I am having a trouble with the existence of "uniform electric field". As far as I know, the electric field is measured with two main formulas:

1) $E=k\frac{\left | Q \right |}{r^{{2}}}$

2) $E=\frac{F}{q}$

However, what I don't stand is that $E$ and $r^{2}$ in this relationship make the eletric field change followed by distance. I have read and get mentioned about infinite line of charge but I still don't understand. I just know that:

1) The uniform with electric field (don't know the specific formula) use $r$ instead of $r^{2}$. That's why the eletric field doesn't change according to distance.

2) The Coulomb's law however applying on sphere charged object. And we get $r^{2}$. But I don't know why we have such formula (where the formula derive from ?)

Is there a simple way to comprehend this matter?

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  • $\begingroup$ An approximately uniform electric field can be produced between two oppositely charged parallel plates. $\endgroup$
    – Farcher
    Commented Oct 21, 2016 at 15:46
  • $\begingroup$ It would help answering this: Where do you want it to be unform? Over what distance scale? $\endgroup$
    – snulty
    Commented Oct 22, 2016 at 19:56

1 Answer 1


Maybe this analogy might help: The earth's gravitational field is almost constant near the surface, even though we know its equation depends on r^2:

$$ \vec{g} = \frac{\vec{F}}{m} = -GM\frac{\hat{R}}{R^2} $$

Very close to the surface it seems that the surface is infinitely long. When you look at the gravitational field of tiny masses and take their sum (keeping in mind that your surface extends infinitely), all components will cancel except for the vertical component. This is why gravity is pretty much constant on the surface of the earth.

The same can be said about two parallel plates. If we use the simple model that electrons are particles on the surface of such plates, the same model applies. We can sum the electric field of each particle to find that only the vertical components remain, and the field is then uniform.

  • $\begingroup$ I tried it with 3 charges $\endgroup$ Commented Oct 22, 2016 at 2:56
  • $\begingroup$ q1 = 1.602x10^-19 (C) q2= 1.602x10^-19(C), qo is a random positive charge put in the middle of q1 and q2. Well I tried 2 cases (both cases q1 and q2 are 2 meters far away from each other). The first case I put qo 1m away from the q1. The second case I put qo 1.5m away from the q1. The difference between the electric field strength in two cases have very big differences. $\endgroup$ Commented Oct 22, 2016 at 3:03
  • $\begingroup$ I don't understand. My charged objects setup is just like the two panels setup. I don't get why two-panel setup gets to have uniform electric field. While mine cannot do such a thing !? $\endgroup$ Commented Oct 22, 2016 at 3:05
  • $\begingroup$ In theory you need an infinite number of charges to obtain a straight field. $\endgroup$
    – michael b
    Commented Oct 22, 2016 at 3:49
  • $\begingroup$ -But what I don't understand is why the straight field cannot only made through an infinite line of number of charges. Could you $\endgroup$ Commented Oct 24, 2016 at 3:48

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