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I'm trying to solve an exercise that asks me the field strength of an uniform electric field where equipotential planes differ by $1\ \mathrm{V}$ and are $2.5\ \mathrm{cm}$ apart.

First of all, I tried to draw the situation so I have a sheet of charge with equally spaced field lines going through it parallel to the unit normal to the sheet's surface. Then I calculated the field by invoking gauss and I know that it isn't dependent on the distance from the source (the sheet of charge). It's constant and equal to $\frac{\sigma}{2\epsilon_0}$ with $\sigma$ the plane charge density. (1/2 because it's isolated, from $\text{flux}=2EA$ for a cylindrical Gaussian surface).

It's also possible to see that the field strength is constant only by knowing that the field lines are equally spaced of course.

My question is how can you say that you have equipotential planes that differ in potential while having an uniform field?

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  • $\begingroup$ uhmm It's constant but the sheet is not infinite, so from a big distance the field strength would be that of a point charge. But that doesn't explain equipotentials with different potentials in an uniform field either $\endgroup$ – Yuri Borges May 16 '15 at 16:13
  • $\begingroup$ btw right answer is just $E={1V\over{2,5x10^{-2}m}}=40{V\over{m}}$ $\endgroup$ – Yuri Borges May 16 '15 at 16:23
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To answer the first question, consider that if the electric field acts in one dimension, then

$\mathbf{E} = -\frac{d\mathbf{V}}{dx}$

Hence, in a uniform field the potential changes linearly across the direction parallel to the field. Like this image, where the potential increases by a constant amount at each equipotential as we go left (because field is the negative gradient of potential)

enter image description here

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  • $\begingroup$ You mean (in one dimension) $E_x=-\frac{dV}{dx}$. If the field is constant, then $E_x=-\frac{\Delta V}{\Delta x}$. $\endgroup$ – Richard H Downey May 16 '15 at 17:23

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