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So the electric field is defined as having units of newtons per coulomb. I dont understand the concept of like a unit test charge. I understand that multiplying the electric field by the test charge cancels out units but I don't conceptually understand the units. I get the concept intuitively. It's basically saying that if you introduce a positive test charge to some point P that depending on what type of charges are around that you will have some sort of vector with both magnitude and direction going away or towards that test charge. Is newtons per coulomb basically saying that we have no idea what the charge in the area actually is but once we find it out that we can figure out the magnitude?

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  • $\begingroup$ It's force per unit charge E=F/q since the Coulomb force is F=Eq=kQq/r^2. It does not matter what the value of q is. What is it that you are not understanding? You need to explain a little more. $\endgroup$
    – joseph h
    Jun 17, 2021 at 20:39
  • $\begingroup$ I don't understand the phrase "you will have some sort of vector with both magnitude and direction going away or towards that test charge". What does that vector represent to you, the E field due to the test charge, the E field at the location of the charge, or the force on the charge? $\endgroup$
    – garyp
    Jun 18, 2021 at 0:09

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Given a configuration of charges, what would be the force on an arbitrary charge placed at some point $P$? It would depend on both the position of $P$ relative to all source charges (and their charges) as well as the charge of the particle we are trying to calculate the force upon.

Lets replace our initial charge at $P$ with some other charge. What would be the force now? Would it be wise to repeat the above calculation again, one charge pair at a time?

It seems redundant to calculate the force on the new charge by going through the entire procedure again. From Coulomb's law we know that the force must have increased proportionally for every source-'new charge' pair. Also, with superposition, that implies the total force on our new charge has also increased proportionally. So, the effect of changing just the charge at $P$ while keeping the source config. same was to increase the overall force proportionally.This implies that all the information regarding the source config. is contained in something that is independent of the charge at $P$ since it was the same for both our cases. Also this "something" must exist at every $P$.

This charge-at-$P$-independent something is called the electric field. To make it independent of the charge-at-$P$ aka test charge, we simply divide the force by it - this leads to the definition:

$$\vec{E}=\vec F/q_{test}^1$$

Thus the benefit at this stage is that by defining the field we can calculate the force on any charge.

But their is an additional advantage. By dealing in terms of only this construct of field, we need not bother ourselves with what generated that field. As mentioned, all the information regarding the individual influences of source charges are included in the field.

While the mathematical expressions for the field of the point charges may seem trivial sums of their individual formulae, the field first point of view allows one to predict the behavior of test charges in arbitrary electric fields even if we don't have a model of the test charges that produced it.

The point of the electric field is to define it everywhere and then forget about the rest.


In the above discussion, the field may seem a mere syntactic simplification to make calculations easier. But the field of any charge is as real as the charge itself. It reaches out and touches others. It has energy and inertia. It can carry currents posses impedance and create and destroy particles. The notion of fields is of fundamental importance in physics.


$^1$ Since the field is test charge independent, we might as well take it to be unit or $1C$. The field then becomes numerically equal to the force.

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