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Electric Potential Due to a Solid Sphere of Charge

Charge on the...

Question

Charge on the outer sphere is $q$ and the inner sphere is grounded. Then the charge $q^{'}$ on the inner sphere is, for $(r_{2} > r_{1})$

zero

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q^{'} = q

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q^{'} = - \frac{r_{1}}{r_{2}} q

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q^{'} = \frac{r_{1}}{r_{2}} q

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Solution

The correct option is C $q^{'} = - \frac{r_{1}}{r_{2}} q$
Potential at the surface of inner sphere = 0 (grounded)
$\frac{1}{4 π ϵ_{0}} \frac{q}{r_{1}} + \frac{1}{4 π ϵ_{0}} \frac{q^{'}}{r_{2}} = 0$
$\Rightarrow q^{'} = \frac{- q r_{2}}{r_{1}}$

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Charge on the outer sphere is $q$ and the inner sphere is grounded. Then the charge $q^{'}$ on the inner sphere is, for $(r_{2} > r_{1})$

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