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Coulomb’s law and electric field – the maths

Electricity and magnetism

Coulomb’s law and electric field – the maths

There is more the Coulomb’s law says of charge interactions than just the inverse square dependence. We explore the mathematical expression of the law here.

The mutual attractiveness between unlike charges, and the repulsive behaviour between the like ones, becomes quite evident when we analyze the mathematical expression for the Coulomb’s law, besides making clear the central nature of the electric force. What required several paragraphs to explain without the mathematics in the previous article, can be made obvious with a single formula, as we will see below.

The qualitative Coulomb’s law

We summarize the facts made evident by Coulomb’s law, qualitatively, as follows –

  • Charges of the same kind (both positive or both negative) repel one another, while those of the opposite kind, attract.
  • The force with which charges attract or repel each other depends directly on the product of the charge magnitudes.
  • The electrostatic force between two charges varies inversely as the square of the distance between them.
  • The force, being a vector quantity, has a direction associated with it, and is found to act along the line joining the two charges. This direction, in fact, even distinguishes the attractive forces from the repulsive ones.
  • Also, in the presence of more than two charges, the resultant force on any of the them is simply the vector sum of the forces due to each of the other charges. We then say, the Coulomb’s force obeys the superposition principle.

The quantitative Coulomb’s law

We write the Coulomb’s law for two charges, in its full mathematical glory as

    \begin{align*} \mathbf{F}&=\frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \mathbf{\hat{r}} \end{align*}

This single equation makes all the inferences we earlier made about the nature of electric forces, apparent.

Clearly we can see that the force \mathbf{F} is directly proportional to the product of the charges that we call q_1 and q_2, here; and is inversely proportional to the distance between them squared, r^2.

The awkward looking fraction \frac{1}{4\pi\epsilon_0} in the middle, is just the proportionality constant that turns the proportionality we have been speaking of so far, into the equality we see in the equation. This constant isn’t unique and arises due to our use of the SI units. (To remind you of the units SI, \mathbf{F} is in newtons, N; q_1 and q_2 are in coulombs, C; distance r is in metres, m; and as you can guess, the constant \frac{1}{4\pi\epsilon_0} is in m^2 C^-2)

The \epsilon_0 in the constant is termed the permittivity of free space and is a measure of the resistance vacuum offers to the “transmission of the force” through it. The factor 4\pi is included only because there are many formulas that have the factor, too, such as the surface area of a sphere 4\pi r^2, the volume of a right cylinder \pi r^2 h, etc, that can appear in the expression and get cancelled, making our calculations simpler.

Finally as for the vector nature of the force, the quantity \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} specifies its magnitude, while the unit vector \mathbf{\hat{r}} shows us the direction in which the force acts. For our convenience we introduce a Cartesian coordinate system with the charges q_1 and q_2 lying on the x-axis, as you can see in the figure below. So far we have been speaking of the electrostatic force between charges, in general, but to be precise, we must be specific about the charge on which the force acts, and the charge which exerts the force. You might argue that both the charges apply force on each other. That’s true, but when analyzing the physical situation we have to consider the effects on the charges one after the other. So we might consider \mathbf{F} to be the force that q_2 experiences due to the presence of q_1. Then the unit vector \mathbf{\hat{r}} will have its tail on q_2, and will point either directly towards or directly away from q_1, depending on the sign of charges on q_1 and q_2 (either way, \mathbf{\hat{r}} lies along the x – axis, which is essential for the central nature of the force). If q_1 and q_2 are like charges (i.e., both positive or both negative), \mathbf{\hat{r}} points away from q_1, as you can see below, indicating the push exerted by q_1. On the other hand, if \mathbf{\hat{r}} points toward q_1, attraction is indicated.

Coulomb's law

Here q_1 and q_2 are like charges, hence the \mathbf{\hat{r}} points away from q_1.


The electric field

The electric field at a point due to a charge is a quantity derived from the Coulomb force law, and is defined to be equal to the electrostatic force that a unit positive charge experiences when placed at that point. Or in a more general way, we say that the electric field is the ratio of force a test charge experiences to the magnitude of the charge (of the test charge). i.e.,

    \begin{align*} \mathbf{E}&=\frac{\mathbf{F}}{q_0} \end{align*}

q_0 here is the test charge magnitude.

Expressing the force \mathbf{F} using the Coulomb’s law, we get

    \begin{align*} \mathbf{E}&=\frac{1}{4\pi\epsilon_0} \frac{q q_0}{q_0 r^2} \mathbf{\hat{r}} \end{align*}

    \begin{align*} \mathbf{E}&=\frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \mathbf{\hat{r}} \end{align*}

Once we get this expression for \mathbf{E}, we needn’t any longer bother of actually bringing in a test charge, and measuring the force it experiences, etc, to calculate the electric field. We simply need to plug in the values for the magnitude of the charge whose electric field we want to calculate, and the distance of the point from the charge, we are interested in, in the above equation. The direction of the field is again given by the \mathbf{\hat{r}}.

On a side note, if you’re calculating the electric field due to a point charge, by actually bringing in a test charge and calculating the force it experiences, we need to be careful, for, the test charge, besides experiencing a force from our point charge, itself exerts the same force back on the point charge, hence displacing it, and all in all changes the very electric field we’d set out to measure! To avoid this complication we define the electric field instead as the force per unit charge in the limiting case as the magnitude of the test charge tends to zero.

    \begin{align*} \mathbf{E}&=\lim_{q_0\to 0}\frac{\mathbf{F}}{q_0} \end{align*}

Also, we have, throughout the article, spoken of just two charges applying forces on one another, but most of the problems in physics aren’t so simple, and have more than two charges  present. Then if we are to calculate the force that one of the charges experiences due to all the other charges, we must calculate the forces due to each charge on our single charge, one after the other, and then add up the results to get the net force acting on our charge (note that vector addition is involved here). This is simply the application of the principle of superposition.

Cover image by Bert Kaufmann

Roshan Sawhil

Roshan Sawhil is a Physics postgraduate who rejoices both doing and explaining Physics. He also finds doing Philosophy as a leisure activity quite interesting. You can find and connect with him on Facebook and Twitter.

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  1. Pingback: The electric potential - part 2Physics Capsule

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