Two conduction spheres of radii \(R_1\) and \(R_2\) are kept widely separated from each other. If the spheres are connected by a metal wire, what will be the capacitance of the combination? Think in terms of series-parallel connections.
\(4 \pi \epsilon_0 \frac{R_1 R_2}{R_1 + R_2}\)
\(4 \pi \epsilon_0 \frac{R_1 R_2}{R_1 - R_2}\)
\(4 \pi \epsilon_0 \sqrt{R_1 R_2}\)
\(4 \pi \epsilon_0 (R_1 + R_2)\)
Solution:
When widely separated conductors are connected, they are effectively in parallel as they share a common potential. The equivalent capacitance is the sum of individual capacitances: \(C_{eq} = C_1 + C_2 = 4 \pi \epsilon_0 R_1 + 4 \pi \epsilon_0 R_2 = 4 \pi \epsilon_0 (R_1 + R_2)\).
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