the electric field is defined as $\vec E= \dfrac {d\vec F}{dq}$ in my text book. So we are using an infinitely small test charge to probe the electric field. But why does the test charge have to be very small? According to Coulombs Law the force between two point charges is $\vec F= \gamma \cdot \dfrac {Q_1 \cdot Q_2}{r^2} \cdot \vec e_r$ . So by dividing Q2 out, I should end up with the electric field of point charge Q1: $\vec E= \gamma \cdot \dfrac {\vec F}{Q_2} = \gamma \cdot \dfrac {Q_1}{r^2} \cdot \vec e_r$ . So, I can use a charge of arbitrary size and correctly measure the electric field yielding the following definition $\vec E= \dfrac {\vec F}{Q}$
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2$\begingroup$ I think that the “methaphisical” reason is that, the perturbation on the sources of the field due to the test charge is negligible when the test charge is “infinitesimal”. I find this way to define physical objects quite dangerous from a pedagogical perspective, as it creates a bad confusion between physical notions and mathematical instruments. $\endgroup$– Valter MorettiCommented Aug 18 at 13:53
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$\begingroup$ The textbook is simply being careful. At the early stages you are only seeing fixed point charges, fixed where they are, and so you do not see the point of taking the small test charge limit. However, once you have materials coming in, this will become important. It is so important that Jackson decided to start his treatment of electrodynamics with this, so that all students of his must understand it before they can continue with other stuff. $\endgroup$– naturallyInconsistentCommented Aug 18 at 15:17
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2$\begingroup$ What I am saying is that $d\vec{F}/dq$ is not the right way to mathematically implement a clear physical argument: here there is no well defined function $\vec{F}(q)$ whose derivative is the electric field. Trying to introduce it would make even more obscure a clear physical argument. What is worse is that this attitude creates the illusion that using abstract mathematics where it is not necessary is a theoretical worth. $\endgroup$– Valter MorettiCommented Aug 18 at 15:30
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$\begingroup$ Probably worth noting point chargers don't actually exist. Typically, charges accumulate in chunks. Further, you can't really say a particle occupies space smaller than its Compton Wavelength. In practice then, you always have a dipole moment. $\endgroup$– R. RomeroCommented Aug 19 at 15:49
2 Answers
The test charge must be very small because, according to Coulomb's Law, charges exert forces on each other. If the test charge $ Q_2 $ is large, it will influence the configuration of the source charges (such as $ Q_1 $) and distort the electric field that you are trying to measure. The definition of the electric field $ \mathbf{E} = \frac{\mathbf{F}}{Q} $ assumes that the test charge does not alter the existing field; this is only true when the test charge is infinitesimally small. When the test charge is small enough, it doesn't affect the field, and the force $ \mathbf{F} $ is directly proportional to the electric field produced by the source charge alone, allowing you to use $ \mathbf{E} = \frac{\mathbf{F}}{Q_2} $ correctly.
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1$\begingroup$ So the test charge could potentially change the arrangement of elementary charges that make up Q1? But if they are all fixed in space, it really shouldnt matter $\endgroup$ Commented Aug 18 at 13:57
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2$\begingroup$ @user2276094, we prefer to have one definition of the electric field that applies both when the source charges are fixed in space and when they are freely moving. $\endgroup$ Commented Aug 18 at 14:24
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$\begingroup$ "So the test charge could potentially change the arrangement of elementary charges that make up Q1?" I had asked a question about the usefulness of the E field concept given that any real-world charge could change the configuration of charges creating the field: physics.stackexchange.com/questions/589313/… $\endgroup$ Commented Aug 18 at 23:26
The actual idea behind that formula is not about derivatives, but the following: the ratio $$ \mathbf E_1:=\frac{\mathbf F_q}{q},\tag{1} $$ motivated by the Coulomb law, is not generally applicable enough, because even while it works for fixed charges, if the source charge distribution producing the field reacts to presence of the test charge $q$ (which it always does in practice), the ratio is a vector that depends on $q$, and thus gives something we did not want; we wanted to define electric field in the situation where the source charge distribution is not affected, and this does not depend on $q$.
To get rid of this problem, and get value of electric field that is not affected by the test charge, this smart-looking definition of electric field strength has been proposed:
$$ \mathbf E_2 := \lim_{q\to 0} \frac{\mathbf F_q}{q}.\tag{2} $$
Your formula using $d\mathbf F/dq$ is a different expression of this, but it lacks an important part - the derivative still depends on $q$, and one must additionally state that the derivative is to be evaluated at $q=0$. Thus we have
$$ \frac{d\mathbf F_q}{dq}(q=0) = \lim_{q\to 0} \frac{\mathbf F_q}{q}. $$
This definition is believed to produce unique result (because it is assumed the limit exists, and the derivative exists), and it corresponds to a situation where the electric field of the system is not changed in any way by the presence of the test charge.
The mentioned belief is true in a fairy-tale world of macroscopic electrostatics, where we can decrease value of charge $q$ continuously to zero, and the influence of this charge on the source distribution can be made arbitrarily small.
But for a long time, from experiments showing granular nature of matter and charge, we have strong reasons to believe that there is a smallest electric charge $q^*$(whether it has value $e$, or $e/3$ does not matter here), and we cannot get a test charge below this value. Thus the limit in the definition (2) refers to a process that cannot be realized (continuous decrease of charge to 0). This turned out to be a bogus theoretical definition, admittedly stemming from valid concern (how to get electric field that is there before we put test charge there), but it is based on erroneous idea that cannot be realized (test charge going to zero).
So the original problem - any measurement of electric field depending on the test charge $q$ - remains unsolved. Any measurement of electric field depends also on the thing we measure it with. However, this theoretical annoyance of how to operationally define electric field that would be there were we not measuring is not really some important problem to be solved. For practical measurements of macroscopic electric field, it is completely sufficient to measure force $\mathbf F_q$ for a small enough charge $q$ and then use the definition (1) - usually, the ratio depends on $q$ negligibly weakly already when $q$ is many times greater than $q^*$. Or, more practically today, we can measure voltage on a small enough FET transistor and calculate the corresponding electric field component, and neglect the effect the charge in the transistor has on the measured electric field.
In cases where the influence cannot be neglected, there is no real problem anyway. We can still measure electric field in action, when the test particle/test device is present and affects the result. Using the ratio (1) with the actual value of the test charge to define the actual electric field in action is a perfectly fine definition of the actual electric field there.
We can't measure electric field in vacuum if the test charge is not present, or if it has zero charge; we can only define it there based on other knowledge, and estimate its value. How? For example, by measuring the force $\mathbf F_q$, and subtracting from it the estimated effect of the test particle/test device on this force, and then using the corrected force in the definition (1).
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$\begingroup$ Thank you for your insights. One question, though. I'd like to track your logic on the claim: $\frac{d\mathbf F_q}{dq}(q=0) = \lim_{q\to 0} \frac{\mathbf F_q}{q}$. Can I think of this formally as follows: $F: \mathbb R \to \mathbb R$, such that $F(Q) \mapsto n$. Therefore, "$\frac{d\mathbf F_q}{dq}(q=0)$" is really your way of saying: $\lim_{q \to 0}\frac{F(Q+q)-F(Q)}{q}$ evaluated at $Q=0$. And if $Q=0$, $F(Q)=0$. So we end up with $\lim_{q \to 0}\frac{F(Q+q)-F(Q)}{q}=\lim_{q \to 0}\frac{F(q)}{q}$. Is that correct? $\endgroup$– S.C.Commented Nov 21 at 16:15
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$\begingroup$ @S.C. I don't see your point. My argument and notation seem clear to me. I don't know what $n$ is, and why you are introducing $Q$. But if I put $Q=0$ in your formula, I obtain expression with lim which I wrote. $\mathbf F_q$ is force on body with net charge $q$, so $\mathbf F_q$ is function of single variable $q$, there is no reason to introduce another variable $Q$. $\endgroup$ Commented Nov 21 at 16:56
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$\begingroup$ To clarify the comment, it is unclear to me what you mean by $F_q$. Are you saying that $F$ is a function of $q$? If so, the formal definition of $\frac{dF_q}{dq} |_{q=0}$ is $\lim_{q \to 0}\frac{F(0+q)-F(0)}{q}$. Are you defining $F(0):=0$? Q is not another variable. It is THE variable, for which you are interested in its derivative evaluation at $Q=0$. $\endgroup$– S.C.Commented Nov 21 at 17:01
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1$\begingroup$ @S.C. Yes, of course, $\mathbf F_q$ is force on body with net charge $q$, and I assume force on body with zero net charge is zero: $\mathbf F_0=0$. $\endgroup$ Commented Nov 21 at 17:02
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$\begingroup$ Cool - sorry for the confusion. Thank you. $\endgroup$– S.C.Commented Nov 21 at 17:03