Electric Force and Coulomb's Law

Electromagnetism constitutes one of the four fundamental forces in the universe, and it encompasses both electricity and magnetism. Electromagnetic interactions are those that involve particles with an electric charge, a fundamental attribute of all matter much like mass. Just as objects with mass are accelerated by gravitational forces, so electrically charged objects are accelerated by electric forces.

We begin our study of electromagnetism by examining the nature of electric charge. The shock you feel when you scuff your shoes across a carpet and then reach for a metal doorknob is due to charged particles leaping between your finger and the doorknob. Electric currents are simply streams of charged particles flowing within wires (or your body) in response to electric forces. Even the forces that hold atoms together to form solid matter, and that keep the atoms of solid objects from passing through each other, are fundamentally due to electric interactions between the charged particles within atoms.

When charges are at rest in our frame of reference, they exert electrostatic forces on each other. These forces are of tremendous importance in chemistry and biology and have many technological applications. Electrostatic forces are governed by a simple relationship known as Coulomb’s law and are most conveniently described through the concept of an electric field. Later, we will expand our discussion to include electric charges in motion. This will lead us to an understanding of magnetism and, remarkably, of the nature of light.

Electric charge is the fundamental quantity in electrostatics. It is a physical property of matter that causes it to experience a force when placed in an electromagnetic field. There are two kinds of charge; positive and negative. Charges of the same sign repel each other, charges of opposite signs attract. Charge is conserved which means that the total charge in an isolated system is constant.

All ordinary matter is made of protons, neutrons, and electrons. The positive protons and electrically neutral neutrons in the nucleus of an atom are bound together by the nuclear force; the negative electrons surround the nucleus at distances much greater than the nuclear size. Electric interactions are chiefly responsible for the structure of atoms, molecules, and solids. Conductors are materials in which charge moves easily; in insulators, charge does not move easily. Most metals are good conductors, while most nonmetals are insulators.

The unit of electric charge is the meter–kilogram–second and, in SI systems, is the coulomb and is defined as the amount of electric charge that flows through a cross section of a conductor in an electric circuit during each second when the current has a value of one ampere. One coulomb consists of 6.24 × 10¹⁸ natural units of electric charge, such as individual electrons or protons. From the definition of the ampere, the electron itself has a negative charge of 1.602176634 × 10⁻¹⁹ coulomb.

For point charges, or charged bodies that are very small in comparison with the distance r between them, Charles Augustin de Coulomb found that the electric force they exert is proportional to 1/r² . That is, when the distance r doubles, the force decreases to one-quarter of its initial value; when the distance is halved, the force increases to four times its initial value.

Coulomb’s law states that: For charges q₁ and q₂ separated by a distance r, the magnitude of the electric force on either charge is proportional to the product q₁q₂ and inversely proportional to r² and can be expressed as where is called the electric constant or the permittivity of free space which is approximately and so,

The force on each charge is along the line joining the two charges—repulsive if q₁ and q₂ have the same sign, attractive if they have opposite signs. As previously stated, in SI units, the unit of electric charge is the coulomb, abbreviated C. When two or more charges each exert a force on a charge, the total force on that charge is the vector sum of the forces exerted by the individual charges. To better model them, we frequently put these point charges into a vector graph. The sign convention assigns forces pointing upward or to the right of the charge as positive and forces pointing downward or to the left as negative.

Let’s do a quick review on vectors, notations, and constants that will be used in this tutorial:

• Vector quantities can be written in bold letters or with arrows above, so these two representations are the same:
  They are defined by their magnitude or length (without the arrow because length is a scalar quantity) and direction described by the unit vectors , and .
• The symbol (i-hat) is the unit x vector, so i = (1,0,0). Similarly is the unit y vector and the unit z vector and so = (0,1,0) and = (0,0,1). Unit vectors have a magnitude of 1.
• Vectors can be written in the form where a and b are the lengths along x and y respectively. The magnitude of this vector is given by while the direction is described by , measured from the x-axis. Trigonometric properties are useful tools in vector analysis.
• From the figure, if we let , then we can write as a sum of two vectors and , where and .
• Force is a vector quantity and is usually represented by , with a magnitude .
• the distance from point 1 to point 2
• the acceleration due to gravity; acceleration is a vector:
  (negative as it points downward). The gravitational force is given by
• to simplify calculations

Two point charges, q₁ = +25 nC and q₂ = -75 nC, are separated by a distance r = 3.0 cm. Find the magnitude and direction of the electric force (a) that q₁ exerts on q₂ and (b) that q₂ exerts on q₁.

Solution. This problem asks for the electric forces that two charges exert on each other. We use Coulomb’s law to calculate the magnitudes of the forces. The signs of the charges will determine the directions of the forces. In this problem, q₁ and q₂ have opposite signs and so the force acting on each charge is attractive, or pointing toward the other charge.

(a) The question asks for the magnitude of the force that q₁ exerts on and so the charge that we must examine is q₂ and how the presence of q₁ affects it. The figure on the right illustrates the force that q₁ exerts on q₂ and we can see that it points to the left of q₂, or attracting it to move to the left.

Converting the units of r to meters and the units of q₁ and q₂ to coulombs, Coulomb’s law gives us

Note that the term |q₁q₂| is the absolute value of the product of the charge values. The sign of the force vector describes its direction, which in turn is determined by whether the interaction between the two charges is attractive or repulsive. In this problem q₁ and q₂ have opposite signs, so the force is attractive; that is, the force that acts on q₂ is directed toward q₁ and vice versa along the line joining the two charges. According to the sign convention, the correct sign for the vector is (-) since the force points to the left of q₂.

(b) The charge that we now examine is q₁. The force that acts on q₁ is to the right, toward q₂.

Proceeding as in part (a), we have Since is to the right of q₁, is positive.

As shown, Newton’s third law applies to the electric force.

Even though the individual charges have different magnitudes, the magnitude of the force that q₂ exerts on q₁ is the same as the magnitude of the force that q₁ exerts on q₂, and these two forces are in opposite directions. Hence, they will have opposite signs.

Two point charges are located on the x-axis of a coordinate system: q₁= 1.0 nC is at x = +2.0 cm, and q₂ = -3.0 nC is at x = +4.0 cm. What is the total electric force exerted by q₁ and q₂ on a charge q₃ = 5.0 nC at x = 0?

Solution. This problem asks for the total electric force on q₃ due to q₁ and q₂.

To sketch a diagram for the forces acting on q₃, we need to figure out how it interacts with the other charges. Since q₁ is positive, q₃ (which has the same sign) is repelled by it;

is in the -x-direction and is therefore negative. Meanwhile, q₂ attracts q₃ because they have opposite signs;

is in the +x-direction and is therefore positive. The directions of all forces lie along the same line. We use Coulomb’s law to calculate the magnitudes of the forces.

The net force on q₃ is

As a check, note that the magnitude of q₂ is three times that of q₁, but q₂ is twice as far from q₃ as q₁. Coulomb’s Law then says that must be 3/2² = 3/4 = 0.75 as large as

This agrees with our calculated values:

Because is the stronger force, the direction of the net force is that of —that is, in the negative x-direction.

A small spherical ball of mass 8.00 x 10^-2 kg and charge +0.600 μC is hung by a thin wire of negligible mass. A charge of −0.900 μC is held 0.150 m away from the sphere and directly to the right of it, so the wire makes an angle θ with the vertical. Find (a) the angle θ and (b) the tension in the wire.

Solution. This problem assumes that the system is in equilibrium as pictured. Basically the charge −0.900 μC is held in place such that its distance from the sphere is 0.150 m and the sphere makes an angle θ with the vertical, confirming that there is attractive force between the charges. Examining the sphere, the forces acting on it are

(1) the tension

(2) gravitational force

and the (3) electric force due to the charge −0.900 μC.

Since the three force vectors do not all lie on a line, the best way to represent the forces is to use components. With the system in equilibrium, the sum of the forces along both the x and y-axis is zero.

ΣF_x: The forces acting along the x-axis are the electric force F and the x-component of the tension, which is this case.

F_e is positive as it points to the right while is negative as the wire pulls the sphere back to the left.

ΣF_y: The forces acting along the y-axis are gravitational force and the y-component of the tension, which is

Gravity (mg) points down hence it is negative while holds up the sphere hence it is positive.

(a) From the two equations we have:

Using Coulomb’s Law,

(b) After obtaining , we can use

to solve for the magnitude of the tension

In vector form,

Recall that

is negative and

is positive.

Two equal positive charges q₁ = q₂ = 2 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. What are the magnitude and direction of the total electric force that q₁ and q₂ exert on a third charge Q = 4.0 μC at x = 0.40 m, y = 0?

Solution. This problem asks for the net force on Q due to the surrounding charges q₁ and q₂. All interactions between the charges are repulsive as they all have the same sign. Again, we use components to calculate the vertical and horizontal forces.

Since q₁ = q₂ = 2 μC and r_1Q = r_2Q = 0.50 m, (F_{1 on Q})_x = (F_{2 on Q})_x and

Again, q₁ = q₂ = 2 μC and r_1Q = r_2Q = 0.50 m,

From symmetry we can already see that the y-components of the two forces are equal and opposite:

Hence their sum is zero and the total force on Q has only an x-component:

The total force on Q is in the +x-direction, with magnitude 0.46 N.

When two electrically charged particles in empty space interact, how does each one know the other is there? In the next tutorial we can begin to answer this question, and at the same time reformulate Coulomb’s law in a very useful way, by using the concept of electric field.