Coaxial cable problem, electric field strength?

[roccofitz]

Homework Statement

A coaxial cable has an inner core diameter 6mm and inner diameter of the outer sheath is 20mm. There are two layers of insulation between the conductors, the inner layer from a radius of 3mm to 9mm has relative permittivity of 8 while the outer layer from radius 9mm to 10mm has a relative permittivity of 1.5. The voltage between the conductors is 1.5kV
a. Plot the electric field strength in the insulation as a function of radius
b. Determine the max field strength in both insulators.

Homework Equations

Permittivity equation: epsilon= epsilon_0*epsilon_r

Electric field strength: E= q/4*pi*epsilon*r^2

The Attempt at a Solution

Not sure how to approach this question.

 

[gneill]

Not sure how to approach this question.

Use Gauss' Law to find the form of the field for an assumed length of coax. Write the expressions for the field with respect to radial distance (note that you'll need separate expressions for the two dielectric domains). Then determine the expression for the potential difference (voltage) between the conductors for a given charge.

Next determine the capacitance of the device as C = charge/voltage , where the voltage is given by the potential difference. You should be able to use this to find an expression in terms of Voltage to replace Q/L in the field strength equations. Rewrite the field strength equations accordingly. You can now plug in your given values and plot the field strength vs radius.

Good Luck!

 

[roccofitz]

Ok wondering if this is right or even heading in the right direction.

E=Electric field strength

E= q/2*pi*epslion*L

For the two different dielectric domains, 1: r=3mm epsilon=7.08*10^-11 2: r=9mm epsilon=1.33*10^-11

V=Potential difference

V=(-q/2*pi*epsilon*L)(ln10/3)

Capacitance= q\v = (2*pi*epsilon*L)/(ln10/3)

 

[gneill]

Ok wondering if this is right or even heading in the right direction.

E=Electric field strength

E= q/2*pi*epslion*L

E(r) = q/(2*pi*k*epsilon0*L*r)
for relative permittivity k. so to cover the whole radius,

E(r) = q/(2*pi*k1*epsilon0*L*r) for a <= r < b
E(r) = q/(2*pi*k2*epsilon0*L*r) for b <= r < c

For the two different dielectric domains, 1: r=3mm epsilon=7.08*10^-11 2: r=9mm epsilon=1.33*10^-11

Okay, you can use two different epsilons, too. Just remember that the equation describes the field at a given radial distance, r. You had left out the r in the denominatior.

V=Potential difference

V=(-q/2*pi*epsilon*L)(ln10/3)

Capacitance= q\v = (2*pi*epsilon*L)/(ln10/3)

There are two different regions to integrate over in order to get the potential difference. What value of epsilon were you planning to plug into your voltage equation to get the overall effect?

 

[roccofitz]

Ok so I am guessing that the V equation isn't the right equation then because there are two different dielectric regions. Or is there a way of calculating an overall dielectric constant??

 

[roccofitz]

Or do you integrate it between radius 3 and 9 with dielectric constant 8 and then integrate it between 9 and 10 with dielectric 1.5?? then how do these two equations combine?? thanks for all the help by the way

 

[gneill]

Or do you integrate it between radius 3 and 9 with dielectric constant 8 and then integrate it between 9 and 10 with dielectric 1.5?? then how do these two equations combine?? thanks for all the help by the way

Yes. They sum. (An integral is a sum, after all)