# Coulomb’s Law: explanation, formula and units, exercises, experiments

The **Coulomb’s law** is the physical law governing the interaction between electrically charged objects. It was enunciated by the French scientist Charles Augustin de Coulomb (1736-1806), thanks to the results of his experiments using the torsion balance.

In 1785 Coulomb experimented numerous times with small electrically charged spheres, for example, moving two spheres closer or farther apart, varying the magnitude of their charge and also their sign. Always carefully watch and record each response.

These small spheres can be considered *point charges* , that is, objects whose dimensions are insignificant. And they fulfill, as has been known since the times of the ancient Greeks, that charges of the same sign repel and charges of different signs attract.

With that in mind, Charles Coulomb found the following:

The force of attraction or repulsion between two point charges is directly proportional to the product of the magnitude of the charges.

-This force is always directed along the line that joins the loads.

-Finally, the magnitude of the force is inversely proportional to the square of the distance separating the loads.

__Coulomb’s law formula and units__

__Coulomb’s law formula and units__

Thanks to these observations, Coulomb concluded that the magnitude of the force *F* between two point charges *q _{1}* and

*q*, separated by a distance

_{2}*r*, is given mathematically as:

Since force is a vector quantity, to express it completely, a unit vector ** r **is defined in the direction of the line joining the charges (a unit vector has magnitude equal to 1).

In addition, the constant of proportionality needed to transform the previous expression on equal is called k _{and} or simply k: a *constant **electrostatic* or *Coulomb constant* .

Finally, Coulomb’s law for point loads is established, given by:

Force, as always in the International System of Units, comes in newton (N). As for loads, the unit is called coulomb (C) after Charles Coulomb and, finally, the distance r comes in meters (m).

Looking closely at the above equation, it is clear that the electrostatic constant must have units of Nm ^{2} / C ^{2} , to get newtons as a result. The constant value was experimentally determined as:

k _{e} = 8.89 x 10 ^{9} Nm ^{2} / C ^{2} ≈ 9 x 10 ^{9} Nm ^{2} / C ^{2}

Figure 1 illustrates the interaction between two electrical charges: when they are of the same sign, they repel, otherwise they attract.

Note that the third law conforms to Coulomb’s law or Newton’s law of action and reaction, therefore the magnitudes of **F **_{1} and **F **_{2} are the same, the direction is the same, but the directions are opposite.

**How to apply Coulomb’s law**

To solve problems of interactions between electrical charges, the following must be taken into account:

– The equation applies exclusively in the case of point charges, ie electrically charged objects but with very small dimensions. If the uploaded objects have measurable dimensions, you need to break them down into very small charges and add up the contributions of each of those charges, for which an integral calculation is required.

– Electric force is a vector quantity. If there are more than two interacting charges, the net force on charge q _{i} is given by the superposition principle:

**F **_{net} = **F **_{i1} + **F **_{i2} + **F **_{i3} + **F **_{i4} +… = ∑ **F **_{ij}

Where the subscript *j is* worth 1, 2, 3, 4 … and represents each of the other charges.

– You must always be consistent with the units. The most common is to work with the electrostatic constant on the SI units, so you can be sure that the charges are in coulomb and the distances are in meters.

Finally, the equation is applied when the loads are in static equilibrium.

__solved exercises__

__solved exercises__

**– Exercise 1**

In the following figure, there are two point charges + q and + 2q. A third point charge – q is placed on P. We are asked to find the electrical force in that charge due to the presence of the others.

**Solution**

The first thing is to establish a suitable frame of reference, which in this case is the horizontal axis or the x axis. The origin of this system can be anywhere, but for convenience, it will be placed in P, as shown in Figure 4a:

A diagram of the forces at -q is also shown, taking into account that it is attracted by the other two (figure 4b).

Let’s call **F **_{1} the force exerted by charge q on charge -q, they are directed along the x axis and point in the negative direction, so:

**F **_{2} is calculated analogously:

Note that the magnitude of **F **_{2} is half the magnitude of **F **_{1} , although the load is double. To find the net force, we finally add **F **_{1} and **F **_{2} vectorally :

_{Net }**M** = (k + k / 2) (Q. ^{2} / d ^{2} ) ( **x** ) = N -. (K / 2) (Q ^{2} / d ^{2} ) ( **x** ) N

**– Exercise 2**

Two polystyrene spheres of equal mass m = 9.0 x ^{10-8} kg have the same positive charge Q and are suspended by a silk thread of length L = 0.98 m. The spheres are separated by a distance of d = 2 cm. Calculate the value of Q.

**Solution**

The status of the declaration is depicted in figure 5a.

We choose one of the spheres and draw a diagram of the isolated body on it, which includes three forces: weight **W** , tension on the chord **T** and electrostatic repulsion **F,** as shown in figure 5b. And now the steps:

**Step 1**

The value of θ / 2 is calculated with the triangle in figure 5c:

θ / 2 = arc (1 x 10 ^{-2 /} 0.98) = 0.585º

**Step 2**

Next, Newton’s second law must be applied and equal to 0 since the charges are in static equilibrium. It is important to note that the voltage **T** is skewed and has two components:

∑F _{x} = -T.sen θ + F = 0

∑F _{y} = T.cos θ – W = 0

**step 3**

We clear the voltage magnitude from the last equation:

T = W / cos θ = mg / cos θ

**Step 4**

This value is substituted into the first equation to find the magnitude of F:

F = T sin θ = mg (sin θ / cos θ) = mg. tg θ

**Step 5**

Since F = k Q ^{2} / d ^{2} , Q is clean:

Q = 2 × 10 ^{-11 } C.

__Experiences__

__Experiences__

It’s easy to verify Coulomb’s law using a torque balance similar to the one Coulomb used in his lab.

Two small elderberry spheres are available, one of which, the one in the center of the scale, is suspended by a string. The experiment involves touching the discharged elderberry spheres with another metal sphere charged with charge Q.

Immediately the charge is shared equally between the two elderberry spheres, but later, as they are charges of the same sign, they repel each other. A force acts on the suspended sphere causing the string from which it hangs to twist and immediately pull away from the fixed sphere.

So we see that it wobbles a few times until it hits balance. Then, the twist of the bar or wire that holds it is balanced by the electrostatic repulsion force.

If the spheres were originally at 0°, now the moving sphere will have rotated an angle θ. Around the scale is a tape graduated in degrees to measure this angle. By determining the torque constant in advance, the repulsion force and the amount of charge acquired by the elder balls are easily calculated.