Two charged spherical conductors of radii $R_1$ and $R_2$ are connected by a wire. The ratio of surface charge

Question

Two charged spherical conductors of radii $R_1$ and $R_2$ are connected by a wire. The ratio of surface charge densities of spheres $(\sigma_1/\sigma_2)$ is:

Accepted Answer

1. **Common Potential:** When two charged conductors are connected by a wire, charge flows between them until they reach the same electrostatic potential ($V$). Thus, $V_1 = V_2$.
2. **Formula for Potential:** The potential of a charged sphere of radius $R$ and charge $Q$ is $V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R}$.
3. **Equating Potentials:** 
   $\frac{Q_1}{R_1} = \frac{Q_2}{R_2} \implies \frac{Q_1}{Q_2} = \frac{R_1}{R_2}$.
4. **Surface Charge Density ($\sigma$):** $\sigma = \frac{	ext{Charge}}{	ext{Area}} = \frac{Q}{4\pi R^2}$. Thus, $Q = \sigma(4\pi R^2)$.
5. **Substitute and Solve:**
   $\frac{\sigma_1(4\pi R_1^2)}{\sigma_2(4\pi R_2^2)} = \frac{R_1}{R_2}$
   $\frac{\sigma_1}{\sigma_2} \cdot \frac{R_1^2}{R_2^2} = \frac{R_1}{R_2}$
   $\frac{\sigma_1}{\sigma_2} = \frac{R_1}{R_2} \cdot \frac{R_2^2}{R_1^2} = \frac{R_2}{R_1}$.

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