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## Homework Statement

:[/B]

Co-axial cable, relative permittivity, capacitance, internal energy

A long straight co-axial cable of length 1 consists of an inner conductor of radius r

_{1}and a thin outer conductor or radius r

_{2}. The dielectric between the conductors has a relative permittivity ε

_{r}.

- (a) Find the strength of the electric field E(r) between the conductors (r1 < r < r2) for equal and opposite charges ±Q on the conductors.

- (b) Calculate the total potential difference V between the conductors if they were to carry equal and opposite charges ±Q.

## Homework Equations

:[/B]Gauss' Law ∫E.ndA = Q/ε

_{r}

dV=-∫E.dl

## The Attempt at a Solution

:Part a)[/B]

I'm not sure if i'm being really stupid here but are you able to work this question out like this? first find the electric field produced at the inner and outer surface of the outer and inner conductors respectively and then sum them together?:

for the inner cable:

∫E.ndA = Q/ε

_{r}⇒ E(2πr

_{1}l)=Q/ε

_{r}⇒ (l=1) ⇒ E = Q/2πr

_{1}ε

_{r}

For the outer cable:

Used same method as before but when taking into account the normal I got the same answer but with r

_{2}in place of r

_{1}

_{1}

E=2πr

_{2}ε

_{r}

Then I summed these together to get:

E=(Q/2πε

_{r})(1/r

_{1}+1/r

_{2})

I thought this was correct but i am not doubting myself as I am struggling to do part b).

**Part B)**

To complete this part of the question surely i need my expression for a) to be in terms of an arbitrary value r rather than r

_{1}and r

_{2}. Any help would be hugely appreciated as i've been going round in circles with this for a while now.

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