So, I've been reading an introductory book to physics. I've gotten to the point where I understand Coulomb's law, and now the book is introducing electric fields.

I'm having a hard time understanding why that unit is useful, and how to apply it to charges. (Force per Coulomb)

Here's a "practical" example...

Supposing the dielectric between two charges is air, $q_1$ is a hydride anion, $q_2$ is a hydrogen cation, and the ions are 1 μm apart.

$k = 9 * 10^9Nm^2/C^2$
$q_1 = -1.6 * 10^{-19}C$
$q_2 = 1.6 * 10^{-19}C$
$r = 1μm$

Using Coulomb's Law we can get the force $f$ in newtons:

$$\frac{(9 * 10^9Nm^2/C^2)(1.6 * 10^{-19}C)(1.6 * 10^{-19}C)}{(1*10^{-6}m)^2}$$

Barring any errors, this works out to $2.304*10^{-16}N$

The equation for an electric field is:


This unit looks remarkably similar to weight. (Another unit that seems useless and arbitrary to me.)

My book says that knowing an electric field, we can get the force on any charge within it. I would assume that would be done using Algebra and getting:


I understand how one might get $E$ for a single charge. (I.E. one of the ions) But, the book also says electric fields are applicable for more than one charge, yet shows no example of that.

I'm at a loss as to what I would plugin for the variables to get the electric field in my practical example.

In addition, I don't understand how this proportion would be maintained without distance.

  • $\begingroup$ are you talking about the resultant field of two or more charged particles? $\endgroup$
    – inya
    Commented Jun 6, 2015 at 23:37
  • $\begingroup$ I think so. I.E., the book insinuates that in my example, I could define an electric field "E" with my two ions, then get the force applied to any charge placed in E. For arguments sake, how would I apply an electric field derived from my ions to, say, a proton placed in that electric field? $\endgroup$
    – Allenph
    Commented Jun 6, 2015 at 23:40

1 Answer 1


Say you had two equal charges, $q$ a distance $x$ apart. enter image description here

$E=kq/r^2$ in general. You are right. The electric field can be defined as the force per unit charge. So a $100N/C$ field is one where 100 newtons of force would be exerted on a 1 Coulomb charged particle.

so if we find the field due to the two particles right at the middle of $x$, at a point $x/2$ from each other, we can do some algebra and find that $E_1=4kq/x^2$ and $E_2=4kq/x^2$

They are the same, as they are the same magnitude and polarity of charge!

The resultant field in the middle will be one field, minus the other field. In this case, $E_1$ - $E_2$ $= 0$

Note how this is surprisingly intuitive - same charges repel, and so if they are the same distance apart, then it is intuitive to think that the exact middle will have zero resultant electric field.

Note the example of the parallel plates, which lead to a uniform electric field. This electric field is given by $E=V/d$ where V is the potential difference between the two plates and d is the separation of the plates. Wherever the charge is in this field, the field will be of the same magnitude, unless the potential difference across (usually from a power supply) changes, or the separation of the plates changes. Hence, $F=QE$ anywhere in this field.

  • $\begingroup$ But it seems useless to use the electric field instead of Columb's law. You said that kq/r^2 = E. Where are we getting the radius from if not the distance to another charge? If that's the case we can simply use Coulombs law. $\endgroup$
    – Allenph
    Commented Jun 6, 2015 at 23:57
  • $\begingroup$ r is not the radius. it is the distance between two charges. $\endgroup$
    – inya
    Commented Jun 6, 2015 at 23:59
  • $\begingroup$ Oops. Well, still. Thag reiterates my point... $\endgroup$
    – Allenph
    Commented Jun 7, 2015 at 0:00
  • $\begingroup$ you could use coulomb's law to find the force betwen the two charges, yes. the point of finding the resultant electric field is so that you can find the force acting on another charge which is in the field, by using F=QE. This is some arbitrary charge you decided to add. $\endgroup$
    – inya
    Commented Jun 7, 2015 at 0:00
  • $\begingroup$ I'm sorry if I'm being brick-headed, but how do is it possible to know that force without knowing the position of the arbitrary charge in the field? $\endgroup$
    – Allenph
    Commented Jun 7, 2015 at 0:02

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