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Two metal spheres A and B have their capacities in the ratio 3:4. They are put in contact with each other and an amount of charge 7×10^-6C is given to the combination. Next, the two spheres are separated and kept wide the apart so that one has no electrical influence on the other. The potential due to the smaller sphere at a distance of 50m from it is

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Detailed Solution

$q_{1} = (\frac{r_{1}}{r_{1} + r_{2}}) q; V_{1} = \frac{1}{4 {πε}_{0}} \frac{q_{1}}{r}$

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