Why a limit in the definition of Electric field

My professor defined electric field as follows: $E=\lim_{q\to0 }F/q$. He said that we take $q$ to be small because we don't want it to distort the original field. Here is where I'am confused. What has the charge got to do with distortion. If we want to test the intensity if the field at a point by putting a charge there, it does not matter how big the charge is. The charge we would be putting will not have any influence on itself. In other words we will arrive at the same answer regardless of the magnitude of the test charge. So why are then we putting that limit there.

If you use a large charge as the test charge, it'll exert a force on the original charged body which creates the field. But if the test charge is indefinitely small (close to zero), using Coulomb's law, you'll see that the force between the two bodies is zero.

Essentially, what the professor means by 'distort the original field' is that there'll be a noticeably large force between the bodies, which could lead to a change in state (position or, under certain circumstances, charge) of the creator of the field, and that leads to a change in the field itself.

• What if you hold the source charge creating the field in place? – M Shehzad Jun 27 '18 at 9:01
• @MShehzad Remember we're not talking about something large and physical like a tennis ball or something. The charges can be indefinitely tiny point particles, so it'll be a pain to hold them. We're also not trying to observe the behavior of a charge in a field in order to measure the field: we're trying to provide a theoretical definition. Considering an indefinitely small charge is excellent for a definition because it makes the math very easy and allows you to consider a completely isolated system with nothing in it other than the electric fields and the charges. No 'holders' needed. – user191954 Jun 27 '18 at 9:08
• One could also think of a conducting half room and when calculating the induced surface charge with the method of mirror charges, one has a pretty.explicit example how a finite charge creates an additiona fiel – lalala Jun 27 '18 at 19:28

This all depends on you believing that an electric field exists around a point charge. A point charge $q_1$ has an electric field

$$\vec{E}_1 = \frac{1}{4\pi\epsilon_0}\frac{q_1}{\|\vec{r} - \vec{r}_1\|^3} (\vec{r} - \vec{r}_1)$$

Imagine the field line diagram. If you bring in another charge $q_2$, the total electric field becomes

$$\vec{E}_{\text{total}} = \vec{E}_1 + \frac{1}{4\pi\epsilon_0}\frac{q_2}{\|\vec{r} - \vec{r}_2\|^3} (\vec{r} - \vec{r}_2)$$

We have changed the electric field in space by bringing in another point charge (we have a dipole configuration if the charges are opposite). However, if the second point charge is sufficiently small, we won't change the electric field in space

$$\lim_{q_2\to 0} \vec{E}_{\text{total}} = \vec{E}_1$$

$\vec{E}_1$ doesn't have to be the field due to a single point charge. It can be any configuration. By bringing in a charge with almost $0$ size, the field in space won't be distorted.

Everything above is the easy answer. I will try to explain a more complete/difficult answer below

So now we ask ourselves, where does the definition of $\vec{E}$ come from? Is $\vec{E}$ purely a definition or something you can measure in the lab? The only thing you can measure directly in the lab is force. We start with Coulomb's law. The force on a charge $q$ at position $\vec{r}$ due to a configuration of charges $q_1, \dots , q_n$ is given by

$$\vec{F}_{\text{total}} = \frac{1}{4\pi\epsilon_0}\frac{qq_1}{\|\vec{r} - \vec{r}_1\|^3} (\vec{r} - \vec{r}_1) + \frac{1}{4\pi\epsilon_0}\frac{qq_2}{\|\vec{r} - \vec{r}_2\|^3} (\vec{r} - \vec{r}_2) + \dots$$

Notice that for fixed position $\vec{r}$, fixed configuration of the charges $\vec{r}_i$, and size of the charges $q_i$, the force rises linearly with the test charge $q$. Therefore, whatever this vector field $\vec{F}_{\text{total}}/q$ is, it is independent of the test charge $q$. Defining the electric field as

$$\vec{E} = \frac{\vec{F}_{\text{total}}}{q} \\ \text{or as the one-sided limit}\\ \vec{E} = \lim_{q\to0}\frac{\vec{F}_{\text{total}}}{q}$$

are completely equivalent. There is no difference due to the linear nature of Coulomb's law ($F$ plotted vs. $q$ is an upward sloping line from the origin)

So to your point, you are correct. It doesn't matter whether you take the limit or not.

Everything I said above assumed that the source distribution was static and remained static no matter the size of the test charge. However, you can imagine that a large test charge would cause movement of the source charges. If this happens, you run into the problem of never being able to measure the force of a particular source distribution. And subsequently, never being able to calculate the electric field of a certain source distribution. Why? Because even if the configuration is just right, as you bring your test charge into place to measure the force at that location, by the time you get there, the configuration has changed. Source charge moved around (imagine bringing in different size test charges to a conductor - whose source charges are free to move around). If you nail your source charges in place, you don't have this problem. If you can teleport a test charge, no matter the size, into place instantaneously, you also don't have this problem. Because then you could measure the force instantaneously before charge moved around. Otherwise, as you bring in test charge (say it takes 10 seconds for you to move a test charge in place). In that time, the source will definitely feel its presence and change before you can measure it.

However I still think you are correct. The division $\vec{F}/q$ is technically independent of $q$ and therefore $\lim_{q\to 0} \vec{F}/q$ is not needed (the limit of a constant is that constant). What textbooks should say is that $\vec{E}$ is defined as $\vec{F}/q$ where the division is taken at a particular instant in time. If you want to measure a certain charge distribution's force and subsequently calculate it's electric field, you need to make sure $q$ is close to $0$. Otherwise by the time you get there, everything will have moved around. The source distribution is a function of time. $\vec{F}/q$ would depend on the time in which you do the division. If $q$ is close to zero, you can assume a static (or time dependent) distribution (but such time dependence wouldn't be affected by the test charge - which is what you want if you are sending $q \to 0$).

• Even if $q$ is big, the electric field of the source is still $\vec{F}/q$. It's just that in this case, the dynamics of the source are partly affected by the test charge. Therefore, the $\vec{E}$ field defined as $\vec{F}/q$ is the electric field of this "altered time-dependent distribution". When $q$ is small, no alternation will take place and you'll get the field of just the "original time-dependent distribution" – DWade64 Jun 27 '18 at 20:11
• The only problem with sending $q$ to $0$ (so you don't put a force on the source distribution) is that you also send the force that you are trying to measure to $0$! Which means you can't make the measurement! You always have to disturb the system! – DWade64 Jun 28 '18 at 13:03