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Electric Field for a cylinder without using Gauss's law?

  1. Oct 18, 2011 #1
    I recently learned how to calculate the field from a cylinder (inside and outside the cylinder) using Gauss's law. I was wondering how I would be able to derive the same formula without using Gauss's law (just for practice). My idea is that you would need to integrate the electric field from cylindrical shells of a radius 0 to R (where R is the radius of the cylinder), and to find the electric field of a cylindrical shell, you would need to integrate the field from rings along the length of the cylinder. The field from a ring is easy to calculate along its axis, but how would you calculate it in a less symmetric situation? Or is there a better way of doing this?
  2. jcsd
  3. Oct 18, 2011 #2


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    How about integrating the fields from a collection of line charges arranged as a cylindrical shell?

    I haven't tried either approach myself, but I do know that the field of a ring at off-axis points is difficult to solve exactly. A line charge is easy, and its field is radial (no longitudinal or azimuthal components) which makes it easier (than for a ring) to integrate a bunch of them together.
  4. Oct 19, 2011 #3


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    The direct solution with the electrostatic Maxwell equations is usually simpler than using the integral form or the solution with the Green's function.

    In your case you just write down the Poisson equation (Heaviside-Lorentz units),

    [tex]\Delta \Phi=-\rho[/tex]

    in cylinder coordinates, assuming cylinder symmetry, i.e., you assume that [itex]\Phi[/itex] and [itex]\rho[/itex] are both functions of [itex]r[/itex] alone. Then you have an ordinary differential equation to solve,

    [tex]\frac{1}{r}[r \Phi'(r)]'=-\rho(r).[/tex]

    For you problem of a cylindrical shell, you have

    [tex]\rho(r)=\frac{\lambda}{2 \pi r} \delta(r-a),[/tex]

    where [itex]a[/itex] is the radius of the cylinder and [itex]\lambda[/itex] the charge per unit length on the surface.

    Now except for [itex]r=a[/itex] your charge density is 0. So we first solve the homogeneous equation

    [tex][r \Phi'(r)]'=0 \; \Rightarrow \; r \Phi'(r)=A \; \Rightarrow \; \Phi(r)=A \ln(r/a)+B.[/tex]

    Here, I've written the integration constant such that the argument of the logarithm becomes dimensionless as it must be.

    The above solution is valid for [itex]r<a[/itex] and [itex]r>a[/itex], but the integration constants have to be chosen differently in the two regions such as to solve the equation including the charge density. For [itex]r<a[/itex] we must have [itex]A=0[/itex], and then we can choose also [itex]B=0[/itex]. So we have

    [tex]\Phi(r)=0 \quad \text{for} \quad r<a.[/tex]

    To find the solution for [itex]\quad r>a[/itex] we must make sure that the potential is continuous at [itex]r=a[/itex], leading to [itex]B=0[/itex]. To find [itex]A[/itex], we integrate the equation,

    [tex][r \Phi'(r)]'=\frac{\lambda}{2 \pi} \delta(r-a)[/tex]

    over an infinitesimally small interval around [itex]r=a[/itex]. Using the solution for [itex]r<a[/itex] this gives

    [tex]A=\frac{\lambda}{2 \pi},[/tex]

    i.e., the solution reads

    [tex]\Phi(r)=\frac{\lambda}{2 \pi} \ln(r/a) \quad \text{for} \quad r>a.[/tex]

    The electric field is

    [tex]\vec{E}=\vec{e}_r \frac{\lambda}{2 \pi r} \Theta(r-a),[/tex]

    where [itex]\Theta[/itex] is the Heaviside-unitstep function.
  5. Oct 19, 2011 #4


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    I think OP is interested in the case where the charge density is not zero inside the cylinder.
  6. Oct 19, 2011 #5


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    Ah, I've misunderstood the question since he wrote about cylindrical shells.

    For a given cylinder-symmetric charge distribution it's also easily evaluated in the way of my previous posting. Let's take the most simple case of a homogeneous charge density inside the cylinder. In cylindrical coordinate it reads

    [tex]\rho(r)=\rho_0 \Theta(a-r).[/tex]

    Then you can simply integrate:

    [tex]\frac{1}{r} [r \Phi'(r)]'=-\rho(r).[/tex]

    For [itex]r<a[/itex] we have

    [tex][r \Phi'(r)]'=-r \rho_0, \; \Rightarrow \; \Phi'(r)=-\rho_0 \frac{r}{2}+\frac{A}{r} \; \Rightarrow \; \Phi(r)=-\rho_0 \frac{r^2}{4} + A \ln(r/a)+B.[/tex]

    Since the potential should not be singular at [itex]r=0[/itex] we must have [itex]A=0[/itex].

    For [itex]r>a[/itex] we can use the previous solution

    [tex]\Phi(r)=A' \ln(r/a)+B', \quad \Phi'(r)=\frac{A'}{r}.[/tex]

    To find the integration constants, we have to fulfill the boundary conditions that [itex]\Phi[/itex] and [itex]\Phi'[/itex] are continuous at [itex]r=a[/itex]. This leads to

    [tex]-\rho_0 \frac{a^2}{4}+B=B', \quad -\rho_0 \frac{a}{2}=\frac{A'}{a}.[/tex]


    [tex]B=\rho_0 \frac{a^2}{4}[/tex]

    leads to [itex]B'=0[/itex], and we finally have

    [tex]\Phi(r)=\frac{\rho_0}{4}(a^2-r^2) \Theta(a-r)-\frac{\rho_0 a^2}{2} \ln(r/a) \Theta(r-a).[/tex]

    The electric field reads

    [tex]\vec{E}=-\vec{\nabla} \Phi=\vec{e}_r \frac{\rho_0}{2} \left[r \Theta(a-r) + \frac{a^2}{r} \Theta(r-a) \right ].[/tex]
    Last edited: Oct 19, 2011
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