Capacitance Of A Spherical Conductor

Posted on By
Spread the love

Capacitance Of A Spherical Conductor

 

Capacitance Of A Spherical Conductor :- When an isolated spherical conductor (solid or hollow) is given a charge, this charge is distributed uniformly over the outer surface of the conductor. The electric potential will be the same at every point on the surface of the spherical conductor. Let’s assume a spherical conductor of radius R  is given a charge Q. The electric potential at every point on the surface of the sphere is :

\displaystyle V=\frac{Q}{4\pi {{\varepsilon }_{0}}R}

Capacitance,

\displaystyle C=\frac{Q}{V}

\displaystyle C=\frac{Q}{\frac{Q}{4\pi {{\varepsilon }_{0}}R}}

\displaystyle \Rightarrow C=4\pi {{\varepsilon }_{0}}R     …..(1)

Equation (1) is the required formula for the capacitance of a spherical conductor. Here, \displaystyle C\propto R, that is, the larger the radius of the sphere, the greater will be the capacitance of the spherical conductor.

Note that a capacitor generally consists of two conductors. Here, by assuming the second conductor to be at infinity, we have derived the formula for the capacitance of an isolated spherical conductor.

Earth’s Capacitance

(Capacitance Of A Spherical Conductor)

If we consider the Earth to be a spherical conductor of radius 6400 km, then the capacitance of the Earth,

\displaystyle C=4\pi {{\varepsilon }_{0}}R=\left( \frac{1}{9\times {{10}^{9}}} \right)\times \left( 6.4\times {{10}^{6}} \right)

\displaystyle \Rightarrow C=711\mu F

 

Example 1.

If the radius of an isolated spherical conductor is doubled, how many times will its capacitance become compared to its previous value ?

Solution :

The capacitance of a spherical conductor is proportional to its radius, hence

\displaystyle \frac{{{C}_{2}}}{{{C}_{1}}}=\frac{{{R}_{2}}}{{{R}_{1}}}

As R2 = 2R1 , therefore

\displaystyle \frac{{{C}_{2}}}{{{C}_{1}}}=\frac{\text{2}{{R}_{1}}}{{{R}_{1}}}

\displaystyle \Rightarrow {{C}_{2}}=2{{C}_{1}}

Example 2.

If a spherical conductor has a diameter of 1 m, then find its capacitance.

Solution :

\displaystyle C=4\pi {{\varepsilon }_{0}}R=\left( \frac{1}{9\times {{10}^{9}}} \right)\times \left( \frac{1}{2} \right)

\displaystyle \Rightarrow C=\text{55}\text{.5}\times {{10}^{-12}}=\text{55}\text{.5}pF

 

Class 12th NCERT
error: Content is protected !!