Cable capacitance is the formation of capacitance between the charged core or conductor and the earthed lead sheath. Therefore it is quite important to have an idea of how the cable capacitance varies or the factors affecting its value.

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A single core cable is similar to two long co-axial cylinders having dielectric in between the space. In such model of cable, the core or conductor is the inner cylinder and the lead sheath is the outer cylinder. Let us consider a single core cable having core diameter d and inner sheath diameter D as shown in figure below.

Let the charge per unit axial length of the conductor is Q columb and the permitivity of the insulating material be ξ. Obviously, ξ = ξ

Let us again consider a cylinder of radius x as shown in figure above and apply Gauss Law to find the value of Electric Filed Intensity vector at point P.

Electric Field Intensity E at point P = Q / 2πξx ...................(1)

But to find the capacitance first of all, we need to find the potential difference between the conductor and the earthed sheath. Well, we know that potential at a point is work done in bringing a unit test charge from infinity to that point. Therefore the potential difference between the conductor and the sheath shall be equal to the work done in bringing the unit test charge from sheath to the conductor. As the electric filed at any point x in between is E, therefore work done in moving a unit test charge by some infinitesimally small distance dx = -Edx. Notice the negative sign. As the direction of E and dx are opposite therefore work done W = EdxCos(180°) = -Edx

Hence,

Potential Difference V = -∫Edx

= -∫ (Q / 2πξx) dx ..........[From equation (1)]

Integrating from D/2 (lower limit) to d/2(upper limit of integration), we get

V = (Q / 2πξ)log(D/d)

Wow..we have calculated the potential difference between the core and earthed sheath but cable capacitance,

C = Q / V

Therefore,

C = Q / (Q / 2πξ)log(D/d)

= 2πξ / log(D/d)

= 2π ξ

But the above capacitance is for cable of unit length. Cable Capacitance for axial length L is

C = 2πL ξ

From the above expression, we observe that cable capacitance depends on the diameter of conductor as well as earthed sheath.

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**Calculation of Cable Capacitance**

Let the charge per unit axial length of the conductor is Q columb and the permitivity of the insulating material be ξ. Obviously, ξ = ξ

_{0}ξ_{r}where ξ_{0 }is the permitivity of free space and ξ_{r }is the relative permitivity.Let us again consider a cylinder of radius x as shown in figure above and apply Gauss Law to find the value of Electric Filed Intensity vector at point P.

Electric Field Intensity E at point P = Q / 2πξx ...................(1)

But to find the capacitance first of all, we need to find the potential difference between the conductor and the earthed sheath. Well, we know that potential at a point is work done in bringing a unit test charge from infinity to that point. Therefore the potential difference between the conductor and the sheath shall be equal to the work done in bringing the unit test charge from sheath to the conductor. As the electric filed at any point x in between is E, therefore work done in moving a unit test charge by some infinitesimally small distance dx = -Edx. Notice the negative sign. As the direction of E and dx are opposite therefore work done W = EdxCos(180°) = -Edx

Hence,

Potential Difference V = -∫Edx

= -∫ (Q / 2πξx) dx ..........[From equation (1)]

Integrating from D/2 (lower limit) to d/2(upper limit of integration), we get

V = (Q / 2πξ)log(D/d)

Wow..we have calculated the potential difference between the core and earthed sheath but cable capacitance,

C = Q / V

Therefore,

C = Q / (Q / 2πξ)log(D/d)

= 2πξ / log(D/d)

= 2π ξ

_{0}ξ_{r}/ log(D/d) FaradBut the above capacitance is for cable of unit length. Cable Capacitance for axial length L is

C = 2πL ξ

_{0}ξ_{r}/ log(D/d) FaradFrom the above expression, we observe that cable capacitance depends on the diameter of conductor as well as earthed sheath.

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