Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm in doubt in the application of Gauss' Law to find electric fields when the charge distribution is symmetric. Well, first of all: I know how to find the magnitude of the field - we just enclose the charge distribution with a gaussian surface on which the electric field will not change it's magnitude, and then using Gauss's Law we can write it in terms of the total charge inside and the area of the gaussian surface.

My problem is: how do I find the direction of the field? I mean, in a spherical symmetric distribution it's easy, because we know what's the vector that points radially outwards (it's simply one of the unit vectors from spherical coordinates). But what about a cylindrical symmetric distribution ? Would I need to use the unit position vector of cylindric coordinates ?

In the general case I would need to switch to more appropriate coordinates to write the field ? Is there a general way of treating this ?

Sorry if this question is to silly or too basic. I'm just trying to understand how to use properly this law.

share|cite|improve this question
Use it in its proper, vector form, that should give you the vector field, not just the magnitude! – Schlomo Mar 21 '13 at 20:54

In cylindrical coordinates, you can choose the unit vector field that points radially outward from and perpendicular to the axis of symmetry of the distribution.

Generally, use a unit vector field that respects the symmetry of the problem. In other words, if the distribution possesses a certain symmetry, then the unit vector field you choose should as well. In the cylindrical example, the unit vector field needs to be both rotationally symmetric (invariant) around the axis of symmetry, and it needs to be symmetric in translations along the axis.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.