equal ($$V_A = V_B$$), we can write:

$$\frac{1}{4\pi\epsilon_0} \frac{Q_A}{R_A} = \frac{1}{4\pi\epsilon_0} \frac{Q_B}{R_B}$$

Canceling out the constants gives the ratio of their charges:

$$\frac{Q_A}{R_A} = \frac{Q_B}{R_B}$$

$$\Rightarrow \frac{Q_A}{Q_B} = \frac{R_A}{R_B}$$

The magnitude of the electric field at the surface of a spherical conductor is given by:

$$E = \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2}$$

We need the ratio of the electric fields, $$\frac{E_A}{E_B}$$:

$$\frac{E_A}{E_B} = \frac{\frac{1}{4\pi\epsilon_0} \frac{Q_A}{R_A^2}}{\frac{1}{4\pi\epsilon_0} \frac{Q_B}{R_B^2}}$$

$$\Rightarrow \frac{E_A}{E_B} = \left(\frac{Q_A}{Q_B}\right) \left(\frac{R_B}{R_A}\right)^2$$

Substituting our previous finding ($$\frac{Q_A}{Q_B} = \frac{R_A}{R_B}$$) into this equation:

$$\frac{E_A}{E_B} = \left(\frac{R_A}{R_B}\right) \left(\frac{R_B}{R_A}\right)^2$$

$$\Rightarrow \frac{E_A}{E_B} = \frac{R_B}{R_A}$$

We are given the radii of the two spheres:

$$R_A = 1\text{ mm}$$

$$R_B = 2\text{ mm}$$

Substituting these values into the ratio:

$$\frac{E_A}{E_B} = \frac{2}{1}$$