Three concentric conducting spherical shells | vedclass

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(D) Let the potentials of the shells be $V_1, V_2, V_3$. Since the innermost $(r)$ and outermost $(3r)$ shells are earthed,$V_1 = 0$ and $V_3 = 0$.The

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$\frac{Q_3}{Q_2} = -\frac{1}{3}$

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(D) Let the potentials of the shells be $V_1, V_2, V_3$. Since the innermost $(r)$ and outermost $(3r)$ shells are earthed,$V_1 = 0$ and $V_3 = 0$.The potential of a shell is given by the sum of potentials due to all charges: $V = \frac{kQ_{self}}{R_{self}} + \sum \frac{kQ_{other}}{R_{other}}$.For shell $1$ $(r)$: $V_1 = k \left( \frac{Q_1}{r} + \frac{Q_2}{2r} + \frac{Q_3}{3r} \right) = 0 \implies \frac{Q_1}{1} + \frac{Q_2}{2} + \frac{Q_3}{3} = 0 \implies 6Q_1 + 3Q_2 + 2Q_3 = 0$  (Eq. $1$)For shell $3$ $(3r)$: $V_3 = k \left( \frac{Q_1}{3r} + \frac{Q_2}{3r} + \frac{Q_3}{3r} \right) = 0 \implies Q_1 + Q_2 + Q_3 = 0 \implies Q_3 = -(Q_1 + Q_2)$ (Eq. $2$)Substitute Eq. $2$ into Eq. $1$: $6Q_1 + 3Q_2 + 2(-(Q_1 + Q_2)) = 0 \implies 6Q_1 + 3Q_2 - 2Q_1 - 2Q_2 = 0 \implies 4Q_1 + Q_2 = 0 \implies Q_1 = -\frac{Q_2}{4}$.Now find $Q_3$: $Q_3 = -(Q_1 + Q_2) = -(-\frac{Q_2}{4} + Q_2) = -(\frac{3Q_2}{4}) = -\frac{3Q_2}{4}$.Check options:$(A)$ $Q_1 + Q_3 = -\frac{Q_2}{4} - \frac{3Q_2}{4} = -Q_2$ (Correct).$(B)$ $Q_1 = -\frac{Q_2}{4}$ (Correct).$(C)$ $\frac{Q_3}{Q_1} = \frac{-3Q_2/4}{-Q_2/4} = 3$ (Correct).$(D)$ $\frac{Q_3}{Q_2} = \frac{-3Q_2/4}{Q_2} = -\frac{3}{4}$ (Incorrect,as it is given as $-1/3$).

Three concentric conducting spherical shells have radii $r, 2r$ and $3r$ and $Q_1, Q_2$ and $Q_3$ are final charges respectively. The innermost and outermost shells are earthed as shown in the figure. Choose the wrong statement.

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