Solution:
When connected, their electric potentials become equal, i.e., \(V_1 = V_2\). Since \(V = \frac{\sigma R}{\varepsilon_0}\), we get \(\sigma_1 R_1 = \sigma_2 R_2\), which gives \(\frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}\).
Two charged spherical conductors of radius \(R_1\) and \(R_2\) are connected by a wire. Then the ratio of final surface charge densities of the spheres \(\sigma_1 / \sigma_2\) is
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