A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Figure). Show that the capacitance of a spherical capacitor is given by
$C=\frac{4 \pi \varepsilon_{0} r_{1} r_{2}}{r_{1}-r_{2}}$
where $r_{1}$ and $r_{2}$ are the radii of outer and inner spheres, respectively.
A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Figure). Show that the capacitance of a spherical capacitor is given by
$C=\frac{4 \pi \varepsilon_{0} r_{1} r_{2}}{r_{1}-r_{2}}$
where $r_{1}$ and $r_{2}$ are the radii of outer and inner spheres, respectively.
Radius of the outer shell $=r_{1},$
radius of the inner shell $=r_{2}$
The inner surface of the outer shell has charge $+Q$.
The outer surface of the inner shell has induced charge $-Q$
Potential difference between the two shells is given by,
$V=\frac{Q}{4 \pi \epsilon_{0} r_{2}}-\frac{Q}{4 \pi \epsilon_{0} r_{1}}$
Where, $\epsilon_{0}=$ Permittivity of free space
$V =\frac{Q}{4 \pi \epsilon_{0}}\left[\frac{1}{r_{2}}-\frac{1}{r_{1}}\right]$
$=\frac{Q\left(r_{1}-r_{2}\right)}{4 \pi \epsilon_{0} r_{1} r_{2}}$
Capacitance of the given system is given by,
$C=\frac{\text { Charge }(Q)}{\text { Potential difference }(V)}$
$=\frac{4 \pi \epsilon_{0} r_{1} r_{2}}{r_{1}-r_{2}}$
Hence, proved.
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- [IIT 2001]
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