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In Raymond Serway's physics textbook, the definition of the electric field vector, was that it's force vector acts on a positive test charge, given as force divided by the test charge:

$$\overrightarrow{E}=\frac{\overrightarrow{F}}{q_0} \tag{1}$$

Then he provided another equation which is:

$$q\overrightarrow{E}=\overrightarrow{F} \tag{2}$$

where $q$ is any charge, no matter if it is positive or negative.

My question is: the book defined the electric field as the force acting on a unit positive test charge, so where has equation $(2)$ come from? We should say that:

$$\overrightarrow{E}=\frac{\overrightarrow{F}}{q}$$

not

$$\overrightarrow{E}=\frac{\overrightarrow{F}}{q_{0}} \ \ ?$$

Qmechanic
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amin
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4 Answers4

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Good question. You say $q_0$ is a unit positive test charge. The actual definition of electric field involves the limit $F/q_0$ as $q_0 \rightarrow 0$. The reason for the limit is that we don't want $q_0$ to be large enough to change the charge distribution that is producing the field. As we approach the limit, we find that the ratio $F/q_0$ remains more or less constant so indeed $F=qE$ for another charge $q$. But $q$ must be small enough, so as not to change the charge configuration that is producing the field, or means are provided to maintain that charge distribution, e.g. by maintaining a potential difference.

Brendan Darrer
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Not_Einstein
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    so you say the actual definition is that $\lim {q_0\to 0}\left(\frac{\overrightarrow{F}}{q_0}\right)$$=$$\overrightarrow{E}$ and it doesn't matter where $q{0}$ is positive or negative what matters it must be small enough or what do you mean by ''As we approach the limit, we find that the ratio $F/q0$ remains more or less constant so indeed $F=qE$ for another charge $q$'' – amin Jul 10 '22 at 17:56
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    No, $q_0$ is positive in the definition. As you approach the limit, the test charge becomes smaller and smaller and so has little effect on the charge distribution that is producing the field i.e. E is more and more constant for such test charges so the ratio F/$q_0$ is more and more constant. So that ratio holds for other charges but with the condition that those other charges are small enough so as not to disturb the charge distribution that is producing the field. Or some external factor is maintaining that charge distribution. – Not_Einstein Jul 10 '22 at 18:38
  • so you're saying that the ratio will be the same for any charge but the definition is considering only the positive test charges ? – amin Jul 11 '22 at 01:08
  • The use of a + test charge is just by definition and arbitrary. If the charge distribution that is creating the E field can be maintained in the presence of a charge q then the ratio F/q will be the same for any q. But consider the example of a single positive charge in deep space at rest in our frame of reference. It is surrounded by a field E, as found using the definition involving the limit as the test charge goes to 0. If you bring another equal charge q in the neighborhood of the original charge, the original charge will be repelled so the originally calculated E field no longer holds. – Not_Einstein Jul 11 '22 at 01:22
  • well then if in general F=qE the (q) must also be small enough or it can be arbitrary ? – amin Jul 11 '22 at 02:45
  • Well I'm getting confused because the definition says that q0 must be small enough then how after this we came to say that F=qE for an arbitrary q ? – amin Jul 11 '22 at 03:08
  • $\lim_{q_0 \downarrow 0} \frac{F}{q_0} = + \infty$. – Jannik Pitt Jul 11 '22 at 08:32
  • @Jannik Pitt But F decreases also as $q_0$-> 0 and F/$q_0$ has a finite limit. – Not_Einstein Jul 11 '22 at 14:49
  • @amin I had similar reservations. Check out this thread (Usefulness of Electric Field): https://physics.stackexchange.com/questions/589313/usefulness-of-electric-field – Not_Einstein Jul 11 '22 at 14:52
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The first formula is a definition. This is how we define the electric field.

The second formula is the effect of the field. It's we put a negative charge in a field pointing to the left, it will experience a force to the right. This is why we call it a "negative" charge: because it reacts to fields in the opposite way from a positive charge.

we should say that $\overrightarrow{E}=\frac{\overrightarrow{F}}{q}$

If we did this, we wouldn't have a unique definition of the electric field. If we tested with a positive charge we'd think the field points one way and if we tested with a negative charge we'd think the field points the opposite way. If someone asked you, "what's the electric field at this point?", you could only answer "I don't know, it depends what charge you are going to place there.", and this wouldn't be very helpful for writing down a single vector that can predict how any charge will be affected when placed in the field.

The Photon
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    How we concluded the second formula then ? – amin Jul 10 '22 at 16:14
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    @amin, by observing how different charges (positive and negative) react when placed in the same field. – The Photon Jul 10 '22 at 16:15
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    @amin, physics starts with observations of the real world. Math is only used to describe and model what we observe. – The Photon Jul 10 '22 at 16:18
  • No what I meant is $\overrightarrow{E}=\frac{\overrightarrow{F}}{q_{0}}$ where $q_{0}$ is a positive test charge so this equation work for positive test charges how the same equation work for negative charges also ? which is $\overrightarrow{F}=q\overrightarrow{E}$ – amin Jul 10 '22 at 16:21
  • I mean the second formula which is the effect of field all we know that the $\overrightarrow{E}=\frac{\overrightarrow{F}}{q_{0}}$ what we've done to find formual ② and all we've is just formual ① – amin Jul 10 '22 at 17:03
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    @amin - think of it like this. If you wanted to define the electric field, you can't use the "raw" notion of the force $\vec F$, because it can have different magnitude depending on the charge, and it points one way or the other depending on the sign of the charge, and so on. So you need some way to make the field unambiguous. So you say: OK, let's just take the field to be the force as it would appear for a positive unit charge at that point. Not for any charge, just for a positive unit charge. That's what $F/q_o$ calculates ①. – Filip Milovanović Jul 10 '22 at 20:16
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    @amin - Now you can take that quantity that says "this is the force for a positive unit charge" and multiply it by any amount of charge (positive or negative) to get the force (magnitude and direction) for that amount of charge. It lets you produce the force for any other charge. So, ② is not derived form ①, it just uses ①. It's really: $\vec F(\vec r, q) = q \vec E(\vec r) = q \vec F(\vec r, q_0)/q_0$ – Filip Milovanović Jul 10 '22 at 20:16
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    @amin the value of $\vec F$ is different in the two cases. In the first case $\vec F$ could be written more clearly as $\vec F_0,$ which is the force felt by the test change $q_0$. In the second case we can drop the ${}_0$ and generalize from the test charge to any charge $q$ – hft Jul 10 '22 at 20:16
  • @FilipMilovanović in order to do so we must know that the ratio between $\overrightarrow{F}$ and $q$ always constant, how did we know that ?? – amin Jul 11 '22 at 01:05
  • @amin - The force is directly proportional to $q$; you know from Coulomb's law that $\vec F = k_e qQ \hat r/r^2$. Take that, and fix $Q$ (the other charge). Then $\vec F/q = k_e Q \hat r/ r^2 = (\text{const}) \hat r / r^2$. So the ratio doesn't depend on q, and the dependence on $\vec r$ is already baked into $\vec E$. – Filip Milovanović Jul 11 '22 at 04:12
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    @amin, that's where observations of the real world come in. There's no way to prove it mathematically, you have to actually do experiments to determine that it's true. – The Photon Jul 11 '22 at 04:27
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The precise definition can be written in many ways. Your question seems to focus on the definition based on Lorentz force, in other words when the system (the object on which the force is applied) is a point-like charge (with the additional assumption that it's at rest, or that there's no magnetic field).

Let $q$ be this point-like charge. Start with the experimental observation that the Lorentz force is proportional to $q$. This can be written: $$\frac{\vec{F}}{q}\text{ doesn't depend on $q$}$$ Therefore it can be useful to study this ratio, because it's independent of the system on which the force is applied. Let's call this ratio the electrical field $\vec{E}$.

So far this definition is very limited, because it was built on the special case of a point-like charge. For a system with a finite size, this definition is awkward, so physicists decided to replace it with something else that doesn't require mentioning anything about the system: Maxwell's equations.

Summary: defining the electric field as $\vec{F}/q$ is mostly an introduction to show that it's possible to decouple the source and the object.

Miyase
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    I think you are mixing two unrelated things: i) the relationship between fields and forces, and ii) the fields and their sources. From Maxwell's equations alone, you can't extract information on i). – GiorgioP-DoomsdayClockIsAt-90 Jul 10 '22 at 16:31
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    I'm not mixing anything, but the initial question is. Defining the electric field from the electric force (as is stated in the question) is mostly a historical approach, but it's a useful one to start a course on electromagnetism. It is not, however, a general definition for the electric field. I was trying to show how the two relate. – Miyase Jul 10 '22 at 16:34
  • well I don't know Maxwell's equations I'm in my first course in electromagnetism as I said the textbook write that we measure electric field according to positive test charge if though how we came with $\overrightarrow{F}=q\overrightarrow{E}$ this equation ? – amin Jul 10 '22 at 16:38
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    As I said in my answer, at first it was observed from experiment. Seeing that the force was proportional to charge gave people the idea that the ratio was a useful vector to study. – Miyase Jul 10 '22 at 16:41
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    I maintain my comment. Relation with the sources and equations for the fields do not explain how the fields act on charges. Speaking at a higher level of the original question, you are focusing on the free field lagrangian instead of taking into account the full lagrangian of charges and fields. The action of the field on the charges is not a problem to be abandoned after the introductory Physics courses. – GiorgioP-DoomsdayClockIsAt-90 Jul 10 '22 at 17:38
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    I never said that it had to be abandoned. I only explained that defining the electric field from the Lorentz force is limited in scope. It's a good definition at first, but it has to be replaced later on. – Miyase Jul 10 '22 at 17:40
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$$\vec F=\frac 1{4\pi\epsilon_0}\frac {qq'}{r^2}\hat r$$

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 $q'$, we get the expression $q\hat r/4\pi r^2$, 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 point charge. The definition that you mention is related to the experimental difficulties of measuring an electric field whereas this one tries to establish why we need electric field at first place.

GedankenExperimentalist
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