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Distance between centers of two spheres A and B, each of radius R as shown in figure. Sphere B has a spherical cavity of radius R/2 such that distance of centre of cavity is (r – R/2) from the centre of sphere A and R/2 from the centre of sphere B. Di-electric constant of material of each sphere is K = 1 and material of each sphere has a uniform charge density $ρ$ per unit volume. The interaction potential energy of the two spheres is found to be $\frac{π ρ^{2} R^{6} (7 r - 4 R)}{K ε_{0} r (2 r - R)}$ . The value of K is

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answer is D.

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Detailed Solution

$U = \frac{K q_{1} q_{2}}{r}$

$U_{0} = U_{12} + U_{13}$

$U_{12} = \frac{K (\frac{4}{3} π R^{3} ρ) (\frac{4}{3} π R^{3} ρ)}{r}$

$= \frac{4 π ρ^{2} R^{6}}{9 ε_{0} r}$

$U_{13} = \frac{K (\frac{4}{3} π R^{3} ρ) [\frac{4}{3} π . {(\frac{R}{2})}^{3} (- ρ)]}{(r - \frac{R}{2})} = \frac{- π ρ^{2} R^{6}}{9 ε_{0} (2 r - R)}$

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