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 a radius r between R1 and R2 will be 1/4(πε0) times :

The correct option is B Q1R1+Q2rHere, potential at r = potential due to Q1 for which r is inside + potential due to Q2 for which r is outside.Vr=kQ1R1+kQ2r=k(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 a radius r between $R_{1}$ and $R_{2}$ will be $1 / 4 (π ε_{0})$ times :

The correct option is B $\frac{Q_{1}}{R_{1}} + \frac{Q_{2}}{r}$
Here, potential at r $=$ potential due to $Q_{1}$ for which r is inside $+$ potential due to $Q_{2}$ for which r is outside.
$V_{r} = k \frac{Q_{1}}{R_{1}} + k \frac{Q_{2}}{r} = k (\frac{Q_{1}}{R_{1}} + \frac{Q_{2}}{r})$