1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric Field and Potential Difference in a Coaxial Cable

  1. Jan 8, 2015 #1
    1. The problem statement, all variables and given/known data:

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

    A long straight co-axial cable of length 1 consists of an inner conductor of radius r1 and a thin outer conductor or radius r2. The dielectric between the conductors has a relative permittivity εr.

    1. (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.

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

    Gauss' Law ∫E.ndA = Q/εr


    3. The attempt at a solution:

    Part a)

    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/εr ⇒ E(2πr1l)=Q/εr ⇒ (l=1) ⇒ E = Q/2πr1εr

    For the outer cable:

    Used same method as before but when taking into account the normal I got the same answer but with r2 in place of r1

    Then I summed these together to get:


    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 r1 and r2. Any help would be hugely appreciated as i've been going round in circles with this for a while now.
  2. jcsd
  3. Jan 8, 2015 #2


    User Avatar
    Homework Helper
    Gold Member

    Hello 12x4,

    Welcome to PF! :)

    Just a note on your version of the equation: In my experience, it is conventional to express the overall permittivity [itex] \varepsilon [/itex] as
    [tex] \varepsilon = \varepsilon_0 \varepsilon_r [/tex]
    where [itex] \varepsilon_0 [/itex] is the permittivity of free space and where [itex] \varepsilon_r [/itex] is the relative permittivity. The relative permittivity is greater than or equal to 1: being 1 for a vacuum, and greater than 1 for dielectrics.*

    *[Edit: Actually, it can be more complicated than that. [itex] \varepsilon_r [/itex] can be a function of frequency, and might even take on complex values. But that's more than what we need to worry about here. For now just treat it is a real number, greater than or equal to 1.]

    I wanted to point this out, because your notation is different from the convention that I've found to be most common. In the convention that I'm used to, you'll still need to throw an [itex] \varepsilon_0 [/itex] in there somewhere [Edit: together with [itex] \varepsilon_r [/itex]]. I suggest checking the conventions of your textbook/coursework before continuing, to make sure we're all on the same page regarding the notation.

    Um, shouldn't that be [tex] \Delta V = -\int_a^b \vec E \cdot \vec {d \ell} ?[/tex]

    As in [itex] \Delta V = V_b - V_a [/itex]?

    I fear leaving the left hand side in differential form (i.e., "dV") is not the right way to put it. It should be "Delta" V. The left hand side of the equation is the potential difference ("difference" as in subtraction of [possibly] large quantities, not an infinitesimal change in potential).
    It's difficult to give a yes/no answer on this question. Technically the answer is 'yes', but probably not the way you are thinking. More on this below.

    I'll repeat this below, but remember, the 'Q' in Gauss' law is the charge enclosed within the Gaussian surface.
    The electric field in between the two conductors is not constant. By that I mean sure, the electric field is constant on the Gaussian surface for a given radius r, but it is not constant as r changes. Your expression for the electric field between [itex]r_1[/itex] and [itex] r_2 [/itex] should be a function of [itex] r [/itex].

    I'll stop you here on this part. Rethink the meaning of Gauss' law for a cylinder. If you removed the center conductor and all of its charge, what would the electric field be within a long, hollow, charged cylinder?

    You are correct that the electric field between [itex]r_1[/itex] and [itex] r_2 [/itex] should be a function of [itex] r [/itex]. Write a general expression for Gauss' law somewhere between [itex]r_1[/itex] and [itex] r_2 [/itex]. The expression should be a function of [itex]r[/itex], not [itex] r_1 [/itex].

    Once you have that, you can use your other equation in your "relevant equations" section to get the potential.
    Last edited: Jan 8, 2015
  4. Jan 9, 2015 #3
    Thank you! its good to finally join PF, I have been using it for the entire length of my degree but only signed up yesterday to actually post. Hopefully I will be in a position to help others as well after these exams. Both of the notation points you explained sound right to me, i fear that i may have got a little sloppy after a few days spending far too much time in the library.


    Okay so my understanding is that the outer cable will produce no electric field inside it as, if we were to ignore the inner cable, there would be no charge contained within a gaussian surface with a radius smaller than that of the outer cable. Therefore, the electric field within the outer cable will be due to the inner cable only.

    For the Inner cable:

    I think I have been thinking about the dA part of ∫E.ndA incorrectly. As i now understand it the dA refers to a segment on the gaussian surface and therefore would be in terms of r rather than r1 or r2. This leads me to work through the question like this.

    ∫E.ndA = Qεrε0

    E(2πrl) = Qεrε0

    E = Q/2πrεrε0


    ΔV = -Q/2πεrε0 ∫1/r⋅dr (going from r1 to r2)

    and gives me the result: ΔV = -Q/2πεrε0 * ln(r2/r1)

    Thank-you very much for your help and hopefully I have got it right this time. Could I ask you how you did your equations so nicely? I'm really struggling to make them look good.
  5. Jan 9, 2015 #4


    User Avatar
    Homework Helper
    Gold Member

    Your final answer looks okay to me. :) (Although I think you forgot to type a couple division signs in earlier equations. And I don't think the negative sign is required in your final answer. [On second though, maybe a ± sign would be appropriate.])

    The equations I wrote in my last post were in [itex] \LaTeX [/itex]. Here's a tutorial, part of an FAQ:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted