- A single core cable is shown in the Figure.
- It is equivalent to two long co – axial cylindrical. The core of the conductor is inner cylinder whereas the lead sheath is outer cylinder.
- The lead sheath is at earth potential.
- Let us consider that the core diameter is d meter and inner sheath diameter is D meter.
- The charge per axial length of the cable is Q coulombs and e is permittivity of the insulation material between core and lead sheath.

Core diameter = d

Sheath diameter = D

Charge per axial length of cable = Q

Permittivity of insulating material between core and
lead sheath = ε

Permittivity ε = ε_{0}ε_{r}

Where ε_{0} = Absolute permittivity

=
8.854 × 10 ^{– 12} Farad / meter

ε_{r}
= Relative permittivity

- Consider a cylinder of radius x meter and axial length
1 meter. The surface area of cylinder = ( 2πx )
× ( 1 ) = 2πx meter
^{2} - Electric
^{ }flux density at any point P on the cylindrical surface is

D_{x} = Q / 2πx Coulomb / meter^{2}

Electric Intensity at point P

E_{x} = D_{x} / ε

= Q / 2πx ε

= Q / 2πx ε_{0}ε_{r}

- If unit positive charge moves from point P through
distance dx in the direction of electrical field, the work done is E
_{x}dx. - Therefore the work done for unit positive charge moves from conductor to sheath is given by

V = ( Q / 2π ε_{0}ε_{r} ) Log_{e}
( D / d )

**Capacitance of cable**

C = Q / V

= Q / { ( Q Log_{e} ( D / d ) / 2π ε_{0}ε_{r}
) }

= ( 2π ε_{0}ε_{r} ) / Log_{e} ( D / d )

= ( 2π × 8.854 × 10 ^{– 14} × ε_{r} )
/ Log_{e} ( D / d )

= ( ε_{r}
× 10 ^{– 9} ) / 41.4 Log_{e} ( D / d ) Farad / meter

If the length of cable is L meter, the capacitance of
cable is

** C = ( ε _{r }L
× 10 ^{– 9} ) / 41.4 Log_{e} ( D / d ) Farad **

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