Two concentric, thin metallic spheres of radii R1and R2 (R1>R2) bear charges Q1 and Q2 respectively. Then the potential at radius r between R1 and R2 will be k=1/4πε0

The potential due to the large sphere will be given by V1=14πϵ0Q1R1V2 due to smaller sphere, V2=14πϵ0Q2rV=V1+V2=14π∈0Q2r+Q1R1=KQ2r+Q1R1

Two concentric, thin metallic spheres of radii R1and R2 (R1>R2) bear charges Q1 and Q2 respectively. Then the potential at radius r between R1 and R2 will be k=1/4πε0

The potential due to the large sphere will be given by V1=14πϵ0Q1R1V2 due to smaller sphere, V2=14πϵ0Q2rV=V1+V2=14π∈0Q2r+Q1R1=KQ2r+Q1R1

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

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

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

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

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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Two concentric, thin metallic spheres of radii R1and R2 (R1>R2) bear charges Q1 and Q2 respectively. Then the potential at radius r between R1 and R2 will be k=1/4πε0

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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 $(k = 1 / 4 {πε}_{0})$