How To Calculate Electric Potential Energy

When you first undertake a study of the motion of particles in electric fields, there's a solid chance that you have already learned something about gravity and gravitational fields.

As it happens, many of the important relationships and equations governing particles with mass have counterparts in the world of electrostatic interactions, making for a smooth transition.

You have perhaps learned that energy of a particle of constant mass and velocity ‌v‌ is the sum of kinetic energy ‌_E_K‌, which is found using the relationship ‌mv_‌²/2, and gravitational potential energy ‌E‌_P, found using the product ‌mgh‌ where ‌g‌ is the acceleration owing to gravity and ‌h‌ is the vertical distance.

As you'll see, finding the electric potential energy of a charged particle involves some analogous mathematics.

A charged particle ‌Q‌ establishes an electric field ‌E‌ that can be visualized as a series of lines radiating symmetrically outward in all directions from the particle. This field imparts a force ‌F‌ on other particles with electric charge ‌q‌. The magnitude of the force is governed by Coulomb's constant ‌k‌ and the distance between the charges:

\(F = \frac{kQq}{r^2}\)

‌k‌ has a magnitude of ‌_9 × 10⁹ N m²/ C²‌, where ‌C‌ stands for Coulomb, the fundamental unit of charge in physics. ‌r_‌ is the distance (often measured in meters). Recall that particles with positive charge attract particles with negative charge while like charges repel.

Coulombs are an SI unit describing electric charge. They are useful in understanding measurements like amps; one amp is equivalent to one coulomb of moving charges that pass by every second.

You can see that the force decreases with the inverse square of increasing distance, not merely "with distance," in which case the ‌r‌ would have no exponent.

The force can also be written ‌F‌ = ‌qE‌, or alternatively, the electric field can be expressed as ‌E‌ = ‌F‌/‌q‌.

Relationships Between Gravity and Electric Fields

A massive object such as a star or planet with mass ‌M‌ establishes a gravitational field that can be visualized in the same manner as an electric field. This field imparts a force ‌F‌ on other objects with mass ‌m‌ in a manner that decreases in magnitude with the square of the distance ‌r‌ between them:

\(F = \frac{mGg}{r^2}\)

where ‌G‌ is the universal gravitational constant.

The analogy between these equations and those in the previous section are evident.

The formula of electrostatic potential energy, written ‌U‌ for charged particles, accounts for both the magnitude and polarity of the charges and their separation:

\(U = \frac{kQq}{r}\)

If you recall that work (which has units of energy) is force times distance (F·d), this explains why this equation differs from the force equation only by an ‌r‌ in the denominator. Multiplying the former by distance ‌r‌ gives the latter.

At this point you may be wondering why there has been so much talk of charges and electric fields, but no mention of voltage. This quantity, ‌V‌, is simply electric potential energy per unit charge. This is given by dividing out the unit charge ‌q‌ from the electrostatic potential energy. This gives the electric potential ‌V‌ for a point charge ‌q:‌

\(V = \frac{U}{q} = \frac{kQ}{r}\)

Electric potential difference represents the work that would have to be done against the electric field to move a particle ‌q‌ against the direction implied by the field. That is, if ‌E‌ is generated by a positively charged particle ‌Q‌, ‌V‌ is the work necessary per unit charge to move a positively charged particle the distance ‌r‌ between them, and also to move a negatively charged particle with the same charge magnitude a distance ‌r‌; away from ‌Q‌.

Electric potential is in units of joules (energy) per coulomb (electric charge), so it describes the amount of energy exerted per unit of charge. We can also describe small amounts of energy (often stored as electric potential energy) through a units called the electron volt (eV). This units is one volt multiplied by the charge of an electron, equivalent to 1.602 x 10^-19 J.

A particle ‌q‌ with a charge of +4.0 nanocoulombs (1 nC = 10^-9 Coulombs) is a distance of ‌r‌ = 50 cm (i.e. 0.5 m) away from a charge of –8.0 nC. What is its potential energy?

\(\begin{aligned}\)
\(U &= \frac{kQq}{r}\)
\(&= \frac{(9 × 10^9 \;\text{N} \;\text{m}^2/\text{C}^2)×(+8.0 × 10^{-9} \;\text{C})×(–4.0 × 10^{-9} \;\text{C})}{0.5 \;\text{m} }\)
\(&= 5.76 × 10^{-7} \;\text{J}\)
\(\end{aligned}\)

The negative sign results from the charges being opposite and therefore attracting each other. The amount of work that must be done to result in a given change in potential energy has the same magnitude but the opposite direction, and in this case positive work must be done to separate the charges (much like lifting an object against gravity).

Calculating voltages and electrical energy becomes much more difficult with a larger system of charges. Voltage also helps untangle measurements like the capacitance of a capacitor or the total energy of a system, not just the potential of a point charge.