Consider two concentric spherical metal shell | vedclass

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(C) Let the charge on the inner shell be $q^{\prime}$.Since the inner shell is grounded,its electric potential must be zero.The potential at the surfa

Vedclass · Accepted Answer

$\frac{-r_{1}}{r_{2}} q$

Vedclass · Answer

(C) Let the charge on the inner shell be $q^{\prime}$.Since the inner shell is grounded,its electric potential must be zero.The potential at the surface of the inner shell is due to its own charge $q^{\prime}$ and the charge $q$ on the outer shell.The potential $V_{1}$ at the inner shell is given by $V_{1} = \frac{k q^{\prime}}{r_{1}} + \frac{k q}{r_{2}}$.Setting $V_{1} = 0$,we get $\frac{k q^{\prime}}{r_{1}} + \frac{k q}{r_{2}} = 0$.Solving for $q^{\prime}$,we get $\frac{q^{\prime}}{r_{1}} = -\frac{q}{r_{2}}$.Therefore,$q^{\prime} = -\left(\frac{r_{1}}{r_{2}}\right) q$.

Consider two concentric spherical metal shells of radii $r_{1}$ and $r_{2}$ $(r_{2} > r_{1})$. If the outer shell has a charge $q$ and the inner one is grounded,the charge on the inner shell is

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