Unit 10 · Topic 10.7

10.7 Conservation of Electric Energy

When a charge moves through a potential difference, its electric potential energy changes by qΔV, and energy conservation says its kinetic energy changes by the opposite amount. This is how particle accelerators and old TV tubes speed up electrons. It's often the fastest way to find a charged particle's speed.

Key terms

conservation of energy
potential difference
kinetic energy
electron volt (eV)
accelerating charges

Energy conservation with charges

When a charge q moves from one point to another, the electric potential energy of the charge–field system changes by:

$\Delta U_E = q\Delta V$

If only electric forces do work, the total energy stays constant, so ΔK + $\Delta U_E$ = 0 and ΔK = −qΔV. A drop in potential energy shows up as a gain in kinetic energy, and the reverse.

The path doesn't matter, only the starting and ending potentials. That makes energy methods much easier than tracking forces along a curved path.

Which way do charges speed up?

Positive charges released from rest move toward lower potential, like a ball rolling downhill. Negative charges do the opposite: they move toward higher potential, because for a negative q, an increase in V means a decrease in qV.

Positive charge, moving to lower V: $U_E$ drops, K rises.
Positive charge, moving to higher V: $U_E$ rises, K drops (it slows down).
Negative charge, moving to higher V: $U_E$ drops, K rises.
Negative charge, moving to lower V: $U_E$ rises, K drops.

The electron volt

Energies of single particles are tiny in joules, so physicists often use the electron volt. One electron volt (1 eV) is the energy a particle with charge e gains when it moves through a potential difference of 1 V: 1 eV = 1.60 × 10⁻¹⁹ J.

This makes some problems almost instant. An electron accelerated through 500 V gains 500 eV. An alpha particle, with charge 2e, gains 1000 eV through the same 500 V. You'll see eV again in modern physics.

Energy and forces agree

In a uniform field, like the one between capacitor plates, you can check the energy method with forces. The field pushes a charge with a constant force qE, and over a distance d along the field it does work qEd. Since ΔV = Ed there, that work is exactly qΔV, the same kinetic energy the energy method gives. Use whichever is quicker, but energy is the only easy route when the field isn't uniform.

Energy bar charts

A bar chart for a charge moving in a field shows K and $U_E$ at the start and end. For a proton released from rest near a fixed positive charge, the start has $U_E$ only; at the end, far away, $U_E$ has dropped to nearly zero and K has grown by the same amount. The total height of the bars stays the same, because no external work is done on the system.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Accelerating an electron
An electron starts from rest and is accelerated through a potential difference of 500 V. Find its final kinetic energy (in eV and in joules) and its speed. (m = 9.11 × 10⁻³¹ kg)
Show the solutionHide the solution
1. Step 1: Kinetic energy gained = |q|ΔV = e × 500 V = 500 eV.
2. Step 2: In joules: 500 × 1.60 × 10⁻¹⁹ = 8.0 × 10⁻¹⁷ J.
3. Step 3: $v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2(8.0 \times 10^{-17})}{9.11 \times 10^{-31}}} \approx 1.3 \times 10^7$ m/s.
Answer: 500 eV = 8.0 × 10⁻¹⁷ J; v ≈ 1.3 × 10⁷ m/s
Example 2Calculator allowed
Proton pushed away by a charge
A proton (m = 1.67 × 10⁻²⁷ kg) is released from rest 1.0 cm from a fixed +1.0 nC charge. How fast is it moving when it's very far away?
Show the solutionHide the solution
1. Step 1: Potential where it starts: $V = \dfrac{kq}{r} = \dfrac{(9.0 \times 10^9)(1.0 \times 10^{-9})}{0.010} = 900$ V. Far away, V = 0.
2. Step 2: $\Delta U_E$ = qΔV = (1.60 × 10⁻¹⁹)(0 − 900) = −1.44 × 10⁻¹⁶ J.
3. Step 3: Energy conservation: K = 1.44 × 10⁻¹⁶ J.
4. Step 4: $v = \sqrt{\dfrac{2(1.44 \times 10^{-16})}{1.67 \times 10^{-27}}} \approx 4.2 \times 10^5$ m/s.
Answer: About 4.2 × 10⁵ m/s
Example 3
Which way does it go? (classic trap)
An electron is released from rest at a point where V = 20 V, between equipotentials of 10 V and 30 V. Which way does it move, and what is its kinetic energy when it reaches the 30 V line?
Show the solutionHide the solution
1. Step 1: A negative charge moves toward higher potential, so it heads for the 30 V line, not the 10 V line.
2. Step 2: $\Delta U_E$ = qΔV = (−e)(30 − 20 V) = −10 eV, so the potential energy drops by 10 eV.
3. Step 3: Energy conservation: K = +10 eV = 1.6 × 10⁻¹⁸ J.
Answer: Toward the 30 V line, arriving with 10 eV (1.6 × 10⁻¹⁸ J)

Common mistakes

Assuming all charges move toward lower potential. Only positive charges do; electrons move toward higher potential.
Forgetting the sign of q in $\Delta U_E$ = qΔV. A negative charge moving to a higher potential loses potential energy.
Mixing up eV and V. The volt is potential (J/C); the electron volt is energy.

On the exam

Expect to combine energy conservation with ΔV, often with energy bar charts, and sometimes to compare a proton and an electron through the same ΔV (same energy, very different speeds).
If a question asks only about speed, use energy rather than forces and kinematics; it's quicker and path doesn't matter.

Connected topics

Videos

AP Physics 2 - Unit 10 - Lesson 11 - Conservation of Energy
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Change in Electric Potential Energy in a Uniform Electric Field
Flipping PhysicsWatch on YouTube (opens in a new tab)
Electron Volt Explained, Conversion to Joules, Basic Introduction
The Organic Chemistry TutorWatch on YouTube (opens in a new tab)
Uniform Electric Field (7 of 9) Kinetic Energy of an Electron thru a Potential Difference
Step by Step ScienceWatch on YouTube (opens in a new tab)
Electric Potential Difference in a Uniform Electric Field
Flipping PhysicsWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 10.7 Conservation of Electric Energy. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

An electron starts from rest and is accelerated through a potential difference of 500 V. What is its final speed?

Question 2 of 4Calculator allowed

A proton (charge +e) and an alpha particle (charge +2e, about 4 times the proton's mass) both start from rest and are accelerated through the same potential difference. How does the alpha particle's final kinetic energy compare with the proton's?

Question 3 of 4Calculator allowed

A −3.0 μC charge moves from a point where the potential is 20 V to a point where the potential is 80 V. What is the change in the electric potential energy of the system?

Question 4 of 4Calculator allowed

An electron is released from rest in a region with an electric field. Which describes its motion and energy?

0 of 4 answered