The concentric, conducting spherical shells X, Y and Z with radii r, 2r and 3r, 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

$q_{x} = \frac{- Q}{4}, q_{z} = \frac{Q}{4}$

$q_{x} = \frac{Q}{4}, q_{z} = \frac{- Q}{4}$

$q_{x} = \frac{Q}{4}, q_{z} = \frac{Q}{4}$

$q_{x} = \frac{- Q}{6}, q_{z} = \frac{Q}{4}$

Solution

$q_{x} = \frac{Q}{4}, q_{z} = \frac{- Q}{4}$

Given:- spherical shells X , Y and Z having radii r, 2r and 3r.

Let charges on X and Z be q_x and q_z

Law of conservation says that net charge in an isolated system will always remain constant.

By conservation of charge

V_x = $\frac{1}{4 {πε}_{0}} [\frac{q_{x}}{r} + \frac{Q}{2 r} + \frac{qz}{3 r}]$

V_z = $\frac{1}{4 {πε}_{0}} [\frac{q_{x}}{3 r} + \frac{Q}{3 r} + \frac{qz}{3 r}]$

Since X and Z are connected.

V_x = V_z

$\frac{q_{x}}{r} + \frac{Q}{2 r} + \frac{qz}{3 r} = \frac{qx}{3 r} + \frac{Q}{3 r} + \frac{qz}{3 r}$

$\frac{2 q_{x}}{3} = \frac{- Q}{6}$

∴ q_x = $\frac{- Q}{4}$

$∵$ q_x + q_z = 0

⇒ q_z = -q_x

⇒ q_z = $\frac{Q}{4}$

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