# The electric field formula $E=F/q$

If we follow the formula $$E=F/q$$ it says that when the force is bigger the electric field is bigger as well but if the charge on which the force is being exerted to bigger the electric field is somehow smaller? How does this make sense?

I'd say that you are looking at it backwards; it is better to imagine that eletric fields generate forces on charges. For a given eletric field, the bigger the charge, the greater the force; to produce an increase on the force exerted on a charge, the eletric field needs to be increased as well; so, for a given force, the greater the charge the smaller the eletrical field. It is analogous to mass in that different masses(A greater than B, say) subject to the same force will have different accelerations (B greater than A).

This is a great example of how causal understanding of physics is not manifestly obvious if you naively look at the mathematical expression.

What you say is true but is correctly formulated in the following way: for a given force $$F$$ on a charge $$q$$, the electric field $$E$$ (at the location of the charge) would have to get bigger as the charge $$q$$ gets smaller. In other words, what it says is that in order to produce the same amount of force on a smaller charge, you need a stronger electric field. As you can see, there is no mystery here at all when you understand it this way.

Of course, as you already understand (as implied by your question), the electric field acting on a charge doesn't change if you only change the test charge because it is determined by the external configuration of charges. Physically speaking, when you only vary the test charge, only the force acting on it will change. However, what $$E\propto 1/q$$ tells you is that if you want to maintain a constant force on a varying test charge, you'll need to vary the electric field in inverse proportion to the value of the test charge (via changing the external configuration of charges that produces the electric field).

If you, say, double the charge $$q$$, then you would naturally also see a doubled force $$F$$. Because for the field $$E$$, there would now be "double as much" charge to "pull" in.

But if you require the force to be the same, then you will have to reduce the field strength in some way. For example by changing the source that causes the field. Otherwise its impossible to change the charge and keep the force constant.

Your confusing may have arrived from the mistaken assumption that the force can be kept constant under normal, unrestricted circumstances. And thus, changing the charge would appear to change the field. Sure, this is mathematically possible - but not physically possible. Not without you actively changing the field in some way. Rather, it would be the field that is constant under such circumstances and the force would change when the charge changes.

The electric field definition uses E = F/q as you show, but you are missing the rest of the definition, namely that it is the ratio of F/q in the limit as q -> 0. This is done so that the test charge does not disturb the charge distribution that is creating the electric field.

I think you are confused by what $$F$$ and $$q$$ mean in the equation for the electric field:

$$E = \frac{F}{q} \tag 1$$

where $$q$$ is the test charge. The test charge is a charge of very small magnitude, so small that it does not affect the electric field but could be used to measure its strength. This is the definition from textbooks, now let's develop some intuition about electric field.

According to the Coulomb's law, the electric force between two charged particles is defined as:

$$\vec{F} = k \frac{q_1 q_2}{\vec{r}^2} \hat{r} \tag 2$$

where $$k$$ is a constant. Now assume that you have only one charged particle. With only one charged particle there is no electrostatic force, i.e. there must be a second charge particle in order for force to exist. But we could define what that force would be in the presence of the second charged particle for every point in space:

$$\vec{E}(\vec{p}) = k \frac{q_1}{\vec{p}^2_\star} \hat{p}, \qquad \forall \vec{p} \in \Omega$$

where $$\Omega$$ is the set of all points. Now if you actually put the second charge particle $$q_2$$ at some position in space $$\vec{p}_0$$, the force between the two charged particles will be:

$$\vec{F} = q_2 \cdot \vec{E}(\vec{p}_0)$$

Note that the above equation is the same as the force defined in Eq. (2).

Having this in mind, how can we measure strength of an electric field? The idea is simple - bring a test charge at certain position in space, measure the force and then normalize the measured force, i.e. divide it by the test charge. This is exactly what Eq. (1) means!