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Dipole Moment

The concentri...

Question

The concentric, conducting spherical shells $X$ , $Y$ and $Z$ with radii $r$ , $2 r$ and $3 r$ , respectively. $X$ and $Z$ are connected by a conducting wire and $Y$ is uniformly charged to charge $Q$ as shown in figure. Charges on shells $X$ and $Z$ will be :

q_{X} = \frac{Q}{4}

q_{Z} = \frac{- Q}{6}

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q_{X} = \frac{- Q}{4}

q_{Z} = \frac{Q}{4}

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q_{X} = \frac{Q}{4}

q_{Z} = \frac{- Q}{4}

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q_{X} = \frac{- Q}{6}

q_{Z} = \frac{Q}{4}

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Solution

The correct option is B $q_{X} = \frac{- Q}{4}$ , $q_{Z} = \frac{Q}{4}$
Let charges on $X$ and $Z$ be $q_{X}$ and $q_{Z}$
By conservation of charge,
$q_{X} + q_{Z} = 0 \Rightarrow q_{Z} = - q_{X}$
$∴ V_{X} = \frac{1}{4 π ε_{0}} [\frac{q_{X}}{r} + \frac{Q}{2 r} + \frac{q_{Z}}{3 r}]$
$V_{Z} = \frac{1}{4 π ε_{0}} [\frac{q_{X}}{3 r} + \frac{Q}{3 r} + \frac{q_{Z}}{3 r}]$
Since, $X$ and $Z$ are connected.
$∴ V_{X} = V_{Z}$
$\Rightarrow \frac{q_{X}}{r} + \frac{Q}{2 r} + \frac{q_{Z}}{3 r} = \frac{q_{X}}{3 r} + \frac{Q}{3 r} + \frac{q_{Z}}{3 r}$
$\Rightarrow \frac{2 q_{X}}{3} = \frac{- Q}{6}$
$∴ q_{X} = \frac{- Q}{4}$
$∵ q_{X} + q_{Z} = 0 \Rightarrow q_{Z} = - q_{X}$
$= \frac{Q}{4}$

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