# Capacitance of a Co-axial cable

A Co-axial cable is a Transmission line, in which two conductors are placed co-axially and are separated by some dielectric material with dielectric constant (or) permittivity  ($\epsilon$ ).

a conductor is in the form of a cylinder with some radius, let the radius of inner conductor is ‘a’ meters and that of outer conductor be ‘b’ meters.

Now connect this co-axial conductor to a supply of ‘V’ volts , after applying ‘V’ assume positive charges are distributed on $M_{2}$ and negative charges on $M_{1}$ .

Now, a field is induced $\overrightarrow{E}$ between $M_{2}$  and $M_{1}$ because of flux lines, to find out $\overrightarrow{E}$ at any point P  between these two conductors

location of P is out of the conductor $M_{2}$ an inside the conductor $M_{1}$.

$\therefore&space;\overrightarrow{E}_{at&space;P}&space;=&space;\overrightarrow{E}_{\&space;due&space;\&space;to&space;\&space;inner&space;\&space;conductor&space;\&space;M_{1}}$ .

assume a cylindrical co-ordinate system $\rho&space;,\&space;\phi&space;,&space;\&space;z$  and axis of cable coincides with z-axis this is similar to a line charge distribution $\rho_{L}$ placed along the z-axis.

$\rho&space;_{L}=\frac{Q}{L}$ .

$\therefore&space;\overrightarrow{E}_{at&space;P}&space;=&space;\frac{\rho_&space;{L}}{2\pi&space;\epsilon&space;_{o}\rho&space;}\overrightarrow{a}_{\rho&space;}$ .

$\therefore&space;V&space;=&space;-\int_{1}^{2}\overrightarrow{E}.\overrightarrow{dl}$ .

$V&space;=&space;-\int_{1}^{2}&space;\frac{\rho_&space;{L}}{2\pi&space;\epsilon&space;_{o}\rho&space;}\overrightarrow{a}_{\rho&space;}.(d\rho&space;\overrightarrow{a}_{\rho&space;}+d\phi&space;\overrightarrow{a}_{\phi&space;}+dz&space;\overrightarrow{a}_{z&space;})$ .

$V&space;=&space;-\int_{b}^{a}&space;\frac{\rho_&space;{L}}{2\pi&space;\epsilon&space;_{o}\rho&space;}\overrightarrow{a}_{\rho&space;}.(d\rho&space;\overrightarrow{a}_{\rho&space;})$ .

$V&space;=&space;-&space;\frac{\rho_&space;{L}}{2\pi&space;\epsilon&space;_{o}&space;}(\ln&space;a-\ln&space;b)$ .

$V&space;=&space;\frac{\rho_&space;{L}}{2\pi&space;\epsilon&space;_{o}&space;}(\ln&space;b-\ln&space;a)$ .

$V&space;=&space;\frac{\rho_&space;{L}}{2\pi&space;\epsilon&space;_{o}&space;}\ln&space;(\frac{b}{a})$ .

$V&space;=&space;\frac{Q}{2\pi&space;\epsilon&space;_{o}L&space;}\ln&space;(\frac{b}{a})&space;\&space;\because&space;\&space;\rho&space;_{L}&space;=&space;\frac{Q}{L}$ .

$\therefore&space;C_{co-axial}&space;=\frac{Q}{V}&space;=&space;\frac{2\pi&space;\epsilon_{o}L}{\ln&space;(\frac{b}{a})}$ .

L- length of the conductors.

a- radius of the inner conductor.

$\epsilon$ – permittivity of the medium.

(No Ratings Yet)