Ratio of Electric Fields on Two Connected Conducting Spheres in Equilibrium

To find the ratio of the electric fields on the surfaces of the two conducting spheres at electrostatic equilibrium, we use the principles of electrostatics for connected conductors.

When two charged conducting spheres $S_1$ and $S_2$ are connected by a thin conducting wire, charge flows between them until they reach the same electric potential: $V_1 = V_2 = V$

The electric potential $V$ at the surface of a isolated conducting sphere of radius $R$ carrying charge $Q$ is given by: $V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R}$

Since $V_1 = V_2$, we have: $\frac{1}{4\pi\varepsilon_0} \frac{Q_1}{R_1} = \frac{1}{4\pi\varepsilon_0} \frac{Q_2}{R_2} \implies \frac{Q_1}{Q_2} = \frac{R_1}{R_2}$

The magnitude of the electric field $E$ at the surface of a conducting sphere is given by: $E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R^2}$

Substituting $V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R}$, the electric field can also be written in terms of the potential $V$ as: $E = \frac{V}{R}$

Therefore, the ratio of the electric fields $E_{S1}$ and $E_{S2}$ on the surfaces of $S_1$ and $S_2$ is: $\frac{E_{S1}}{E_{S2}} = \frac{\frac{V_1}{R_1}}{\frac{V_2}{R_2}}$

Since $V_1 = V_2$: $\frac{E_{S1}}{E_{S2}} = \frac{R_2}{R_1}$

Given the radii of the spheres:

• $R_1 = 8\text{ cm}$
• $R_2 = 18\text{ cm}$

Substituting these values into the ratio: $\frac{E_{S1}}{E_{S2}} = \frac{18\text{ cm}}{8\text{ cm}} = \frac{9}{4}$

Thus, the ratio $\frac{E_{S1}}{E_{S2}}$ is $\frac{9}{4}$, which corresponds to Option D.