Electrostatics: Coloumb's law and the Electric Field

[Mayhem]

Homework Statement

How do the charges in Coloumb's law relate to the charge in the equation for an electric field?

Relevant Equations

$F_E = k \frac{q_1 q_2}{r^2}$

$E_E = \frac{F_E}{q}$

Hi.

How does the charge $q$ relate to the charges $q_1, q_2$ (see equations). For example, if $q_1$ is an electron and $q_2$ is a proton, is $q$ just a product or sum or something else of the two? Also, in Columb's law, how would I conceptualize charges $q_1, q_2$ in a system of a nucleus (of protons only) and a single electron? My textbook does not make this particularly clear in my opinion.

Thanks.

 

[etotheipi]

$F_E = k \frac{q_1 q_2}{r^2}$

is the force that $q_1$ exerts on $q_2$, and vice versa [more strictly: the component of force parallel to the line of centres, and directed away from the centre of the two particles]

$E_E = \frac{F_E}{q}$

is expressing that the electric field can be thought of as the electric force acting on a 'test charge" at a certain position in space, divided by the charge of that test charge. In the context you are describing, you could set $q=q_1$ to determine the electric field $q_2$ causes at the position of $q_1$, or vice versa. (You definitely do not take their sum, or product, or anything like that!). More generally, for an arbitrary distribution of "source" charges, we usually say something like$$\mathbf{E} = \lim_{q\rightarrow 0} \frac{1}{q} \left(\sum \mathbf{F}_E \right)$$and think of $q$ as some arbitrarily small test charge, which is not one of the source charges and which is insignificant enough to not influence the original distribution of source charges.

Also, in Columb's law, how would I conceptualize charges $q_1, q_2$ in a system of a nucleus (of protons only) and a single electron? My textbook does not make this particularly clear in my opinion.

Between any two given charged particles, e.g. one particular proton and the electron, or two particular protons, etc., the electric force one particle in the given pair exerts on the other particle in the given pair (and vice versa) is given by your first equation. The total electric force on any given one of the particles is the vector sum of the $\mathbf{F}_E$ due to all of the other charges present, in turn.

Note that, if the protons are sufficiently close together with respect to the nucleus-electron distance, you may as well take $q_{+} = Ne$ to be a lumped point charge at the position of the nucleus, and now you're again stuck with a problem involving only two electric charges.

 

[BvU]

My textbook

Namely ?
$$F_E = q_1 E = q_ 1 \; k \,\frac{q_2}{r^2}$$is the force that a test charge $q_1$ experiences when it is placed -- at a distance $r$ from a charge $q_2$ in a field that is generated by a charge $q_2$

But I bet your book says something identical !

There are not many nuclei that consist of protons only -- a matter of Coulomb's law as well -- but a single electron in the neighborhood of a bare $^4_2He$ nucleus (an alpha 'particle') moves in a field with strength $$E = k\,{Q\over r^2} = k\,{(+2e)\over r^2} $$ and therefore experiences an electrostatic force $$F_E = (-e) E = -k\,{(-e)(+2e)\over r^2} = -k\,{2e^2\over r^2}$$
The minus sign indicating that the force is attracting the electron towards the nucleus (and vv).

It is instructive to calculate the force and look at he masses.$\$