9.1 Electric Potential Energy

Electric potential energy is the energy of a system of charges because of their positions relative to one another. For two point charges, use:

$U_E = k\frac{q_1q_2}{r}$

The sign matters: like charges give positive electric potential energy, while opposite charges give negative electric potential energy. For more than two charges, add the energy for every unique pair.

Why This Matters for the AP Physics C: E&M Exam

AP Physics C: E&M questions may ask you to compare electric potential energy as distance changes, determine whether the energy is positive or negative, or find the total electric potential energy of a multi-charge system. The exam expects you to reason from the system energy equation, not just plug numbers into a formula.

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Electric Potential Energy Overview

Electric potential energy represents the stored energy in a system of charged particles due to their electric field interactions. This energy depends on the charges' magnitudes, their separation distance, and Coulomb's constant. When charges move within an electric field, this potential energy can be converted to kinetic energy or other forms of energy.

Electric Potential Energy of Systems

Work and Potential Energy

When we bring charged particles from infinite separation to a specific configuration, we must do work against the electric force. This work becomes stored as electric potential energy in the system.

The work required to bring charges together depends on:

• The magnitude of each charge
• The final separation distance between charges
• Whether the charges attract or repel each other

For example, bringing two like charges (which repel) closer together requires positive work from an external force, resulting in positive potential energy. Conversely, bringing opposite charges together releases energy, resulting in negative potential energy.

General Form of Potential Energy

The electric potential energy between two point charges is calculated using:

$U_{E}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r}=k \frac{q_{1} q_{2}}{r}$

Where:

• $U_E$ is the electric potential energy measured in joules (J)
• $q_1$ and $q_2$ are the charges in coulombs (C)
• $r$ is the distance between the charges in meters (m)
• $k$ is Coulomb's constant, equal to $8.99 \times 10^9 \frac{\mathrm{N} \cdot \mathrm{m}^2}{\mathrm{C}^2}$

The potential energy has these important characteristics:

• It varies inversely with distance ($\frac{1}{r}$), meaning potential energy decreases as separation increases
• It approaches zero as $r$ approaches infinity (our reference point)
• The sign of $U_E$ depends on the signs of the charges:
  • Like charges (both positive or both negative): $U_E > 0$ (positive)
  • Unlike charges (one positive, one negative): $U_E < 0$ (negative)

Multiple Charge Systems

For systems containing more than two charged particles, the total electric potential energy is found by summing the potential energies of all possible pairs of charges.

This process works as follows:

1. Identify all unique pairs of charges in the system
2. Calculate the potential energy for each pair using $U_{E}=k \frac{q_{1} q_{2}}{r}$
3. Add all these individual potential energies together

The total electric potential energy is given by:

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

Where the sum is over all pairs of charges, and $r_{ij}$ is the distance between charges $q_i$ and $q_j$.

This summation approach works because electric potential energy is a scalar quantity, allowing us to simply add the contributions from each pair of charges to find the total energy of the system.

Practice Problem 1: Two-Charge System

Solution

To find the electric potential energy, we use the formula: $U_E = k\frac{q_1q_2}{r}$

Given:

• $q_1 = +3.0 \times 10^{-6}$ C
• $q_2 = -5.0 \times 10^{-6}$ C
• $r = 0.2$ m
• $k = 8.99 \times 10^9$ N·m²/C²

Substituting these values: $U_E = (8.99 \times 10^9) \frac{(3.0 \times 10^{-6})(-5.0 \times 10^{-6})}{0.2}$ $U_E = (8.99 \times 10^9) \frac{-15.0 \times 10^{-12}}{0.2}$ $U_E = -674.25 \times 10^{-3}$ $U_E = -0.674 \text{ J}$

The negative value indicates that the charges attract each other, and energy would be released if they were allowed to move closer together.

Practice Problem 2: Three-Charge System

Solution

For a three-charge system, we need to calculate the potential energy for each pair and then sum them: $U_{E,total} = U_{12} + U_{13} + U_{23}$

Where: $U_{12} = k\frac{q_1q_2}{r_{12}}$ $U_{13} = k\frac{q_1q_3}{r_{13}}$ $U_{23} = k\frac{q_2q_3}{r_{23}}$

Given:

• $q_1 = +2.0 \times 10^{-6}$ C
• $q_2 = -3.0 \times 10^{-6}$ C
• $q_3 = +4.0 \times 10^{-6}$ C
• All sides of the triangle = 0.3 m
• $k = 8.99 \times 10^9$ N·m²/C²

Calculating each pair: $U_{12} = (8.99 \times 10^9) \frac{(2.0 \times 10^{-6})(-3.0 \times 10^{-6})}{0.3} = -0.180 \text{ J}$

$U_{13} = (8.99 \times 10^9) \frac{(2.0 \times 10^{-6})(4.0 \times 10^{-6})}{0.3} = +0.240 \text{ J}$

$U_{23} = (8.99 \times 10^9) \frac{(-3.0 \times 10^{-6})(4.0 \times 10^{-6})}{0.3} = -0.360 \text{ J}$

Total electric potential energy: $U_{E,total} = -0.180 \text{ J} + 0.240 \text{ J} + (-0.360 \text{ J}) = -0.300 \text{ J}$

The negative total energy indicates that the system would release energy if the charges were allowed to rearrange themselves.