Practicing Success
Practicing Success
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The spheres shown in the figure are connected by a conductor. The capacitance of the system is
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$4 \pi \varepsilon_0 C_1 \frac{a b}{b-a}$ $4 \pi \varepsilon_0 a$ $4 \pi \varepsilon_0 b$ $4 \pi \varepsilon_0 \frac{a^2}{b-a}$ |
$4 \pi \varepsilon_0 b$ |
As there will be no charge on the inner sphere, therefore the capacitance only will exist due to outer sphere. Hence the capacitance of the system is the capacitance due to outer sphere of radius b, therefore C = $4 \pi \varepsilon_0 b$. ∴ (C) is correct |
