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(A) Let $q$ be the charge appearing on the outer shell of radius $2R$ after earthing.Since the outer shell is grounded, its potential must be zero.The

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$\frac{1}{4 \pi \varepsilon_0} \frac{Q}{6R}$

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(A) Let $q$ be the charge appearing on the outer shell of radius $2R$ after earthing.Since the outer shell is grounded, its potential must be zero.The potential at the outer shell is the sum of potentials due to the inner charge $Q$ and the outer charge $q$:$V_{	ext{outer}} = \frac{KQ}{2R} + \frac{Kq}{2R} = 0$Solving for $q$, we get $q = -Q$.Now, we calculate the potential at a distance $r = \frac{3R}{2}$ from the center.Since $R The potential at this point is the sum of the potential due to the inner sphere (as a point charge) and the potential due to the outer shell (constant inside):$V(r) = \frac{KQ}{r} + \frac{Kq}{2R}$Substituting $r = \frac{3R}{2}$ and $q = -Q$:$V = \frac{KQ}{3R/2} + \frac{K(-Q)}{2R} = \frac{2KQ}{3R} - \frac{KQ}{2R} = \frac{4KQ - 3KQ}{6R} = \frac{KQ}{6R} = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{6R}$

Consider two concentric conducting spheres of radii $R$ and $2R$ respectively. The inner sphere is given a charge $+Q$. The outer sphere is grounded. The potential at $r = \frac{3R}{2}$ is

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