Electric field definition

Question

In my book it says that "electric field" is defined as the "force per unit charge". What does the "unit charge" mean? My initial thoughts was that it meant 1 Coulumb, but that wouldn't make sense as then electric field simply equals force. But unit means SI units?!

Answer 1

If we place a small 'test' charge, $q$, in an electric field, we find that the force, $\vec F$, that it experiences due to the field is proportional in magnitude to $q$. Therefore if we divide $\vec F$ by $q$ we get a quantity, $\vec E$, that tells us about the field itself, independently of the value of $q$.

$\vec E$ is sometimes called the force per unit charge. It is numerically equal the force that a unit charge would experience if placed at the same point, but only if the placing of such a huge charge didn't disturb the 'source' charges that give rise to he field! Use of an actual unit charge is best not pursued any further.

Answer 2

What does the "unit charge" mean?

"Per unit charge" means the force is proportional to the charge being acted on. If you double the charge, the force doubles. If you halve the charge, the force halves.

My initial thoughts was that it meant 1 Coulumb, but that wouldn't make sense as then electric field simply equals force.

Not at all. If the field strength is 1 N/C, then if you place 2 C in that field, the force on that charge will be 2 N instead of 1 N.

For every unit of charge you place, the field exerts one unit of force. Most often the units we use are coulombs and newtons, but you could express the field strength in other units if you like to confuse the people you work with.

Note: electric field strength is usually expressed in units of volt per meter. However, this is equivalent (you can do the dimensional analysis as an exercise) to newtons per coulomb.

Answer 3

A better way to define electric field IMO is following:

The problem with the expression of force is that it depends on external factors, that is it is not a fundamental aspect of the charge itself. However, if we remove charge $q'$ from equation $$\vec F=\frac1{4\pi}\frac{qq'}{ r^3}\vec r\equiv\frac1{4\pi}\frac{qq'}{r^2}\hat r $$ (I have set $\epsilon_0=1$) we get the expression $$\frac{q}{4\pi r^2}\hat r$$ where $r$ is the displacement between $q$ and any point in space. This relation is fundamental to the charge itself, independent of any charge other than itself. This expression is the electric field of a single particle.

Answer 4

Suppose we have a certain charge Q and a test charge q (this one is needed to study Q and can in principle have any value, including 1 Coulomb). The force that q feels because of Q is given by $F_{Qq} = \frac{Q\cdot q}{4\pi \epsilon_0 r^2_{Qq}}$ (Coulomb's law). The electric field because of Q is now defined as $E_Q = \frac{F_{Qq}}{q} = \frac{Q}{4\pi \epsilon_0 r^2_{Qq}}$. Note that we construct this definition of $E_Q$ to be independent of the test charge q. If we look to the the relation between electric field and force (given above), we see that the value of the electric field because of Q is the same as the force exerted from Q on q when the last one is a unit charge (1 Coulomb): $E_Q = \frac{F_{Qq_{unit}}}{q_{unit}} = \frac{F_{Qq_{unit}}}{1} = F_{Qq_{unit}}$ .