If a tube has an electrostatic charge, is the charge uniformly distributed across its body or do the charge distributions vary from inside walls and the head edges? If so, how is the charge distributed?

I appreciate if you can name some references elaborating the charge distribution over objects with geometrical (or even physical) non-uniformity.

  • $\begingroup$ At least if the material is conductive (is it assumed @Kama ?) the charge distribution must be uniformly distributed in all geometrically symmetric directions. $\endgroup$
    – Communisty
    Jul 3, 2017 at 12:01
  • $\begingroup$ @Communisty my question is exactly about when there is no geometrical symmetry. Comparing the sharp edges with the tube surface. $\endgroup$
    – Kama
    Jul 3, 2017 at 12:43
  • $\begingroup$ Yeah I got it @Kama. My intuition would say that whenever the material is conductive the charge distribution is uniform. But don't take my word for it, I'm pretty weak in these kind of things. $\endgroup$
    – Communisty
    Jul 3, 2017 at 12:48

2 Answers 2


As pointed out in sammy gerbil's answer, the charge density is quite large near the corners of a conducting surface. If we look near the edges of the cylinder, at length scales much less than the cylinder's radius, the corner will "look like" two planes meeting with an "interior angle" of $3 \pi/2$. It is a general result (see, e.g., Section 2.11 of Jackson) that the charge density at a location where two conducting planes meet with an interior angle of $\beta$ is $$ \sigma(\rho) \propto \rho^{(\pi/\beta) - 1}, $$ where $\rho$ is the distance from the edge. In particular, this implies that the charge density near the edges of the cylinder will diverge: $$ \sigma \propto \rho^{-1/3}. $$

In reality, the charge density is only divergent to the extent that we have the faces meeting at an infinitely sharp edge. Realistic cylinders will have some slight rounding to their edges; and if nothing else, we cannot think of the conductor as a continuum when we're at scales smaller than the atomic spacing of the metal in question. Still, it can be deduced that the charge density will in fact be very large near the edges of a charged conducting cylinder.

(Aside #1: The case of a conical point is also treated in Jackson; see Section 3.4. However, there isn't a nice closed-form expression in this case; rather the solution is expressed in terms of the zeroes of Legendre functions $P_\nu(x)$ with non-integer $\nu$.)

(Aside #2: Although the charge density is infinite along an infinitely sharp edge, the actual amount of charge "on the edge" is negligible. Specifically, the amount of charge within a distance $\delta$ of such an edge will be $$ Q_\delta = 2 \int_0^\delta \rho^{-1/3} d\rho \propto \delta^{2/3}, $$ which goes to 0 as $\delta \to 0$.)


Excess charge is distrubuted on the surface of a conductor. It is distributed symmetrically but not uniformly (except for spherical symmetry), with higher concentration at sharp edges or points. If the cylinder is hollow with open ends then some charge resides on the inner surface. There is no charge inside a closed cavity.

In general, for finite conductors other than spheres it is very difficult to calculate the charge distribution, even when there is a high degree of symmetry. A few special cases have been solved analytically. For example :

  1. Thin conducting disk. The method is to treat the disk as the limiting case of a squashed ellipsoid. Unrealistically, the result is that surface charge density diverges to infinity at the edge of the disk $(R=r)$. A related problem, solved by the same method, is :

  2. Long thin needle. The case of a finite number of discrete charges on a needle is solved in Distribution of point charges on a line of finite length).

  3. Two non-concentric conducting spheres can be solved using the Method of Image Charges. The same method can be used to find the surface charge density on an infinite conducting plane close to a point charge.

  4. Another classic problem is the charge distribution on an infinite conducting cone whose apex is close to an infinite conducting plane. This is discussed by K Battacharya in section 5 of On the Dependence of Charge Density on Surface Curvature of an Isolated Conductor.

  5. Conducting sphere in uniform electrostatic field

Other examples, such as an isolated finite rectangular plate, usually require numerical methods. This involves solving the Laplace Equation for the electrical potential subject to the geometrical boundary conditions, then differentiating to find the electric field $E$ close the the surface and hence the surface charge density $\sigma=\epsilon_0 E$.

Related questions :

Charge distribution on conductors? - why it is not uniform.
The charge surface distribution of a conductor with a non centered charge - spherical conductor with non-concentric cavity.
Why is electric field strong at sharp edges?
Simulate / plot electrostatic field
First- and last-passage Monte Carlo algorithms for the charge density distribution on a conducting surface


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