Two concentric, thin metallic spherical shells of radii $R_{1}$ and $R_{2} (R_{1} > R_{2})$ bear charges $Q_{1}$ and $Q_{2}$ respectively. Then the potential at radius ‘r’ between $R_{1}$ and $R_{2}$ will be $\frac{1}{4 π ϵ_{0}}$ times

Q. Two thin concentric hollow conducting spheres of radii

R_{1}

and

R_{2}

carry charges

Q_{1}

and

Q_{2}

respectively. If

R_{1} < R_{2}

then the potential at a point distant

r

such that

R_{1} < r < R_{2}

Q. Two concentric spheres of radii

R_{1}

and

R_{2} > R_{2} > R_{1}

carry charges

Q_{1}

and

Q_{2}

respectively. The potential at a point in between the two spheres and at a distance

r

from the centre will be

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

The correct option is D Q1R1+Q2rPotential at radius, r,Vr= potential at P due to Q1 + potential atP due to Q2Vr=V1+V2=14πϵo[Q1R1+Q2r]

Two concentric, thin, metallic spheres of radii $R_{1}$ and $R_{2}$ $(R_{1} > R_{2})$ bear charges $Q_{1}$ and $Q_{2}$ respectively. Then the potential at radius r between $R_{1}$ and $R_{2}$ will be $1 / 4 π ε_{o}$ time

The correct option is D $\frac{Q_{1}}{R_{1}} + \frac{Q_{2}}{r}$
Potential at radius, r,
$V_{r} =$ potential at P due to $Q_{1}$ + potential at
P due to $Q_{2}$
$V_{r} = V_{1} + V_{2}$
$= \frac{1}{4 π ϵ_{o}} [\frac{Q_{1}}{R_{1}} + \frac{Q_{2}}{r}]$