Derivation →
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| Capacitance of an Isolated Spherical Conductor |
Let us consider an isolated spherical conductor of radius $a$ is placed in a vacuum or air. Let a charge $+q$ be given to the sphere and this charge is distributed uniformly on the surface of conducting sphere. Then the electric potential at the surface of the conducting sphere is
$V=\frac{1}{4\pi \epsilon_{\circ}} \frac{q}{a} \qquad(1)$
So the capacitance of the sphere
$C=\frac{q}{V} \qquad(2)$
Now substitute the value of $V$ in equation $(2)$ Then we get
$C=\frac{q}{\frac{q}{4\pi \epsilon_{\circ}a}}$
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$C=4\pi \epsilon_{\circ}a$
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Thus, The capacitance of a spherical conductor is directly proportional to its radius. i.e If the radius of conducting sphere is large then the sphere will hold a large amount of the given charge without running up too high a voltage.
Unit:
The unit of capacitance is "farad" i.e. $F$
