9.3 Conservation of Electric Energy

Conservation of electric energy connects how a charged object moves between two points at different electric potentials to its change in energy. The core relationship is $\Delta U_E = q\Delta V$, and because the electrostatic force is conservative, any loss in electric potential energy shows up as a gain in kinetic energy, or the reverse.

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

This topic ties electric potential to energy conservation, which is one of the most useful tools on the exam. You will use $\Delta U_E = q\Delta V$ together with the work-energy theorem to predict how fast a charge moves, how much energy it gains, or how a system changes when a charge shifts position.

Energy reasoning shows up in both multiple-choice and free-response questions. The free-response section includes a Translation Between Representations question, where you might sketch an energy diagram for a charge moving through a region with an electric field and then explain how that diagram is consistent with an equipotential or field map. Being fluent with energy conservation helps you connect graphs, equations, and verbal explanations of the same scenario. Content from any unit can appear here, but the energy ideas in this topic give you strong practice for that style of question.

Key Takeaways

• The change in electric potential energy of a charge-field system is $\Delta U_E = q\Delta V$, measured in joules.
• The electrostatic force is conservative, so energy is conserved: $U_E + K$ stays constant when no nonconservative forces act.
• A drop in electric potential energy means a gain in kinetic energy, and a rise in potential energy means a loss of kinetic energy.
• The sign of the charge decides which way it moves "naturally." Positive charges tend toward lower potential, negative charges toward higher potential.
• Use $\Delta K = -\Delta U_E = -q\Delta V$ to find speed changes from a potential difference.
• Watch units and signs carefully. Coulombs times volts gives joules, and the sign of $q$ and $\Delta V$ both matter.

Changes Due to Electric Potential Difference

System Energy Changes

When a charged object moves through an electric field, the energy of the charge-field system changes in a predictable way. The core relationship is:

$\Delta U_{E}=q \Delta V$

This says the change in electric potential energy depends on both the charge of the object and the potential difference it moves through.

• The change in electric potential energy ($\Delta U_{E}$) is measured in joules (J)
• The charge ($q$) is measured in coulombs (C)
• The electric potential difference ($\Delta V$) is measured in volts (V)

For a positive charge:

• Moving from high to low potential (downhill electrically), the system loses electric potential energy
• Moving from low to high potential (uphill electrically), the system gains electric potential energy

A 1 coulomb positive charge moving through a potential difference of 5 volts has a change in electric potential energy of 5 joules. Whether that energy is gained or lost depends on whether the charge moves to higher or lower potential.

Conservation of energy ties the change in potential energy to an opposite change in kinetic energy:

• Decreasing potential energy means increasing kinetic energy (the charge speeds up)
• Increasing potential energy means decreasing kinetic energy (the charge slows down)

Charged Object Movement

How a charged particle moves in an electric field depends on the sign of the charge. The sign sets both the direction of motion and the energy transformation.

Positive charges (like protons) move naturally from high to low potential:

• They move in the direction of the electric field
• The field does positive work on the charge
• Electric potential energy converts to kinetic energy
• The particle accelerates in the field direction

Negative charges (like electrons) behave the opposite way:

• They move naturally from low to high potential
• They move opposite to the electric field direction
• The field still does positive work on the charge
• Electric potential energy converts to kinetic energy
• The particle accelerates opposite to the field direction

The link between potential difference and kinetic energy change comes from the work-energy theorem:

$\Delta K=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}=-q \Delta V$

This lets you calculate how much a charged particle's speed changes as it moves through a potential difference.

Conservation of Energy

Because the electrostatic force is conservative, energy conservation applies cleanly to a charge moving in an electric field. In an isolated system, energy is not created or destroyed, only transformed.

For a charged particle moving in an electric field:

$U_{E}+K=\text{constant}$

The sum of electric potential energy and kinetic energy stays the same throughout the motion. As one increases, the other decreases by the same amount.

• When electric potential energy decreases, kinetic energy increases
• When electric potential energy increases, kinetic energy decreases
• The magnitude of the change is identical for both forms of energy

In real setups, some energy can convert to thermal energy through resistance or friction. Even then, the total energy of the system (including thermal energy) stays constant, so conservation still holds.

How to Use This on the AP Physics C: E&M Exam

Problem Solving

A reliable approach for energy problems in this topic:

1. Identify the charge $q$ (with its sign) and the potential difference $\Delta V$ between start and end points.
2. Find the change in potential energy with $\Delta U_E = q\Delta V$.
3. Apply conservation of energy: $\Delta K = -\Delta U_E$.
4. If the charge starts from rest, the final kinetic energy equals $\Delta K$, so $\tfrac{1}{2}mv_f^2 = -q\Delta V$.
5. Solve for the unknown (speed, energy, or potential difference) and check units.

Free Response

For the Translation Between Representations question, you may be asked to build an energy diagram for a charge moving through a region with an electric field and then connect it to an equipotential or field map. Show the trade-off between $U_E$ and $K$ clearly, and justify why your energy diagram matches the field or equipotential picture. Tie your reasoning back to $\Delta U_E = q\Delta V$ and energy conservation.

Common Trap

Sign errors are the biggest issue here. Keep the sign of the charge attached to $q$ and the sign of $\Delta V$ attached to the potential difference. Decide direction of motion from the physics (positive charges fall toward lower potential, negative charges toward higher potential) rather than guessing from the math alone.

Practice Problem 1: Electron Acceleration

Solution

Use the relationship between potential difference and kinetic energy change:

$\Delta K = -q \Delta V$

Since the electron starts from rest, its initial kinetic energy is zero, so the final kinetic energy equals the change in kinetic energy:

$K_f = \Delta K = -q \Delta V$

For an electron with charge q = -1.6 × 10^-19 C moving through a potential difference of ΔV = 100 V:

$K_f = -(-1.6 × 10^{-19} \text{ C})(100 \text{ V}) = 1.6 × 10^{-17} \text{ J}$

Now find the final speed using the kinetic energy formula:

$K_f = \frac{1}{2}mv_f^2$

Rearranging to solve for v_f:

$v_f = \sqrt{\frac{2K_f}{m}} = \sqrt{\frac{2(1.6 × 10^{-17} \text{ J})}{9.11 × 10^{-31} \text{ kg}}} = 5.92 × 10^6 \text{ m/s}$

The electron's final speed is approximately 5.92 × 10^6 m/s.

Practice Problem 2: Conservation of Energy in an Electric Field

Solution

By conservation of energy, the decrease in electric potential energy equals the increase in kinetic energy:

$-\Delta U_E = \Delta K$

Since the change in electric potential energy is related to the potential difference by:

$\Delta U_E = q\Delta V$

We can write:

$-q\Delta V = \Delta K$

Rearranging to solve for the potential difference:

$\Delta V = -\frac{\Delta K}{q}$

Substituting the given values:

$\Delta V = -\frac{8.0 × 10^{-18} \text{ J}}{1.6 × 10^{-19} \text{ C}} = -50 \text{ V}$

The negative sign means the proton moved from higher to lower potential. So the potential difference between start and end is 50 V, with the starting point at the higher potential.

Common Misconceptions

• Electric potential and electric potential energy are not the same thing. Potential ($V$) is energy per unit charge at a point, while potential energy ($U_E = q\Delta V$ change) depends on the charge placed there.
• Positive work by the field does not always mean the charge moves toward lower potential in space. For a negative charge, the field does positive work as it moves toward higher potential. Track the sign of $q$ in $\Delta U_E = q\Delta V$.
• A charge speeding up does not break energy conservation. The kinetic energy gained came from a drop in electric potential energy, so the total stays constant.
• $\Delta V$ being negative does not automatically mean potential energy decreases. The sign of $\Delta U_E$ depends on the product $q\Delta V$, so a negative charge in a negative $\Delta V$ gives a positive $\Delta U_E$.
• Electric work is path independent because the electrostatic force is conservative. Only the start and end potentials matter, not the route the charge takes.

Related AP Physics C: E&M Guides

• 9.2 Electric Potential
• 9.1 Electric Potential Energy