Coulomb’s Law and the Electrostatic Force

The “Coulomb’s Law and the Electrostatic Force” has not only expanded our understanding of electric forces but also generated unexpected anecdotes. Benjamin Franklin, during an experiment to stun and cook a turkey with electricity, ended up as the test subject: a discharge left him dazed and with his hair standing up, almost as if illustrating the electric field lines in real life. Now, it’s our turn to study electric forces.

Learning Objectives:
By the end of this class, students will be able to:

1. Model electric phenomena using the superposition principle to calculate the resultant force on a test charge.
2. Simplify the study of electric forces by restricting it to the electrostatic case.
3. Apply Coulomb’s Law to determine the force between two charges in various situations.
4. Analyze charge-centered systems through a simplified formulation of Coulomb’s Law.
5. Solve practical problems related to charge distributions.

TABLE OF CONTENTS:
The Superposition Principle
The Electrostatic Simplification
Coulomb’s Law
Coulomb’s Law for Charge-Centered Systems
Exercises

We will now begin to mathematically model these phenomena, and to do so, we will introduce Coulomb’s Law. But first, it’s necessary to explain some points, namely: the superposition principle and the electrostatic simplification.

The Superposition Principle

The fundamental problem in electrodynamics consists of determining the force that a “cloud” of electric charges q_1, q_2, \cdots exerts on a test charge q_0, when the position of each charge is a known function of time. In general, both the source charges and the test charge are in relative motion.

The solution to this problem is facilitated by the superposition principle, which states that the interaction of the test charge with any one source is entirely independent of its interaction with the other sources. This means that it is always possible to determine the force \vec{F}_1 contributed by charge q_1, the force \vec{F}_2 contributed by q_2, and so on, to ultimately sum them up to obtain the total force:

\vec{F}_{tot} = \displaystyle \sum_{i}\vec{F}_i

The Electrostatic Simplification

If forces only need to be summed, one might argue that it is enough to state the equation describing the force that each source charge exerts on the test charge, and the problem would be solved; however, the issue is not that simple. The problem lies in the fact that the force depends not only on the distance and magnitude of the charges but also on the relative velocity and acceleration of each particle. Additionally, “electric information” about changes in position, velocity, and acceleration of each particle travels at the speed of light, meaning it takes a certain amount of time to reach the test charge and alter its effect.

Therefore, to simplify our study for now, we will restrict ourselves to the electrostatic case, i.e., all source charges will remain stationary, and only the test charge will be allowed to move. It is within this context that Coulomb’s Law emerges.

Coulomb’s Law

Suppose we have a test charge q_0 located at position \vec{r}, and a source charge q located at position \vec{r}^\prime. What will be the force \vec{F}_{q \to q_0}(\vec{r}) exerted by the source charge on the test charge? The answer to this question is provided by Coulomb’s Law, expressed by the formula:

\vec{F}_{q \to q_0} (\vec{r}) = \displaystyle \frac{1}{4\pi \epsilon_0} \frac{q q_0 }{\|\vec{r} - \vec{r}^\prime \|^2} \frac{\vec{r} - \vec{r}^\prime}{\|\vec{r} - \vec{r}^\prime\|}

Coulomb’s Law not only summarizes the rule of signs for electrostatic forces but also establishes that the force between electric charges is inversely proportional to the square of the distance separating them.

The constant \epsilon_0 is called the electric permittivity of free space. Its value in the International System is:

\displaystyle \epsilon_0 = 8.85 \cdot 10^{-12} \left[ \frac{C^2}{N\cdot m^2}\right]

Coulomb’s Law for Charge-Centered Systems

Coulomb’s Law can be expressed in a simpler way if we position the observer at the source of the charges, i.e., by setting \vec{r}^\prime = \vec{0}. In this case, we have:

\displaystyle \vec{F}_{q \to q_0} (\vec{r}) = \frac{1}{4\pi \epsilon_0} \frac{q q_0 }{\|\vec{r}\|^2} \frac{\vec{r} }{\|\vec{r} \|} = \frac{1}{4\pi \epsilon_0} \frac{q q_0 }{\|\vec{r}\|^2} \hat{r}

Where \hat{r}=\vec{r}/\|\vec{r}\| is the unit vector pointing from the source to the test charge.

Exercises

1. Twelve point charges of equal magnitude q are placed at the corners of a regular twelve-sided polygon (similar to the numbers on a clock). What will be the net force on a point charge q placed at the center?
2. One of the twelve charges from the previous exercise is removed. Suppose it is the one located at the 12 o’clock position (if imagined like a clock). What force will the central point charge q experience now?
3. Extend the reasoning from the previous two exercises to a distribution of n source charges arranged in a regular n-sided polygon, with a test charge at the center.
4. There are three point charges: q_1=+3[nC] located at position (0;0)[mm], q_2=-5[nC] at position (0,56;0)[mm], and q_3=+7[nC] at position (1;1)[mm]. Calculate the total force on charge q_3.
5. On a straight line, a charge q_1 = 3[C] is located, and at a distance of 40[mm] another charge q_2 = 7[C]. If a third charge is placed between them so that the sum of the forces on it is zero, what will be the distance of this third charge from the other two?
6. Two small copper spheres, each with a mass of 0.040[kg], are placed at a distance of 2.0[m] apart. Considering that the molar mass of copper is 63.5[g/mol] and its atomic number is 20, answer the following questions:
   1. How many electrons does each sphere have?
   2. How many electrons must be transferred from one sphere to the other to produce an attractive force of approximately 10^4[N] between the spheres?
   3. What fraction of the spheres’ electrons does this represent?