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(B) The capacitance of a spherical conductor is given by $C = 4\pi \epsilon_0 R$.When two spheres are connected by a wire,charge flows until they reac

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$4.5 	imes 10^6 \ V$

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(B) The capacitance of a spherical conductor is given by $C = 4\pi \epsilon_0 R$.When two spheres are connected by a wire,charge flows until they reach a common potential $V$.The common potential is given by $V = \frac{Q_{total}}{C_1 + C_2}$.Here,$Q_{total} = 150 \ \mu C = 150 	imes 10^{-6} \ C$,$R_1 = 0.1 \ m$,and $R_2 = 0.2 \ m$.The capacitances are $C_1 = 4\pi \epsilon_0 R_1$ and $C_2 = 4\pi \epsilon_0 R_2$.Thus,$V = \frac{Q_{total}}{4\pi \epsilon_0 (R_1 + R_2)} = \frac{k \cdot Q_{total}}{R_1 + R_2}$,where $k = 9 	imes 10^9 \ N \cdot m^2/C^2$.Substituting the values: $V = \frac{9 	imes 10^9 	imes 150 	imes 10^{-6}}{0.1 + 0.2} = \frac{9 	imes 150 	imes 10^3}{0.3} = \frac{1350 	imes 10^3}{0.3} = 4500 	imes 10^3 \ V = 4.5 	imes 10^6 \ V$.

There are two conducting spheres of radii $10 \ cm$ and $20 \ cm$. One sphere is given a charge of $150 \ \mu C$ and is connected to the other sphere by a wire. What will be their common potential?

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