The simplified situation is depicted below.

• If x is the displacement of a mass from equilibrium, then x is in the same proportion to L as the repulsive force F to mg.
• We conclude that the force is F = (x / L) * mg.

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Coulomb's law is stated in the figure below.

• q1 and q2 represent two charges separated by a distance r.
• The force F₁₂ is the force exerted by the charge q1 on the charge q2.
• The force F21 is the force exerted by the charge q2 on the charge q1.
• Coulomb's law states to the magnitudes of these two forces are equal, and that the common magnitude is | k q1 q2 / r^2 | , where k = 9 * 10^9 N m^2 / C^2.
• Note the similarity between the form of this law and Newton's Law of Universal Gravitational.
• Coulomb's law also states that the forces are equal and opposite, and that the force is one of attraction if the charges are unlike, repulsive if the charges are like.

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A more concise statement of Coulomb's law is given in the figure below.

• The vector r₁₂ originates at the charge q1 and terminates at the charge q2.
• The vector r₁₂, when divided by its magnitude r₁₂, yields a vector of magnitude 1. This vector is called a unit vector, and is equal to r₁₂ / r₁₂.
• When this unit vector is multiplied by k q1 q2 / r^2, the result is a vector of magnitude | k q1 q2 / r^2 | either in the direction of r₁₂ or opposite to this direction, depending of whether q1 q2 is positive or negative.

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If we have two charges each of 1 `micro C separated by 6 cm, which is a fairly typical separation for the experiment with the charged tape suspended by threads, the force is easily calculated to be about 2.5 N.

• We note that this force is quite large compared with the forces inferred from the observed separations, the thread lengths and the masses of the pieces of tape. These forces were on the order of a few thousandths of a Newton.
• We conclude that the charges in our experiment were somewhat less that 1 `micro C.

A more typical force would be .003 N at a separation of 6 cm.

• Assuming equal charges q1 = q2 = q, the force will be F = k q^2 / r^2.
• We can solve this equation for q as indicated below.
• Substituting the known quantities into the expression for q we see that the charges would be approximately 3.5 * 10^-8 C.

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Suppose now the we have a fixed charge q1 = 3 `micro C at the origin of a coordinate system and a 5 `micro C charge at (1m, 0), as indicated in the figure below.

• We easily calculate the force of repulsion between the two charges; this force is .135 N.
• Now if we wish to move the 5 `micro C charge 1 cm to the point (.99m, 0), we must exert a force equal and opposite to the repulsive force and in so doing we will perform work W = .00135 J, as indicated below.
• The question we now pose is how many Joules of energy per Coulomb are required to move the charge between the two points.

If .00135 J are required to move the .000005 C charge, then the number of Joules per Coulomb is 270.

• A 270 Joule/Coulomb difference is a 270 Volt difference: A Volt is a Joule/Coulomb.
• We thus say that the potential difference between these points is about 270 volts.

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Suppose now that our 5 `micro C charge was located at (.5 m, .3 m). We wish to find the magnitude and direction of the force of the charge at the origin on the 5 `micro C charge.

• To find the magnitude of the force we will use the known charges q1 and q2 and the separation r between the charges.
• The separation is found by the Pythagorean Theorem to be .58 meters.
• To find the angle of the force, we observe first that a force of repulsion will lie along a line whose direction is the same as that of the line joining the two charges.
• Simple trigonometry shows us that the angle of this line is therefore tan^-1 (.3 / .5) = 31 deg (approx.).

University physics students will see that the vector r₁₂ is .5m i + .3 m j, so that r₁₂ / r₁₂ = (.5 i + .3 j ) / .58, where the units are understood to be meters.

• In this case F₁₂ = k q1 q2 / r^2 * r₁₂ / r₁₂ = -.41 N * (.5 i + .3 j ) / .58.
• This expression can be easily simplified to determine the i and j components of this resultant vector.
• The magnitude and angle are equally easy to determine.