← Back to Concepts

Electric Potential Energy

Electric potential energy is the energy stored in a system of charges due to their positions relative to each other. It represents the work done to assemble the charge configuration and is a fundamental concept in electrostatics.

What is Electric Potential Energy?

$$U = k\frac{q_1 q_2}{r}$$
What does zero electric potential energy mean?
The value "zero" for electric potential energy (EPE) is always relative—it depends on where you choose your reference point and the sign of the charges involved. For two charges, we usually define EPE = 0 when they are infinitely far apart. But: Bottom line: Zero EPE can mean "no energy stored" or "all the energy has been released," depending on the sign and arrangement of the charges. Always pay attention to the context and the sign!

Electric potential energy is the work required to bring charges from infinity to their current positions. It depends on the charges involved and their separation distance.

EPE Work is done to bring charges together, storing potential energy.

Relationship to Work

Work-Energy Principle

The electric potential energy equals the work done by an external force to assemble the charge configuration:

$$W = \Delta U = U_f - U_i$$

When charges are moved by an external force:

Sign of Electric Potential Energy

Energy Sign Rules

Multiple Charge Systems

$$U_{\text{total}} = k\sum_{i < j} \frac{q_i q_j}{r_{ij}}$$

For multiple charges, the total potential energy is the sum of all pairwise interactions. Each pair contributes a term to the total energy.

Example: Three Charges

For three charges q₁, q₂, q₃:

$$U = k\left(\frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}}\right)$$

Worked Examples

Example 1: Two Point Charges

Problem: Calculate the electric potential energy of two charges: +3.0 μC and -2.0 μC, separated by 2.0 m.

Solution Steps:

  1. Given: q₁ = +3.0 × 10⁻⁶ C, q₂ = -2.0 × 10⁻⁶ C, r = 2.0 m
  2. Formula: \(U = k\frac{q_1 q_2}{r}\)
  3. Substitute: \(U = (8.99 \times 10^9) \frac{(+3.0 \times 10^{-6})(-2.0 \times 10^{-6})}{2.0}\)
  4. Calculate: \(U = (8.99 \times 10^9) \frac{-6.0 \times 10^{-12}}{2.0}\)
  5. Result: \(U = -2.7 \times 10^{-2} \text{ J}\)

Answer: The electric potential energy is -2.7 × 10⁻² J (negative because unlike charges attract).

Example 2: Work Required to Separate Charges

Problem: Two charges of +1.0 μC each are initially 1.0 m apart. How much work is required to separate them to 3.0 m?

Solution Steps:

  1. Initial energy: \(U_i = k\frac{(1.0 \times 10^{-6})^2}{1.0} = 9.0 \times 10^{-3} \text{ J}\)
  2. Final energy: \(U_f = k\frac{(1.0 \times 10^{-6})^2}{3.0} = 3.0 \times 10^{-3} \text{ J}\)
  3. Work required: \(W = U_f - U_i = 3.0 \times 10^{-3} - 9.0 \times 10^{-3}\)
  4. Result: \(W = -6.0 \times 10^{-3} \text{ J}\)

Answer: -6.0 × 10⁻³ J (negative work means the electric force does the work).

Example 3: Three Charge System

Problem: Three charges are placed at the corners of an equilateral triangle: +2.0 μC at each corner. The side length is 1.0 m. Find the total electric potential energy.

Solution Steps:

  1. Each pair interaction: \(U_{pair} = k\frac{(2.0 \times 10^{-6})^2}{1.0} = 3.6 \times 10^{-2} \text{ J}\)
  2. Three pairs: Total energy = 3 × U_pair
  3. Calculate: \(U_{total} = 3 \times 3.6 \times 10^{-2} = 1.08 \times 10^{-1} \text{ J}\)

Answer: The total electric potential energy is 1.08 × 10⁻¹ J.

Electric Potential Energy vs. Electric Potential

Property Electric Potential Energy (U) Electric Potential (V)
Definition Energy of charge system Energy per unit charge
Units Joules (J) Volts (V = J/C)
Formula U = kq₁q₂/r V = kq/r
Dependence Depends on all charges Depends on source charge only
Use System energy analysis Field strength analysis

Applications of Electric Potential Energy

Conservation of Energy

Energy Conservation Principle

In an isolated system, the total energy (kinetic + potential) remains constant:

$$K_i + U_i = K_f + U_f$$

This principle allows us to:

Practice Problems

Practice Problem 1

Problem: A charge of +5.0 μC is moved from infinity to a point 2.0 m from a fixed charge of +3.0 μC. How much work is done?

Click for solution

Solution:

  1. Initial energy: U_i = 0 (at infinity)
  2. Final energy: \(U_f = k\frac{(5.0 \times 10^{-6})(3.0 \times 10^{-6})}{2.0}\)
  3. Work done: \(W = U_f - U_i = 6.7 \times 10^{-2} \text{ J}\)

Answer: 6.7 × 10⁻² J

Practice Problem 2

Problem: Two electrons are initially 1.0 nm apart. What is their electric potential energy?

Click for solution

Solution:

  1. Electron charge: q = -1.6 × 10⁻¹⁹ C
  2. Distance: r = 1.0 × 10⁻⁹ m
  3. Energy: \(U = k\frac{(-1.6 \times 10^{-19})^2}{1.0 \times 10^{-9}}\)
  4. Result: \(U = 2.3 \times 10^{-19} \text{ J}\)

Answer: 2.3 × 10⁻¹⁹ J

Key Concepts Summary

Quick Reference