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I am going through an EM textbook and on the chapter on Gauss's law for electric fields, it discusses the field around an infinitely long charged cylinder with a linear charge density, say $\lambda$. (EDIT1: uniform linear density)The author calculates the electric flux through a cylinder surface around the charge by assuming that the electric field is always normal to the surface.

I understand the intuitive reasoning behind this - it does indeed seem natural for the field to radiate around the cylinder - however I would like to see this assumption justified mathematically. I tried proving by myself that two points lying on a ring around the cylinder have a field with the same magnitude by integrating Coulomb's law, but I couldn't calculate the resulting integral, so I would be thankful if someone showed how to prove this (even if by using a method other than simply calculating the superposed electric field of the infinite charged cylinder).

EDIT1: I only just realized that I failed to mention that the cylinder is uniformly charged, which probably changes the interpretation of the question. I'm sorry.

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  • $\begingroup$ Are you sure that cylinder is charged with linear charge density $\lambda$ ? $\endgroup$
    – user240696
    Commented Jan 21, 2020 at 14:55
  • $\begingroup$ …" a linear charge density" - probably not. More likely a constant charge density on an infinite line. Is this actually a 2D problem? $\endgroup$
    – D. Halsey
    Commented Jan 21, 2020 at 15:45
  • $\begingroup$ Per the wording of the textbook, it does indeed say linear charge density $\lambda$. The problem is indeed set in 3D, with the cylinder (which is described as an electron beam) having an arbitrary radius $r$. $\endgroup$ Commented Jan 21, 2020 at 15:50
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    $\begingroup$ You don’t need to do the integral. Set it up in polar or cylindrical coordinates and then show the result does not depend on theta by substituting theta -> theta+c and getting the same integral. $\endgroup$ Commented Jan 21, 2020 at 17:07
  • $\begingroup$ Related: physics.stackexchange.com/q/441005/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Jan 21, 2020 at 17:54

3 Answers 3

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The conceptually simplest mathematical proof I know uses a Coulomb's law integral over the charge distribution. This requires some knowledge of multi-dimensional integrals and matrices. It isn't the most condensed proof, but I think it is thorough and requires the least advanced math.

If you have some charge distribution $\rho(\vec{x})$ then the electric field is

$$\vec{E}(\vec{x}) = \int \, d^3 \vec{x}' \, \rho(x') \frac{\vec{x}-\vec{x'}}{|\vec{x}-\vec{x'}|^3}$$

Now suppose that the charge distribution has some symmetry that is a rotation or a reflection or a translation. That means there is a matrix $R$ and vector $\vec{b}$ that acts on $\vec{x}$ such that $\rho(R\vec{x}+\vec{b}) = \rho(\vec{x})$. Then we can use the integral to calculate the vector field at a transformed point:

$$\vec{E}(R\vec{x}+\vec{b}) = \int \, d^3 \vec{x}' \, \rho(x') \frac{R\vec{x}+\vec{b}-\vec{x'}}{|R\vec{x}+\vec{b}-\vec{x'}|^3}$$

We can change the variable of integration to $\vec{x}' = R\vec{y}+\vec{b}$. So

$$\vec{E}(R\vec{x}) = \int \, d^3 [Ry+b] \, \rho(Ry+\vec{b}) \frac{R(\vec{x}-\vec{y})}{|R(\vec{x}-y)|^3}$$

The charge distribution is symmetrical under $R$ so $\rho(Ry+b) = \rho(y)$. I already used the fact that rotations and reflections are linear in the line above, and that the translation cancels out of the vector difference. A reflection or rotation doesn't change the length of vectors so $|R(\vec{x}-\vec{y})| = |\vec{x}-\vec{y}|$. Since it's a linear transformation $d^3 [Ry+b] = |\det(R)| d^3 y$. Putting in all those relations and simplifying we get that

$$\vec{E}(R\vec{x}+\vec{b}) = |\det{R}|\cdot R \int \, d^3y \, \rho(y) \frac{\vec{x}-\vec{y}}{|\vec{x}-y|^3}$$ $$\vec{E}(R\vec{x}+\vec{b}) = |\det{R}|\cdot (R\vec{E}(\vec{x}))$$

So if the charge distribution is symmetrical under a rotation or reflection so is the electric field in the sense captured by the equation above. Rotations and reflections both have $|\det R| = 1$.

The equation above applies to any charge distribution with any set of rotation/reflection/translation symmetries. The rest of the proof uses the symmetries of the specific case to constrain the form of the electric field.

The infinite uniform cylinder has four symmetries: translation along z, reflection in z, reflection in x (or y), and rotation around z. We can go through each symmetry to further constrain the possible form of the electric field.

Translation: Using $\vec{b} = (0,0,z)$ we get $\vec{E}(x,y,z) = \vec{E}(x,y,0)$, so the entire z dependence of $\vec{E}$ is determined by just the $z=0$ plane.

Reflection in z: Using $R = diag(1,1,-1)$ we get $E_z(x,y,0) = -E_z(x,y,0) \Rightarrow E_z(x,y,0) = 0$, so the electric field has zero z-component everywhere.

Rotation around z: Here it helps to use cylindrical coordinates because they don't interchange under rotation. Using $R$ as the matrix for rotation by $\phi$ around the z axis we get $E_{r,\phi}(r,\phi,z) = E_{r,\phi}(r,0,0)$, so the electric field at any point with cylindrical radius $r$ is by the electric field at the point $(r,0,0)$.

Reflection in y: At the point $(r,0,0)$ a reflection the y axis sends $(r,0,0)$ to $(r,0,0)$. It flips the $\phi$ component of the electric field but not the $r$ component, so $E_\phi (r,0,0) = -E_\phi(r,0,0)\Rightarrow E_\phi (r,0,0) = 0$.

The only remaining component that hasn't been determined by the four symmetries is $E_r(r,0,0)$. We can then use this function and the symmetries to find out the electric field at any other point. Ultimately we have determined that the electric field points radially with no $\phi$ component (reflection in y), no $z$ component (reflection and translation in z), and is rotationally symmetric (rotation around z).

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  • $\begingroup$ I have only limited knowledge about vector calculus, so I cannot verify the formulae, but I think this correctly answers my question. You worded it differently, but basically proved the same thing with the symmetries. Thanks! $\endgroup$ Commented Jan 23, 2020 at 12:32
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It's not true in general that the electric field for this charge density has cylindrical symmetry. If you calculate a field for this charge distribution, and it satisfies Gauss's law, then you can always, for example, add a uniform field onto that. There are uniqueness and existence theorems for electrostatics, but they require boundary conditions.

As a simpler example, suppose you're using Gauss's law to find the field of a point charge. The usual Coulomb field is a solution, and that field has the same spherical symmetry as the charge distribution. But you can also add any homogeneous solution, such as a uniform field. However, if you impose a boundary condition that $E\rightarrow0$ at infinity, then you have a unique solution through one of the standard existence/uniqueness theorems for the Poisson equation. If you were looking for a solution and didn't know the Coulomb field, it would be reasonable to look for a solution that had spherical symmetry. You would wonder if that was going to succeed. You'd try it, and it would. Then by imposing the boundary conditions you'd get uniqueness.

Returning to the case of the field of the cylinder, I would imagine that the solution is unique if you impose $E\rightarrow0$ at infinite distance from the cylinder. However, I don't know of any existence-uniqueness theorem that applies for this type of boundary condition.

I tried proving by myself that two points lying on a ring around the cylinder have a field with the same magnitude by integrating Coulomb's law,

There is no physical principle that says an electrostatic field is the same as the result given by integrating Coulomb's law. A counterexample is the one I have above, in which we add a uniform field to an inhomogeneous solution.

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  • $\begingroup$ I do not have the background to fully understand what you're saying, however assuming that by adding a uniform field you mean putting the cylinder, say, between two charged plates, i.e. in the middle of a capacitor, then that is out of the scope of what I asked. I apologize if my wording was unclear. My question is strictly about a "lone" charged cylinder. $\endgroup$ Commented Jan 23, 2020 at 12:26
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Suppose you are looking at the cylinder that is oriented horizontally and suppose for argument's sake that the electric field is pointing down and to the right instead of normal to the surface. Imagine attaching a rod to the cylinder showing the direction of the electric field.

Now imagine standing on your head and what do you see? The cylinder is exactly the same as before and should have the same electric field but the rod from before is now pointing up and to the left. That contradiction implies that the field can't point down and to the right. That kind of reasoning leads you to the fact that the field can only point normal to the surface.

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    $\begingroup$ This is wrong. As a counterexample, take the cylindrically symmetric field and add a uniform field to it. $\endgroup$
    – user4552
    Commented Jan 21, 2020 at 18:51
  • $\begingroup$ What do you mean by adding a uniform field to it? From external sources? $\endgroup$ Commented Jan 22, 2020 at 1:04

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