Can a Charged Particle Move Without Changing Electric Potential Energy?

A charged particle can move a considerable distance without changing its electric potential energy. An electron, for example, could move several centimetres through an electric field while its electric potential energy stays constant. The important detail is not how far it moves, but whether the electric potential at its final position differs from the potential at its starting position.

If the particle moves between two points at the same electric potential, its change in electric potential energy is zero:

\[
\Delta U=q\Delta V=0
\]

This is true for positive and negative charges. However, there is an important distinction between moving along an equipotential and moving through an electric field. Let's work through the difference, including what happens to the particle's kinetic energy.

## 1. The short answer: yes, a charged particle can move without changing its potential energy

Electric potential energy depends on two things:

- The charge of the particle, \(q\).
- The electric potential at its position, \(V\).

The relationship is:

\[
U=qV
\]

Here, \(U\) is electric potential energy in joules (J), \(q\) is signed electric charge in coulombs (C), and \(V\) is electric potential in volts (V).

For a particle moving from an initial position to a final position, we use:

\[
\boxed{\Delta U=q(V_f-V_i)}
\]

The symbols \(V_i\) and \(V_f\) represent the initial and final electric potentials. We define the potential difference as:

\[
\Delta V=V_f-V_i
\]

The crucial point is that the equation uses the signed value of the charge. An electron has a negative charge, while a proton has a positive charge.

If the initial and final potentials are equal, then:

\[
V_f=V_i
\]

Therefore:

\[
\boxed{\Delta U=0\text{ J}}
\]

The particle's electric potential energy has not changed, regardless of whether it is an electron, proton, or another charged particle.

### What does this look like physically?

Imagine two parallel metal plates. One is connected to a positive terminal and the other to a negative terminal of a power supply.

Between the plates, the electric field is approximately uniform in the central region. The equipotential surfaces are parallel to the plates and perpendicular to the electric field.

A particle moving parallel to the plates stays at approximately the same electric potential. A particle moving towards the opposite plate crosses different equipotential surfaces and changes its electric potential energy.

[IMAGE: Annotated scientific diagram of two vertical parallel metal plates, one positive and one negative, with a uniform horizontal electric field directed from positive to negative. Show several vertical equipotential lines between the plates, labelled with progressively decreasing potential values from the positive plate to the negative plate. Include two particle paths: a horizontal path parallel to the plates that stays on one equipotential, and a horizontal? No - show a separate vertical path across the field that crosses equipotential lines. Label the two paths as constant potential and changing potential, without adding force vectors or kinetic-energy conclusions.]

The first path is the important one: the particle changes position, but not electric potential.

There is a useful term for a surface or line on which every point has the same electric potential.

An **equipotential surface** (or equipotential line in a two-dimensional diagram) is a region where the electric potential is constant.

Moving along an equipotential gives:

\[
\Delta V=0
\]

and therefore:

\[
\Delta U=q(0)=0
\]

This is the physical explanation behind the mathematical result.

## 2. Why the direction of movement matters

You might predict that any movement through an electric field must change a charged particle's potential energy. That sounds reasonable because an electric field exerts a force on the particle.

However, a force does not automatically do work. Work depends on the component of displacement in the direction of the force.

In an electrostatic field, the electric field points perpendicular to equipotential surfaces. The electric force on a particle is:

\[
\vec F=q\vec E
\]

where \(\vec F\) is the electric force in newtons (N), and \(\vec E\) is the electric field in newtons per coulomb (N/C).

If the particle moves along an equipotential, its displacement is perpendicular to the electric force. The electric force therefore does no work along that displacement.

The work done by the electric field is related to the change in potential energy by:

\[
W_{\text{electric}}=-\Delta U
\]

Consequently, if \(\Delta U=0\), the net work done by the electric field is also zero.

[IMAGE: Close-up vector diagram showing a positive and a negative charged particle at separate points on the same equipotential line. At each point, draw the electric field perpendicular to the equipotential and the corresponding electric force direction for each charge. Show a small tangential displacement along the equipotential. Make the opposite force directions for positive and negative charges clear.]

There is one important qualification. An equipotential describes the electric field, not necessarily the natural path of a freely moving particle.

The electric force generally pushes a particle perpendicular to an equipotential. To keep it travelling along that surface, an external constraint or another force may be needed. The statement that potential energy stays constant does not mean a particle will automatically follow an equipotential.

### The difference between zero change and constant energy throughout the journey

These two statements are not always equivalent:

- A particle starts and finishes at the same electric potential.
- A particle stays at the same electric potential throughout its journey.

The first guarantees that its net change in electric potential energy is zero. However, the particle could have crossed other potential values along the way.

The second means its electric potential energy remains unchanged at every point along the journey.

This distinction becomes useful when analysing curved paths or particles that move through several regions of an electric field.

## 3. Positive and negative charges: the sign matters

The equation \(\Delta U=q\Delta V\) becomes particularly important when the particle is negatively charged.

Electric potential is defined independently of the particle placed in the field. The potential at a point does not become negative just because an electron is there.

Instead, the sign of the charge determines how potential difference affects potential energy.

| Particle movement | Positive charge | Negative charge |
|---|---|---|
| Towards higher potential | Potential energy increases | Potential energy decreases |
| Towards lower potential | Potential energy decreases | Potential energy increases |
| Between equal potentials | No change | No change |

[IMAGE: Two side-by-side diagrams of a positive and negative charged particle moving between the same two labelled electric potentials, 0 V and +100 V. Use arrows to show movement from 0 V to +100 V in both cases. Annotate that the positive charge gains electric potential energy while the negative charge loses electric potential energy. Include the signed charge symbols, not just generic particles.]

For a positive charge, moving towards a higher electric potential increases its electric potential energy.

For a negative charge, the opposite happens. Moving towards a higher electric potential decreases its electric potential energy.

The easiest way to avoid confusion is to substitute the signed charge directly into the equation rather than trying to memorise separate rules.

With vertical plates, the constant-potential path is vertical, parallel to the plates. The path crossing equipotentials is horizontal, perpendicular to the plates.

A useful analogy is a topographic map. You can walk a long distance around a hillside while staying on the same contour line. Your position changes, but your altitude does not.

Equipotential lines work similarly, except they describe electric potential rather than height. There is one important difference: a positive charge’s potential energy increases with potential, while a negative charge’s potential energy decreases with potential.

04How can we test whether two points have the same potential?

We can investigate this by mapping equipotential lines using conductive paper and a voltmeter.

This experiment does not require us to move an actual charged particle. Instead, we measure the electric potential at different positions and use \(U=qV\) to predict the change in potential energy of a charge moved between those positions.

The experimental setup

Place a sheet of conductive paper on an insulating board. Attach two long, parallel metal strips to opposite sides of the paper.

Connect the strips to a low-voltage DC power supply. For example:

• Left electrode: \(+6.0\text{ V}\).
• Right electrode: \(0\text{ V}\).

Connect the negative or COM terminal of a digital multimeter to the \(0\text{ V}\) electrode. Use the positive probe to measure the potential at different positions on the conductive paper.

A grid drawn on the paper allows you to record the coordinates and voltage at each measurement point.

What should we predict?

In the central region between the long parallel electrodes, the electric field should be approximately uniform.

Therefore, points at the same horizontal distance from the electrodes should have approximately the same potential, even if their vertical positions differ.

If we move a charged particle between two such points, we predict:

\[ \Delta V=0 \]

and consequently:

\[ \Delta U=q\Delta V=0 \]

What would the measurements show?

The following is an illustrative dataset, not a record of an actual experiment.

Points A, B, and C have the same measured potential despite being separated in space.

Points D and E have different potentials because they are further across the electric field.

The observations support the prediction: moving vertically between A, B, and C produces no measurable change in electric potential.

However, moving horizontally from A towards D and E changes the potential.

A small calculation using the measurements

Suppose a particle with charge

\[ q=-2.0\times10^{-6}\text{ C} \]

moves from A to C.

Both points have a measured potential of \(4.2\text{ V}\).

The potential difference is:

\[ \Delta V=V_C-V_A \]

\[ \Delta V=4.2-4.2=0\text{ V} \]

Now calculate the change in electric potential energy:

\[ \begin{aligned} \Delta U&=q\Delta V\\\ &=(-2.0\times10^{-6})(0)\\\ &=\boxed{0\text{ J}} \end{aligned} \]

The particle has moved, but its electric potential energy has not changed.

A limitation of the experiment: A voltmeter has finite resolution. If both readings display \(4.2\text{ V}\) to the nearest \(0.1\text{ V}\), the actual potential difference could be as large as approximately \(0.1\text{ V}\).

For this charge, that means the magnitude of the potential energy change could be up to approximately:

\[ |\Delta U|<2.0\times10^{-7}\text{ J} \]

The experiment therefore supports the conclusion within its measurement uncertainty. It cannot establish mathematically exact equality from rounded readings alone.

What if the first experimental design does not work?

Imagine replacing the long parallel electrodes with two small metal pins.

The voltmeter still measures potential correctly, but the electric field now spreads out around the electrodes. The equipotential lines curve rather than remaining approximately straight.

This makes it harder to test the simple prediction that moving parallel to a plate produces no potential difference.

The solution is to revise the apparatus:

1. Use long, parallel electrodes rather than small point electrodes.
2. Take measurements in the central region, away from the ends.
3. Keep the power supply voltage constant.
4. Repeat measurements to check consistency.

The parallel electrodes reduce the effect of edge fringing in the region being investigated.

The experiment now provides a much clearer test of the relationship between movement, potential difference, and electric potential energy.

05Worked examples: applying the idea

The experiment establishes the central idea. Now let’s apply it to particle calculations.

06Three mistakes to avoid

Before attempting exam questions, make sure you can distinguish these situations.

There is another important distinction involving kinetic energy.

If electric potential energy does not change, the electric field does no net work:

\[ W_{\text{electric}}=0 \]

However, this does not automatically mean the particle’s kinetic energy cannot change.

The work-energy theorem states:

\[ W_{\text{net}}=\Delta K \]

If the electric force is the only force doing work, then zero electric work means zero change in kinetic energy.

If an external motor or another force does work on the particle, its kinetic energy can still change even though its electric potential energy stays constant.

04Questions and solutions

Question 1

A proton with charge \(+1.60\times10^{-19}\text{ C}\) moves from point A to point B.

Both points have an electric potential of \(-25\text{ V}\).

Calculate the change in the proton’s electric potential energy. Explain whether the distance travelled affects your answer.

Solution 1

The proton experiences no change in electric potential energy:

\[ \boxed{\Delta U=0\text{ J}} \]

The potential difference is:

\[ \Delta V=V_f-V_i \]

\[ \Delta V=-25-(-25)=0\text{ V} \]

Therefore:

\[ \begin{aligned} \Delta U&=q\Delta V\\\ &=(1.60\times10^{-19})(0)\\\ &=0\text{ J} \end{aligned} \]

The distance travelled does not affect the answer. Electric potential energy depends on the charge and the potential difference, not the distance between the two positions.

Even if the proton travels a metre, the change in electric potential energy is still zero if the initial and final potentials are equal.

Question 2

Two large parallel plates are separated by \(0.12\text{ m}\). The left plate is maintained at \(+600\text{ V}\), while the right plate is at \(0\text{ V}\).

An electron moves from point A to point B.

• Point A is \(0.040\text{ m}\) from the left plate.
• Point B is \(0.080\text{ m}\) from the left plate.
• Point B is also \(0.030\text{ m}\) lower than point A.

Assume the electric field is uniform between the plates and neglect edge effects.

Calculate:

a. The electric field strength between the plates.

b. The potential difference between A and B.

c. The change in the electron’s electric potential energy.

Use \(q=-1.60\times10^{-19}\text{ C}\).

Solution 2

The electric field strength is \(5.0\times10^3\text{ N C}^{-1}\), the potential difference is \(-200\text{ V}\), and the electron’s electric potential energy increases by \(3.2\times10^{-17}\text{ J}\).

Step 1: Calculate the electric field strength.

For a uniform electric field between parallel plates:

\[ E=\frac{|\Delta V_{\text{plates}}|}{d} \]

Substitute:

\[ E=\frac{600}{0.12} \]

\[ \boxed{E=5.0\times10^3\text{ N C}^{-1}} \]

Step 2: Calculate the potential difference between A and B.

The electric field points from the positive plate towards the negative plate, so potential decreases as we move to the right.

The horizontal displacement is:

\[ \Delta x=0.080-0.040=0.040\text{ m} \]

Therefore:

\[ \Delta V=-E\Delta x \]

\[ \Delta V=-(5.0\times10^3)(0.040) \]

\[ \boxed{\Delta V=-200\text{ V}} \]

Notice that the vertical displacement does not affect the potential difference.

This is because the ideal uniform electric field is horizontal. Moving vertically means moving along an equipotential.

Step 3: Calculate the change in electric potential energy.

\[ \Delta U=q\Delta V \]

\[ \begin{aligned} \Delta U&=(-1.60\times10^{-19})(-200)\\\ &=\boxed{+3.2\times10^{-17}\text{ J}} \end{aligned} \]

The electron gains electric potential energy.

The tempting mistake is to assume that the electron loses potential energy because it moves towards a lower electric potential. That would be correct for a positive charge, but the electron has negative charge. Both signs must be included.

Question 3

A negatively charged particle with charge \(-2.0\times10^{-6}\text{ C}\) is transported slowly through a static electric field.

Its path passes through three points:

• A: \(30\text{ V}\)
• B: \(90\text{ V}\)
• C: \(30\text{ V}\)

The particle travels from A to B and then from B to C.

Calculate:

a. The change in electric potential energy from A to B.

b. The change in electric potential energy from B to C.

c. The total change in electric potential energy from A to C.

d. The total work done by the electric field.

Explain whether the particle’s potential energy remains constant throughout the journey.

Solution 3

The particle first loses \(1.2\times10^{-4}\text{ J}\) of electric potential energy, then gains the same amount. Its total change in electric potential energy is zero, and the total work done by the electric field is also zero.

Step 1: Calculate the change from A to B.

\[ \Delta V_{AB}=90-30=+60\text{ V} \]

Using:

\[ \Delta U=q\Delta V \]

\[ \begin{aligned} \Delta U_{AB} &=(-2.0\times10^{-6})(60)\\\ &=-1.2\times10^{-4}\text{ J} \end{aligned} \]

Therefore:

\[ \boxed{\Delta U_{AB}=-1.2\times10^{-4}\text{ J}} \]

The particle loses electric potential energy.

Step 2: Calculate the change from B to C.

\[ \Delta V_{BC}=30-90=-60\text{ V} \]

Therefore:

\[ \begin{aligned} \Delta U_{BC} &=(-2.0\times10^{-6})(-60)\\\ &=+1.2\times10^{-4}\text{ J} \end{aligned} \]

So:

\[ \boxed{\Delta U_{BC}=+1.2\times10^{-4}\text{ J}} \]

The particle gains electric potential energy.

Step 3: Calculate the total change.

The initial and final potentials are both \(30\text{ V}\).

Therefore:

\[ \Delta U_{AC}=q(V_C-V_A) \]

\[ \Delta U_{AC}=(-2.0\times10^{-6})(30-30) \]

\[ \boxed{\Delta U_{AC}=0\text{ J}} \]

Step 4: Calculate the total work done by the electric field.

The relationship is:

\[ W_{\text{electric}}=-\Delta U \]

Therefore:

\[ \boxed{W_{\text{electric, total}}=0\text{ J}} \]

The electric field first does positive work and then negative work.

The particle’s electric potential energy does not remain constant throughout the journey. It changes at B, even though it returns to its original value at C.

This is the important distinction: zero net change does not mean no intermediate change.

Question 4

An electron is launched vertically upwards between two large parallel plates.

The electric field is uniform, has magnitude \(2.0\times10^4\text{ N C}^{-1}\), and points horizontally to the right.

The electron’s initial velocity is \(1.5\times10^6\text{ m s}^{-1}\), parallel to the plates.

After \(2.0\text{ ns}\), calculate:

a. The electron’s horizontal displacement.

b. The potential difference between its initial and final positions.

c. Its change in electric potential energy.

d. Whether the electron has remained on the same equipotential.

Assume the field extends across the entire region of motion, and neglect gravity.

Use:

• \(q=-1.60\times10^{-19}\text{ C}\)
• \(m=9.11\times10^{-31}\text{ kg}\)

Solution 4

The electron moves approximately \(7.0\text{ mm}\) to the left, its electric potential increases by approximately \(140\text{ V}\), and its electric potential energy decreases by \(2.25\times10^{-17}\text{ J}\). It does not remain on the same equipotential.

Step 1: Determine the electron’s acceleration.

The electric force is:

\[ F=qE \]

Newton’s second law gives:

\[ a=\frac{qE}{m} \]

Substitute:

\[ a= \frac{(-1.60\times10^{-19})(2.0\times10^4)} {9.11\times10^{-31}} \]

\[ a=-3.51\times10^{15}\text{ m s}^{-2} \]

The negative sign means the acceleration is to the left, opposite to the electric field.

Step 2: Calculate the horizontal displacement.

The initial horizontal velocity is zero.

Using:

\[ \Delta x=u_xt+\frac12at^2 \]

we obtain:

\[ \begin{aligned} \Delta x &=0+\frac12(-3.51\times10^{15})(2.0\times10^{-9})^2\\\ &=-7.02\times10^{-3}\text{ m} \end{aligned} \]

Therefore:

\[ \boxed{\Delta x=-7.0\text{ mm}} \]

The electron moves to the left.

Step 3: Calculate the potential difference.

The electric field points right, so moving left increases electric potential.

Using the uniform-field relationship:

\[ \Delta V=-E\Delta x \]

\[ \Delta V=-(2.0\times10^4)(-7.02\times10^{-3}) \]

\[ \boxed{\Delta V\approx+140\text{ V}} \]

Step 4: Calculate the change in electric potential energy.

\[ \Delta U=q\Delta V \]

\[ \begin{aligned} \Delta U &=(-1.60\times10^{-19})(140)\\\ &=\boxed{-2.25\times10^{-17}\text{ J}} \end{aligned} \]

The electron loses electric potential energy.

Step 5: Interpret the motion.

The electron initially travels parallel to an equipotential. However, the electric force acts horizontally to the left.

As a result, the electron accelerates across the equipotential lines rather than remaining on one.

The initial velocity being perpendicular to the electric field does not prevent the particle from accelerating in the direction of the electric force.

Its final speed is approximately \(7.2\times10^6\text{ m s}^{-1}\), which is about \(2.4\%\) of the speed of light. The non-relativistic approximation is therefore reasonable.

Question 5

A proton and an electron enter an ideal frictionless guide at point A. The guide takes each particle through the same route:

\[ A\longrightarrow B\longrightarrow C\longrightarrow D \]

The electric potential at each point is:

The potential changes smoothly and monotonically between consecutive points.

Both particles enter at A with an initial kinetic energy of \(100\text{ eV}\).

Assume:

• The proton has charge \(+e\).
• The electron has charge \(-e\).
• The guide exerts only a normal force and does no work.
• The electric field is static.
• No other forces do work on either particle.

Use \(1\text{ eV}=1.60\times10^{-19}\text{ J}\).

Answer the following:

a. Calculate the proton’s kinetic energy at B, C, and D.

b. Determine whether the electron can reach B.

c. If the electron cannot reach B, determine the potential at which it stops.

d. Explain why the fact that A and D have equal potentials does not guarantee that both particles can complete the journey.

Solution 5

The proton reaches D with its original kinetic energy of \(100\text{ eV}\). The electron cannot reach B: it stops when the potential reaches \(-100\text{ V}\), then reverses direction.

The key is that equal initial and final potentials guarantee zero net change in potential energy only if the particle actually reaches the final point. The intermediate potential values can prevent it from getting there.

Step 1: Calculate the proton’s kinetic energy at B.

For a proton:

\[ \Delta U=q\Delta V \]

From A to B:

\[ \Delta V=-120-0=-120\text{ V} \]

Therefore:

\[ \Delta U=(+e)(-120)=-120\text{ eV} \]

The proton loses \(120\text{ eV}\) of electric potential energy.

Since the guide does no work:

\[ \Delta K=-\Delta U \]

Therefore:

\[ K_B=100+120 \]

\[ \boxed{K_B=220\text{ eV}} \]

The proton gains kinetic energy as it moves towards B.

Step 2: Calculate the proton’s kinetic energy at C.

From A to C, the potential difference is:

\[ \Delta V=80-0=+80\text{ V} \]

Therefore:

\[ \Delta U=(+e)(80)=+80\text{ eV} \]

The proton has gained \(80\text{ eV}\) of electric potential energy relative to its starting point.

Its kinetic energy is:

\[ K_C=100-80 \]

\[ \boxed{K_C=20\text{ eV}} \]

The proton slows down as it moves from B towards C, but it still has positive kinetic energy.

It can therefore reach C.

Step 3: Calculate the proton’s kinetic energy at D.

Point D is at the same potential as A.

Therefore:

\[ \Delta V=0-0=0\text{ V} \]

and:

\[ \Delta U=0\text{ eV} \]

Consequently:

\[ \boxed{K_D=100\text{ eV}} \]

The proton reaches D with the same kinetic energy it had at A.

Its kinetic energy changed substantially during the journey, but its final kinetic energy returned to its initial value.

Step 4: Determine whether the electron reaches B.

For the electron:

\[ q=-e \]

From A to B:

\[ \Delta U=(-e)(-120)=+120\text{ eV} \]

The electron would need an additional \(120\text{ eV}\) of energy to reach B.

However, it only starts with \(100\text{ eV}\) of kinetic energy.

Its predicted kinetic energy at B would be:

\[ K_B=100-120=-20\text{ eV} \]

Negative kinetic energy is impossible in this situation.

Therefore:

\[ \boxed{\text{The electron cannot reach B.}} \]

Step 5: Find the electron’s turning potential.

At the turning point, its instantaneous kinetic energy is zero.

Conservation of mechanical energy gives:

\[ K_i+\Delta U=0 \]

For the electron:

\[ 100+(-1)(V-0)=0 \]

Here, energies are expressed in electronvolts and potentials in volts.

Rearranging:

\[ V=100\text{ V} \]

Wait – this would give the wrong turning direction if the charge sign is not applied consistently. Using the signed charge correctly:

\[ K_i-\Delta U=0 \]

and:

\[ \Delta U=-e\Delta V \]

Therefore, in electronvolts:

\[ 100-(-V)=0 \]

\[ 100+V=0 \]

Hence:

\[ \boxed{V=-100\text{ V}} \]

The electron stops when the potential has fallen to \(-100\text{ V}\). It then reverses direction and returns towards A.

Step 6: Explain the central distinction.

A and D have the same electric potential, so a particle that completes the journey has no net change in electric potential energy.

For the proton, the intermediate changes are manageable. It gains kinetic energy on the way to B and loses some of it on the way to C.

For the electron, the potential drop towards B creates an energy barrier. Its initial kinetic energy is insufficient to overcome that barrier.

The tempting argument is that equal starting and ending potentials guarantee the same outcome for every particle. They do not.

The endpoint relationship tells us the net energy change. The intermediate potentials determine whether the particle can actually reach the endpoint.

This is a useful way to approach more demanding electric-field questions: never check only the initial and final positions if the particle must pass through intermediate regions.

07Where to take this next

When you encounter a charged-particle problem, separate three questions:

1. What are the initial and final electric potentials?
2. What is the signed charge, and therefore the change in electric potential energy?
3. Can the particle actually follow the proposed path, given its kinetic energy and any intermediate changes in potential?

For a particle moving along an equipotential, the electric potential energy stays constant. For a particle crossing equipotentials, use the signed relationship \(\Delta U=q\Delta V\) to track the energy change.

The next step is connecting that energy change to the particle’s motion. The relationship between electric force, acceleration, and trajectory is developed in Charged-Particle Trajectories in Electric Fields: HSC Physics.

For a broader treatment of voltage, work, and energy conservation, see Electric Potential Energy and Work in Electric Fields.