Studying Physics Topics can lead to exciting new discoveries and technological advancements.
What is the Capacitance of a Spherical Conductor of Radius R?
Let us consider a spherical conductor of radius R charged with Q amount of charge. The charge Q is uniformly distributed over the surface of the sphere. Potential at the surface of the sphere, as we know, is the same as that produced by an isolated point charge Q placed at the centre of the sphere.
The potential of the sphere is given by,
V = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{R}\) [ε0 = permittivity of air or vacuum]
∴ Capacitance of the sphere,
C = \(\frac{Q}{V}\) = 4πε0R
We know, \(\frac{1}{4 \pi \epsilon_0}\) = 9 × 109 N ᐧ m2 ᐧ C-2
So, in vacuum or air, capacitance of the sphere,
C = \(\frac{R}{9 \times 10^9}\) farad
In CGS system, replacing ε0 by \(\frac{1}{4 \pi}\), we have, C = R .
Hence the capacitance in CGS unit of a spherical conductor placed in air (or vacuum) is numerically equal to its radius in centimetre. For this reason, the capacitance in CGS unit is some-times expressed in centimetre.
Unit of ε0 : From the relation C = 4πε0R , we have, ε0 = \(\frac{C}{4 \pi R}\)
So, the unit of ε0
\(=\frac{\text { unit of } C}{\text { unit of } R}\) = \(\frac{\mathrm{F}}{\mathrm{m}}\) = fraud/metre(F ᐧ m-1)
Using this simpler unit, we may write,
ε0 = 8.854 × 10-12 F ᐧ m-1
That this unit F ᐧ m-1 is identical to the unit C2 ᐧ N-1 ᐧ m-2 of ε0, used cartier, is shown here:
Numerical Examples
Example 1.
Radius of the earth is 6400 km. Determine its capacitance in µF. [HS ‘03]
Solution:
Radius of the earth = 6400 km = 6400 × 103 m.
Capacitance, C = 4πε0R = \(\frac{1}{9 \times 10^9}\) × 6400 × 103
= \(\frac{64}{9}\) × 10-4 = \(\frac{6400}{9}\) × 10-6 F
= 711.1 µF
Example 2.
A metal sphere has a diameter of 1 m. What will be the amount of charge required to raise its potential by 2.7 × 106 V?
Solution:
Radius, R = \(\frac{1}{2}\) = 0.5 m;
C = 4πε0R = \(\frac{1}{9 \times 10^9}\) × 0.5 = \(\frac{0.5}{9 \times 10^9}\)F
∴ The amount of charge required
Q = CV = \(\frac{0.5}{9 \times 10^9}\) × (2.7 × 106)
= 1.5 × 10-4 = 150 × 10-6C = 150 µC
Example 3.
Is it possible for a metal sphere of radius 1 cm to hold a charge of 1C?
Solution:
Radius, r = 1 cm = 0.01 m
So, capacitance of the sphere,
C = 4πε0r = \(\frac{1}{9 \times 10^9}\) × 0.01 = \(\frac{1}{9}\) × 10-11F
∴ Potential of the sphere,
V = \(\frac{Q}{C}\) = \(\frac{1}{\frac{1}{9} \times 10^{-11}}\) = 9 × 1011V
At this very high potential, the sphere will discharge in the surrounding air, i.e., it will not be able to hold the charge of 1 C.
Example 4.
The diameter of a spherical liquid drop is 2 mm and Its charge is 5 × 10-6 esu.
(i) What is the potential on its surface?
(ii) If two such liquid drops coalesce to form a bigger drop, what will be the potential on its surface?
Solution:
i) In CGS system, radius of the spherical conductor = its capacitance (numerically).
∴ Capacitance of the spherical liquid drop,
C = 0.1 statF [∵ Radius = 1 mm = 0.1 cm]
∴ Potential on the surface of the liquid drop,
V = \(\frac{Q}{C}\) = \(\frac{5 \times 10^{-6}}{0.1}\) = 5 × 10-5 statV
= 5 × 10-5 × 300 V = 0.015 V
ii) Let the radius of the bigger drop be R.
According to the question.
\(\frac{4}{3} \pi R^3\) = 2 × \(\frac{4}{3} \pi(0.1)^3\) or, R3 = 2 × (0.1)3
or, R = 0.1 × 21/3 = 0.1 × 1.26 = 0.126cm
Total charge, Q = 2 × 5 × 10-6 = 10-5 esu of charge.
∴ Potential on the surface of the bigger liquid drop,
V = \(\frac{Q}{C}\) = \(\frac{10^{-5}}{0.126}\) = 7.94 × 10-5 statV
= 7.94 × 10-5 × 300 V = 0.0238 V