Derivation of Spherical Capacitor

A spherical capacitor consists of two concentric conducting spherical shells. Let the radius of the inner sphere be a and the radius of the outer sphere be b.

Let charge +Q be given to the inner sphere and charge -Q to the outer sphere.

For a point at distance r from the centre, where:

a < r < b

Using Gauss’s law:

E × 4πr² = Q / ε₀

So,

E = Q / (4πε₀r²)

Potential difference between the two spheres is:

V = ∫[a to b] E dr

V = ∫[a to b] Q / (4πε₀r²) dr

V = Q / (4πε₀) ∫[a to b] r⁻² dr

V = Q / (4πε₀) [-1/r] from a to b

V = Q / (4πε₀) (1/a - 1/b)

Capacitance is:

C = Q / V

C = Q / { Q / (4πε₀) (1/a - 1/b) }

C = 4πε₀ / (1/a - 1/b)

C = 4πε₀ab / (b - a)

Therefore,

C = 4πε₀ab / (b - a)

If a dielectric medium of relative permittivity K is present between the spheres:

C = 4πε₀K ab / (b - a)

Derivation of Concentric Spherical Capacitor

A concentric spherical capacitor is the same arrangement of two spherical conducting shells having a common centre.

Let:

Inner sphere radius = R₁

Outer sphere radius = R₂

Charge on inner sphere = +Q

Charge on outer sphere = -Q

For any point between the two spheres:

R₁ < r < R₂

By Gauss’s law:

E × 4πr² = Q / ε₀

Therefore,

E = Q / (4πε₀r²)

Potential difference between inner and outer sphere:

V = ∫[R₁ to R₂] E dr

V = ∫[R₁ to R₂] Q / (4πε₀r²) dr

V = Q / (4πε₀) ∫[R₁ to R₂] r⁻² dr

V = Q / (4πε₀) [-1/r] from R₁ to R₂

V = Q / (4πε₀) (1/R₁ - 1/R₂)

Now,

C = Q / V

C = Q / { Q / (4πε₀) (1/R₁ - 1/R₂) }

C = 4πε₀ / (1/R₁ - 1/R₂)

C = 4πε₀R₁R₂ / (R₂ - R₁)

Therefore, capacitance of concentric spherical capacitor is:

C = 4πε₀R₁R₂ / (R₂ - R₁)

Special Case: Isolated Spherical Conductor

If the outer sphere is taken very far away, then:

R₂ = ∞

Using:

C = 4πε₀R₁R₂ / (R₂ - R₁)

or

C = 4πε₀ / (1/R₁ - 1/R₂)

Since:

1/R₂ = 0

So,

C = 4πε₀ / (1/R₁)

Therefore,

C = 4πε₀R₁

This is the capacitance of an isolated spherical conductor.

Derivation of Cylindrical Capacitor

A cylindrical capacitor consists of two long coaxial conducting cylinders.

Let:

Radius of inner cylinder = a

Radius of outer cylinder = b

Length of each cylinder = L

Charge on inner cylinder = +Q

Charge on outer cylinder = -Q

Linear charge density is:

λ = Q / L

For a point at distance r from the axis:

a < r < b

Using Gauss’s law, choose a cylindrical Gaussian surface of radius r and length L.

Electric flux:

Φ = E × 2πrL

By Gauss’s law:

E × 2πrL = Q / ε₀

Since Q = λL,

E × 2πrL = λL / ε₀

So,

E = λ / (2πε₀r)

Potential difference between inner and outer cylinder:

V = ∫[a to b] E dr

V = ∫[a to b] λ / (2πε₀r) dr

V = λ / (2πε₀) ∫[a to b] dr/r

V = λ / (2πε₀) ln(b/a)

Now capacitance:

C = Q / V

Since Q = λL,

C = λL / { λ / (2πε₀) ln(b/a) }

C = 2πε₀L / ln(b/a)

Therefore,

C = 2πε₀L / ln(b/a)

If a dielectric medium of relative permittivity K is filled between the cylinders:

C = 2πε₀K L / ln(b/a)

Capacitance per unit length is:

C/L = 2πε₀ / ln(b/a)

With dielectric:

C/L = 2πε₀K / ln(b/a)