Two concentric, thin metallic spheres of radii R₁and R₂ (R₁>R₂) bear charges Q₁and Q₂ respectively. Then the potential at radius r between R₁ and R₂ will be $(k = 1 / 4 {πε}_{0})$

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$k (\frac{Q_{1} + Q_{2}}{r})$

$k (\frac{Q_{2}}{r} + \frac{Q_{1}}{R_{1}})$

$k (\frac{Q_{1}}{r} + \frac{Q_{2}}{R_{2}})$

$k (\frac{Q_{1}}{R_{1}} + \frac{Q_{2}}{R_{2}})$

answer is C.

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Detailed Solution

The potential due to the large sphere will be given by $V_{1} = \frac{1}{4 {πϵ}_{0}} \frac{Q_{1}}{R_{1}}$

V₂due to smaller sphere, $V_{2} = \frac{1}{4 {πϵ}_{0}} \frac{Q_{2}}{r}$

$V = V_{1} + V_{2} = \frac{1}{4 π \in_{0}} (\frac{Q_{2}}{r} + \frac{Q_{1}}{R_{1}}) = K (\frac{Q_{2}}{r} + \frac{Q_{1}}{R_{1}})$

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