I have been trying to find an equation (or some solution) of how to calculate the electric field strength (in N/C) of a conducting rectangular (nearly flat) plate which has non-zero potential to it, say the plate has a potential of 5V.

I have searched over the internet (and read Griffiths book on the method of images section), and it seems one would need to apply the method of images to solve the problem. However, since almost all examples regarding method of images is about "grounded" plates at 0V, it is very difficult for me to understand how to use method of images to solve for the electric field of a plate at 5V?

Also, I would like to mention that within the electric field of the 5V conducting plate, there is a small freely-moving positively charged particle in the vicinity (I assume because the charge on this small particle is the same as the charge on an electron/proton-- it can be regraded as a test charge and therefore, modification to the combined electric field between the conducting plate and the charge is not necessary?-please correct me here if I am wrong.)

  • $\begingroup$ For a rectangulat plate You have to give the ratio of edges, and of course the ratio of thickness rel to edges. And You should decide about that "nearly" flat. $\endgroup$
    – Georg
    Commented Jul 24, 2011 at 13:52

2 Answers 2


It sounds a bit like you're missing something from the problem description? 5V potential relative to what? It would make sense if you have the field between two conducting plates of different potential for example.

The "0V" they are talking about in the method of mirrors is really a way of saying that any inherent excess or deficit charge in the conductor vanishes (since you assume the plate has a connection to a 0V potential reservoir of vanishing resistance) hence you have a neutral overall charge distribution on the plate.

This allows you to consider the field-distribution when for example a point-charge with charge q is placed in the vicinity of the grounded plate; by the method of images the solution is equivalent to no plate but a charge of charge -q an equivalent distance behind the plate. By differentiating the potential field you can find the equivalent distribution of charge on the plate. Without reading the actual problem, the test charge if it has a specified charge really sounds like it has to be considered in the solution.

Simply saying that a plate has a 5V potential doesn't give you the information needed to calculate a charge distribution on it or a potential field strength. It doesn't even mean that the plate is non-neutral, because Volt is a relative unit. It would have made sense if there was an equivalent plate of, say, 0V in parallel. It might also be a red herring to compound the exercise, which might be a "standard" pointcharge vs. plate problem.

  • $\begingroup$ If this plate is connected to some battery and the other terminal of the battery is connected far away to infinity, then the plate will have a certain voltage V wrt infinity (ground). $\endgroup$
    – Rhea
    Commented Jul 24, 2011 at 13:39
  • $\begingroup$ I would also like to add that I do not have any information on the surface charge s on the plate, and hence, cannot use E=s/E0 (where E0 is permittivity of free space) to calculate the field strength. $\endgroup$
    – Rhea
    Commented Jul 24, 2011 at 13:50
  • $\begingroup$ @Rhea yes it will by that definition, but the electric field strength will be vanishing for that situation (since the strength of the E-field is the gradient of the potential which is varying infinitely slowly as you move from infinity to your plate). A vanishing field means vanishing surface charge density of the 5V plate by Gauss law, and hence no difference from the "grounded at 0V" case. $\endgroup$
    – BjornW
    Commented Jul 24, 2011 at 14:09
  • $\begingroup$ I think I may have found out what I am asking: use E=V/d. I know the voltage and depending on where the point (test) charge is, the distance of separation d can be known. This gives the electric field strength on the test charge at any distance from the plate. $\endgroup$
    – Rhea
    Commented Jul 24, 2011 at 14:40
  • $\begingroup$ That equation is just a simple special case for two infinitely long plates separated by a distance d and a voltage differential V. Since d equals infinity E will be vanishing (as I wrote above) and thus there will be no force on the test-charge coming from the plate potential. There will be a force on the test-charge coming from the charge-mirroring in the plate from the test-charge itself though. $\endgroup$
    – BjornW
    Commented Jul 24, 2011 at 16:01

The plate potential can always be made zero by adding a constant to the potential. It is traditional to make the metal surface grounded when solving problems of this type. Under this convention, the potential of infinity is then -5V. The plate will be positively charged, and you are asking for the electrostatic field of a positively charged infinitesimally thin metal plate of finite size.

This is not trivial to find, and the method of images is useless--- there is no other charge in the problem. If you introduce a test charge, the "image charge" method only works when the plate is infinitely large, or if you have a spherical metal. The image charge method is useless here

Thickness is irrelevant

First, you need to prove that there is a zero-thickness limit. This follows from the fact that a charged infinitesimal sheet has a finite electric field on the surface. If you have a finite thickness grounded plate of width $\epsilon$, and you take the limit of infinitesimal thickness, you only get $E\epsilon$ extra potential (the potential difference is the electric field times the distance).

Approximate description

The approximation to use to get a rough idea is the point-plate approximation. Assume that the radius of the plate is R, and assume that the charge on the plate is Q, and that the Q is uniformly distributed. The electric field near the surface is determined by the surface charge density and Gauss's law, and the surface charge density is $Q/\pi R^2$, the electric field is half this (using units where $\epsilon_0=1$). You assume that this holds until the radius R, and from R on out, the field becomes $Q/R^2$.

This is absurd, but it gets the asymptotic fields roughly right in the two regimes--- very close to the plate, and very far from the plate. The far solution gives a potential difference of $Q/4\pi R$ from infinity to R, and the near solution gives an additional $Q/2\pi R$ (the constant electric field, times the distance R). To the level of accuracy of this description, which is really not much better than order of magnitude, you find that

$$ V = {3Q\over 4\pi\epsilon_0 R} $$

And this gives you a rough estimate (restoring $\epsilon_0$). The result is somewhat less charge, about 30% of the amount, induced on the plate compared to a sphere of the same radius.

If you want a better answer, you need a solution of Laplace's equation for a plate. This can be gotten numerically in a few seconds using a modern computer, and it should be within a factor of 2 of the above rough estimate.


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