Unit 9 · Topic 9.3

9.3 Conservation of Electric Energy

When a charge moves through a potential difference, the system's electric potential energy changes by qΔV, and if nothing else does work that energy turns into kinetic energy. This lets you find speeds and distances with energy conservation, with no vectors or kinematics.

Key terms

conservation of energy
change in potential energy
kinetic energy
accelerating potential difference

The energy bookkeeping

For a charge q moving between two points, the change in the system's potential energy is $\Delta U_E = q\Delta V$ . If the electric force is the only force doing work, total energy is conserved: $\Delta K + \Delta U_E = 0 \quad\Rightarrow\quad \Delta K = -q\Delta V$

If other forces do work too (a push from your hand, friction), include it: $W_{\text{other}} = \Delta K + \Delta U_E$ .

Which way charges go

Left alone, a charge moves in the direction that lowers the system's potential energy. That means the direction depends on the sign of the charge:

Charge	Moves on its own toward	ΔV	ΔU	ΔK
Positive (proton)	lower potential	negative	negative	positive
Negative (electron)	higher potential	positive	negative	positive

The electron-volt

Energies of particles are tiny in joules, so physicists often use the electron-volt: the energy one elementary charge gains moving through 1 V. 1 eV = 1.60 × 10⁻¹⁹ J. An electron accelerated through 500 V gains 500 eV. Convert to joules before using $K = \tfrac{1}{2}mv^2$ with kilograms and meters per second.

Common setups

Accelerating from rest through a potential difference: $\lvert q\Delta V\rvert = \tfrac{1}{2}mv^2$ , so $v = \sqrt{2\lvert q\Delta V\rvert/m}$ . Particle accelerators and old TV tubes work this way.
Closest approach: a charge fired straight at a fixed like charge slows down and stops for an instant when all its kinetic energy has become potential energy: $K_i = \dfrac{kq_1q_2}{r_{\min}}$ .
Released from rest near a fixed charge: like charges fly apart, and the speed far away comes from $\dfrac{kq_1q_2}{r_i} = \tfrac{1}{2}mv_f^2$ .
Two charges both free to move: conserve momentum as well as energy, since each one pushes the other.

Energy in a uniform field

Between two parallel plates the field is uniform, so the potential difference across a gap d is $\lvert\Delta V\rvert = Ed$ . A charge crossing the gap gains kinetic energy $\lvert q\rvert Ed$ . That matches the force picture: a constant force $\lvert q\rvert E$ acting over a distance d does work $\lvert q\rvert Ed$ . The two methods always agree; energy is just quicker when you only need a speed.

If the charge enters the gap moving sideways, only the motion along the field changes its potential energy. The energy method still gives its final speed, but you need forces and kinematics to find where it lands.

Energy versus force

Use energy when a question asks about speed at a position, or position at a speed. You don't need the path or the time, and the field doesn't have to be uniform. Use forces and kinematics (8.3, 10.3) when the question asks about time, acceleration or direction of motion in a uniform field.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Electron through an accelerating voltage
An electron starts from rest and is accelerated through a potential difference of 500 V. Find its final speed. (mₑ = 9.11 × 10⁻³¹ kg)
Show the solutionHide the solution
1. Step 1: Energy gained: $\lvert q\Delta V\rvert = (1.60\times10^{-19})(500) = 8.0\times10^{-17}$ J (that's 500 eV).
2. Step 2: $\tfrac{1}{2}mv^2 = 8.0\times10^{-17}$ , so $v = \sqrt{\dfrac{2(8.0\times10^{-17})}{9.11\times10^{-31}}} \approx 1.3\times10^{7}$ m/s.
3. Step 3: That's about 4% of the speed of light, so ignoring relativity is reasonable.
Answer: About 1.3 × 10⁷ m/s
Example 2Calculator allowed
Closest approach to a nucleus
An alpha particle (charge +2e) with 5.0 MeV of kinetic energy heads straight toward a gold nucleus (charge +79e), which stays fixed. How close does it get?
Show the solutionHide the solution
1. Step 1: Convert: K = (5.0 × 10⁶)(1.60 × 10⁻¹⁹) = 8.0 × 10⁻¹³ J.
2. Step 2: At closest approach K = 0, so $K_i = \dfrac{k(2e)(79e)}{r_{\min}}$ , starting from far away where U ≈ 0.
3. Step 3: $r_{\min} = \dfrac{(9.0\times10^{9})(158)(1.60\times10^{-19})^2}{8.0\times10^{-13}} \approx 4.6\times10^{-14}$ m.
Answer: About 4.6 × 10⁻¹⁴ m
Example 3Calculator allowed
Electron moving to higher potential (classic trap)
An electron starts from rest at a point where V = +20 V and moves to a point where V = +80 V. Does it gain or lose kinetic energy, and what is its final speed?
Show the solutionHide the solution
1. Step 1: ΔV = 80 − 20 = +60 V. With q = −e: $\Delta U = q\Delta V = (-1.60\times10^{-19})(60) = -9.6\times10^{-18}$ J.
2. Step 2: $\Delta K = -\Delta U = +9.6\times10^{-18}$ J (60 eV). The electron speeds up.
3. Step 3: $v = \sqrt{\dfrac{2(9.6\times10^{-18})}{9.11\times10^{-31}}} \approx 4.6\times10^{6}$ m/s.
4. Step 4: The trap is thinking every charge speeds up going to lower potential. Negative charges speed up going to higher potential.
Answer: It gains 9.6 × 10⁻¹⁸ J (60 eV); v ≈ 4.6 × 10⁶ m/s

Common mistakes

Dropping the sign of the charge in ΔU = qΔV. For an electron, q = −e, which flips the sign of ΔU.
Forgetting to convert eV to joules before solving for speed.
Using the potential at one point instead of the potential difference between two points.
Setting the final kinetic energy to zero at closest approach when the target is free to move. If both particles move, use momentum conservation too.

On the exam

Energy-conservation questions often give a V graph or an equipotential map and ask for the speed at a second point. Write the energy equation symbolically first, then substitute.
In free response, justify a speed comparison with energy: say which way the charge moves relative to the potential and how ΔU and ΔK change.

Connected topics

Videos

AP Physics C E&M - Unit 9 - Lesson 7 - Using Work-Energy for Charges
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Topic 9.3 - Conservation of Electric Energy
Lessons With LondotWatch on YouTube (opens in a new tab)
Speed of a Proton in a Uniform Electric Field
Flipping PhysicsWatch 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)
Final electron speed using energy conservation, electron accelerated between charged parallel plates
Zak's LabWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 9.3 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? (Electron mass 9.11 × 10⁻³¹ kg.)

Question 2 of 4Calculator allowed

A proton moves from a point where V = +200 V to a point where V = −100 V, with only the electric force acting. What is the change in its kinetic energy?

Question 3 of 4Calculator allowed

An alpha particle (charge +2e) with 5.0 MeV of kinetic energy heads straight toward a gold nucleus (charge +79e), which stays put. How close does it get before stopping? (1 MeV = 1.6 × 10⁻¹³ J.)

Question 4 of 4Calculator allowed

Two protons are held 1.0 × 10⁻¹² m apart and released from rest. What is the speed of each proton when they are very far apart? (Proton mass 1.67 × 10⁻²⁷ kg.)

0 of 4 answered