Three concentric conducting spherical shells have radii $r, 2 r$ and $3 r$ and charges $q_{1}, q_{2}$ and $q_{3}$ respectively. Innermost and outermost shells are earthed as shown in figure . Select the correct alternatives:

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$q_{1} + q_{3} = - q_{2}$

$q_{1} = \frac{- q_{2}}{4}$

$\frac{q_{3}}{q_{2}} = - \frac{1}{3}$

$\frac{q_{3}}{q_{1}} = 3$

answer is A, B, C.

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Detailed Solution

As shell A is earthed
Question Image
$\Rightarrow V_{A} = 0 = \frac{K q_{1}}{r} + \frac{K q_{1}}{2 r} + \frac{K q_{3}}{3 r} = 0$
Also $V_{c} = 0 = \frac{K q_{1}}{3 r} + \frac{K q_{2}}{3 r} + \frac{K q_{3}}{3 r} = 0$
$\begin{array}{l} \Rightarrow q_{1} + q_{2} + q_{3} = 0 \\ \Rightarrow q_{1} + q_{3} = - q_{2} \end{array}$
Equation (i) is
$q_{1} + \frac{q_{2}}{2} + \frac{q_{3}}{3} = 0$
Solving (ii) and (iii),
We get
$\begin{array}{l} q_{2} = \frac{- 4 q_{3}}{3} \\ \frac{q_{3}}{q_{2}} = \frac{- 3}{4} \end{array}$
Similarly, getting other radius by putting value of (iv) in (ii)
$\frac{q_{3}}{q_{1}} = 3 \Rightarrow \frac{q_{1}}{q_{2}} = \frac{- 1}{4}$