How to Calculate Work Done by an External Agent Moving a Charge

Imagine moving a positively charged particle between two points in an electric field. You know its charge and the potential difference between the points, but you haven’t measured the electric force at every position. How can you calculate the work done by the external agent?

If the charge is moved slowly, so its kinetic energy remains approximately constant, the answer is:

\[ \boxed{W_{\mathrm{ext}}=q\Delta V=q(V_f-V_i)} \]

Here, \(q\) is the signed charge, \(V_i\) is the initial electric potential, and \(V_f\) is the final electric potential.

The important detail is that you must keep the sign of the charge. A positive charge and a negative charge can move through exactly the same potential difference while requiring opposite amounts of work from the external agent.

Let’s build the method and work through the cases that commonly cause mistakes.

01The equation for work done by an external agent

Start with the relationship between electric potential and electric potential energy.

Electric potential, measured in volts (V), describes the electric potential energy available per coulomb of charge at a particular location.

For a charge \(q\), its electric potential energy is:

\[ U=qV \]

Therefore, when the charge moves between two positions, its change in electric potential energy is:

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

We define the potential difference as:

\[ \Delta V=V_f-V_i \]

This means:

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

The charge’s potential energy changes because it has moved to a location with a different electric potential.

Why does the external agent do this much work?

Imagine pushing a positive charge towards the positive plate of a parallel-plate arrangement.

The electric field pushes the charge towards the negative plate. Your external force must act in the opposite direction to move the charge towards the positive plate.

If you move the charge slowly, its kinetic energy doesn’t change significantly. The work done by the external agent must therefore balance the work done by the electric field.

We can express this using the work-energy theorem:

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

If the only forces doing work are the external force and the electric force:

\[ W_{\mathrm{ext}}+W_{\mathrm{electric}}=\Delta K \]

The work done by the electric field is the negative of the change in electric potential energy:

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

Substituting:

\[ W_{\mathrm{ext}}-\Delta U=\Delta K \]

Rearranging:

\[ W_{\mathrm{ext}}=\Delta U+\Delta K \]

For slow movement with negligible change in kinetic energy:

\[ \Delta K\approx0 \]

Therefore:

\[ \boxed{W_{\mathrm{ext}}=q\Delta V} \]

This is the equation you need for the standard HSC question.

Important assumption: Other forces, such as friction, are assumed to do negligible work. If additional forces do work, their contributions must also be included in the energy calculation.

02Getting the signs right

This is where students often lose marks.

You might remember that moving a charge towards a higher potential increases its potential energy. That’s true for a positive charge, but not for a negative charge.

The safest approach is to use the signed value of \(q\) in the equation rather than trying to memorise separate rules.

Consider the following situations.

Let’s interpret these results.

Positive work by the external agent

If \(W_{\mathrm{ext}}>0\), the external agent transfers energy to the charge-field system.

For example, moving a positive charge slowly towards a positively charged plate requires the external agent to push against the electric force.

The charge gains electric potential energy.

Negative work by the external agent

If \(W_{\mathrm{ext}}<0\), the external agent does negative work.

This means energy is transferred from the charge-field system to the external agent. The agent might be restraining the particle rather than supplying energy.

For example, a negative charge naturally experiences an electric force towards higher electric potential. If you move it in that direction slowly, the external agent must oppose its motion.

The difference between external work and electric work

Be careful not to confuse these two quantities.

The work done by the electric field is:

\[ W_{\mathrm{electric}}=-q\Delta V \]

The work done by the external agent during slow movement is:

\[ W_{\mathrm{ext}}=q\Delta V \]

They have equal magnitudes and opposite signs when the kinetic energy remains constant.

If the electric field does positive work, the external agent must do negative work to maintain the slow movement.

03How the potential difference method connects to force and displacement

You might already know the constant-force work equation:

\[ W=Fs\cos\theta \]

This calculates the work done by a force acting through a displacement, where \(F\) is force magnitude, \(s\) is displacement, and \(\theta\) is the angle between them.

You can read more about this approach in Work Done by a Constant Force: HSC Physics Guide.

So why use potential difference instead?

Because calculating the force at every point along a particle’s path can be difficult. The potential difference gives you the total change in electric potential energy without requiring you to track every force along the way.

For a uniform electric field between parallel plates, the relationship is particularly easy to see.

The electric field magnitude is:

\[ E=\frac{|\Delta V|}{d} \]

where:

• \(E\) is electric field strength in newtons per coulomb (N/C).
• \(|\Delta V|\) is the magnitude of the potential difference in volts.
• \(d\) is the perpendicular separation between the plates in metres.

The electric force magnitude on a charge is:

\[ F=|q|E \]

For movement directly against the electric force through the full plate separation:

\[ W_{\mathrm{ext}}=Fd \]

Substituting:

\[ W_{\mathrm{ext}}=|q|Ed \]

And using \(Ed=|\Delta V|\):

\[ W_{\mathrm{ext}}=|q||\Delta V| \]

This gives the magnitude of the work. The direction of movement and the sign of the charge determine whether the actual work is positive or negative.

The potential difference method is more general for electrostatic fields. It works even when the field strength varies along the path.

How could we experimentally test this relationship?

Imagine an idealised experiment using two parallel metal plates.

Connect the plates to a power supply so that one plate is at 0 V and the other is at +300 V. A voltmeter measures the potential difference across them.

Place a small positively charged conducting bead between the plates. Attach it to an insulating rod connected to a force sensor and a motorised stage.

The stage moves the bead slowly from the 0 V plate towards the +300 V plate.

What should we expect?

The positive bead experiences an electric force towards the lower-potential plate. The external agent must therefore apply a force in the direction of the bead’s displacement.

If the field is uniform and the bead moves slowly, the two methods should produce the same work.

Consider these illustrative values, not measurements from an actual experiment:

First, calculate the electric force:

\[ \begin{aligned} F_{\mathrm{electric}}&=|q|E\\ &=(2.0\times10^{-6})(3.0\times10^3)\\ &=6.0\times10^{-3}\text{ N} \end{aligned} \]

The external agent must apply an equal and opposite force to maintain slow movement.

The work calculated from force and displacement is:

\[ \begin{aligned} W_{\mathrm{ext}}&=Fs\\ &=(6.0\times10^{-3})(0.10)\\ &=6.0\times10^{-4}\text{ J} \end{aligned} \]

Now calculate the work using potential difference:

\[ \begin{aligned} W_{\mathrm{ext}}&=q\Delta V\\ &=(2.0\times10^{-6})(300-0)\\ &=6.0\times10^{-4}\text{ J} \end{aligned} \]

Both methods give:

\[ \boxed{W_{\mathrm{ext}}=6.0\times10^{-4}\text{ J}} \]

The agreement is exactly what the energy model predicts.

There is one important experimental limitation. If the motorised stage moves the bead too quickly and its kinetic energy changes, the external work will no longer equal \(q\Delta V\) alone.

A better design controls the movement so the initial and final speeds are equal, or measures the change in kinetic energy and includes it in the calculation.

This is why the slow-movement condition matters.

04Worked examples

05Does the speed of the particle matter?

There are two different ideas to keep separate.

Slow movement means the particle’s kinetic energy remains approximately constant. This is the condition that allows us to use:

\[ W_{\mathrm{ext}}=q\Delta V \]

Non-relativistic movement means the particle’s speed is much smaller than the speed of light.

When calculating kinetic energy using:

\[ K=\frac12mv^2 \]

you are using the non-relativistic approximation.

It is appropriate when:

\[ v\ll c \]

where \(c=3.00\times10^8\text{ m s}^{-1}\) is the speed of light.

As a useful guide, at approximately \(0.1c\), the classical kinetic energy is still within about 1% of the relativistic value. There is no sharp cutoff, but the approximation becomes increasingly inaccurate as the speed approaches \(c\).

At relativistic speeds, the kinetic energy must instead be calculated using:

\[ K=(\gamma-1)mc^2 \]

where:

\[ \gamma=\frac{1}{\sqrt{1-v^2/c^2}} \]

This is an extension beyond the usual non-relativistic calculation.

One final distinction is important: the relationship

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

does not depend on using the classical kinetic energy equation. It describes the change in electric potential energy in an electrostatic field.

A particle can therefore move very quickly and still obey the same potential-energy relationship.

06A reliable method for HSC questions

When you see a question asking for the work done by an external agent moving a charge slowly, follow this process.

1. Identify the signed charge. A proton is positive, while an electron is negative.
2. Identify the initial and final electric potentials.
3. Calculate the potential difference using \(V_f-V_i\).
4. Confirm that the movement is slow, or that the initial and final kinetic energies are equal.
5. Substitute into \(W_{\mathrm{ext}}=q\Delta V\).
6. Include the sign and joule unit in your answer.
7. Explain whether the external agent supplies or removes energy.

Three common mistakes are worth watching for:

Remember:

\[ \boxed{W_{\mathrm{ext}}=q(V_f-V_i)} \]

for slow movement with negligible change in kinetic energy.

For more on what happens when a charge moves between points at the same potential, see Can a Charged Particle Move Without Changing Electric Potential Energy?.

07Questions and solutions

Question 1

A proton with charge \(+1.60\times10^{-19}\text{ C}\) is moved slowly from point A, at an electric potential of 120 V, to point B, at an electric potential of 470 V.

Calculate the work done by the external agent.

Solution 1

The external agent does \(+5.60\times10^{-17}\text{ J}\) of work.

First, calculate the potential difference:

\[ \begin{aligned} \Delta V&=V_f-V_i\\ &=470-120\\ &=350\text{ V} \end{aligned} \]

Since the proton is positively charged:

\[ q=+1.60\times10^{-19}\text{ C} \]

Apply:

\[ W_{\mathrm{ext}}=q\Delta V \]

\[ \begin{aligned} W_{\mathrm{ext}} &=(1.60\times10^{-19})(350)\\ &=5.60\times10^{-17}\text{ J} \end{aligned} \]

The answer is positive because the external agent must supply energy to move a positive charge towards a higher electric potential.

Question 2

A negatively charged ion with charge \(-3.0\,\mu\text{C}\) is moved slowly through an electrostatic field.

It travels along the route A → B → C, where:

• Point A has a potential of \(+150\text{ V}\).
• Point B has a potential of \(-50\text{ V}\).
• Point C has a potential of \(+250\text{ V}\).

The distance travelled from A to B is 0.40 m, and the distance from B to C is 0.25 m.

Calculate:

(a) The work done by the external agent from A to B.

(b) The work done by the external agent from B to C.

(c) The total work done by the external agent from A to C.

(d) Whether the total work would change if the particle travelled directly from A to C along a different path of length 0.12 m.

Solution 2

The external agent does \(+6.0\times10^{-4}\text{ J}\) from A to B, \(-9.0\times10^{-4}\text{ J}\) from B to C, and \(-3.0\times10^{-4}\text{ J}\) overall.

The important idea is that electrostatic work depends on the potential difference between the endpoints, not the distance travelled.

(a) Work from A to B

Calculate the potential difference:

\[ \begin{aligned} \Delta V_{AB}&=V_B-V_A\\ &=-50-150\\ &=-200\text{ V} \end{aligned} \]

Use the signed charge:

\[ q=-3.0\times10^{-6}\text{ C} \]

Therefore:

\[ \begin{aligned} W_{\mathrm{ext},AB} &=q\Delta V_{AB}\\ &=(-3.0\times10^{-6})(-200)\\ &=+6.0\times10^{-4}\text{ J} \end{aligned} \]

The external agent does positive work.

(b) Work from B to C

Calculate the new potential difference:

\[ \begin{aligned} \Delta V_{BC}&=V_C-V_B\\ &=250-(-50)\\ &=+300\text{ V} \end{aligned} \]

Therefore:

\[ \begin{aligned} W_{\mathrm{ext},BC} &=(-3.0\times10^{-6})(300)\\ &=-9.0\times10^{-4}\text{ J} \end{aligned} \]

The external agent does negative work.

This makes sense because the negative charge is moving towards a higher potential, in the direction of its electric force.

(c) Total work from A to C

Add the work done on each segment:

\[ \begin{aligned} W_{\mathrm{ext,total}} &=W_{\mathrm{ext},AB}+W_{\mathrm{ext},BC}\\ &=6.0\times10^{-4}-9.0\times10^{-4}\\ &=-3.0\times10^{-4}\text{ J} \end{aligned} \]

(d) Does a different path change the work?

No. The total work remains \(-3.0\times10^{-4}\text{ J}\).

Calculate directly from A to C:

\[ \begin{aligned} \Delta V_{AC}&=250-150\\ &=100\text{ V} \end{aligned} \]

\[ \begin{aligned} W_{\mathrm{ext},AC} &=(-3.0\times10^{-6})(100)\\ &=-3.0\times10^{-4}\text{ J} \end{aligned} \]

The result is identical.

The tempting mistake is to assume that the longer path requires more work. That would be relevant if the force and displacement information changed the energy transferred, but in an electrostatic field, the total work depends only on the initial and final potentials.

The intermediate potential at B changes the work on each segment, but it cannot change the total between A and C.

Question 3

An idealised experiment uses a positively charged bead with:

• Charge \(q=+4.0\times10^{-6}\text{ C}\).
• Mass \(m=2.0\times10^{-3}\text{ kg}\).

The bead moves between two parallel plates separated by 0.080 m.

The initial point A is at \(+20\text{ V}\), and the final point B is at \(+260\text{ V}\).

During a slow run, a force sensor records an average external force of \(0.012\text{ N}\), parallel to the displacement.

(a) Calculate the work done by the external agent using the potential difference method.

(b) Verify your answer using the measured force and displacement.

A second run uses the same initial and final positions, but the bead’s speed increases from \(0.10\text{ m s}^{-1}\) to \(0.30\text{ m s}^{-1}\).

(c) Calculate the work done by the external agent during the second run, assuming no other force does work.

(d) Explain why the second run gives a different result from the slow run.

Solution 3

The external agent does \(9.6\times10^{-4}\text{ J}\) of work during the slow run and \(1.04\times10^{-3}\text{ J}\) during the second run.

The difference comes from the change in kinetic energy during the faster movement.

(a) Calculate work using potential difference

First:

\[ \begin{aligned} \Delta V&=V_f-V_i\\ &=260-20\\ &=240\text{ V} \end{aligned} \]

Now substitute:

\[ \begin{aligned} W_{\mathrm{ext}}&=q\Delta V\\ &=(4.0\times10^{-6})(240)\\ &=9.6\times10^{-4}\text{ J} \end{aligned} \]

(b) Verify using force and displacement

Since the external force is parallel to the displacement:

\[ W_{\mathrm{ext}}=Fs \]

Substitute:

\[ \begin{aligned} W_{\mathrm{ext}}&=(0.012)(0.080)\\ &=9.6\times10^{-4}\text{ J} \end{aligned} \]

Both methods agree.

This supports the expected relationship for slow movement in an electrostatic field.

(c) Calculate work during the second run

The kinetic energy changes, so use:

\[ W_{\mathrm{ext}}=q\Delta V+\Delta K \]

Calculate the change in kinetic energy:

\[ \Delta K=\frac12m(v_f^2-v_i^2) \]

Substitute:

\[ \begin{aligned} \Delta K &=\frac12(2.0\times10^{-3})(0.30^2-0.10^2)\\ &=(1.0\times10^{-3})(0.08)\\ &=8.0\times10^{-5}\text{ J} \end{aligned} \]

Now calculate the external work:

\[ \begin{aligned} W_{\mathrm{ext}} &=9.6\times10^{-4}+8.0\times10^{-5}\\ &=1.04\times10^{-3}\text{ J} \end{aligned} \]

(d) Explain the difference

The second run produces more external work because the bead finishes with greater kinetic energy.

During the slow run, the change in kinetic energy is negligible, so the external work only accounts for the change in electric potential energy.

During the second run, the external agent must supply energy for both the increase in electric potential energy and the increase in kinetic energy.

The mistake would be to assume that having the same initial and final positions always means the external work must be identical. That is only true when the kinetic energy change and other work contributions are also the same.

Question 4

A proton starts at rest at point A, where the electric potential is 0 V.

An external agent moves it slowly to point B, where the electric potential is \(+2.0\times10^6\text{ V}\).

The proton is then released from rest at B and accelerates back towards A under the influence of the electric field alone.

Assume:

• Proton charge: \(+1.60\times10^{-19}\text{ C}\).
• Proton mass: \(1.67\times10^{-27}\text{ kg}\).
• No energy is lost through collisions or other dissipative effects.

Calculate:

(a) The work done by the external agent while moving the proton from A to B.

(b) The work done by the electric field as the proton travels back to A.

(c) The proton’s final speed at A using non-relativistic kinetic energy.

(d) Whether the non-relativistic approximation is appropriate.

Solution 4

The external agent does \(+3.2\times10^{-13}\text{ J}\) of work moving the proton to B. The electric field then does \(+3.2\times10^{-13}\text{ J}\) of work as the proton returns to A, giving it a final speed of approximately \(1.96\times10^7\text{ m s}^{-1}\).

(a) Work done by the external agent

Calculate the potential difference:

\[ \Delta V=2.0\times10^6-0 \]

\[ \Delta V=+2.0\times10^6\text{ V} \]

Apply:

\[ W_{\mathrm{ext}}=q\Delta V \]

\[ \begin{aligned} W_{\mathrm{ext}} &=(1.60\times10^{-19})(2.0\times10^6)\\ &=3.2\times10^{-13}\text{ J} \end{aligned} \]

The external agent does positive work because it moves the positive proton against the electric force.

(b) Work done by the electric field on the return journey

The proton now moves from B to A.

The potential difference is:

\[ \begin{aligned} \Delta V&=V_A-V_B\\ &=0-2.0\times10^6\\ &=-2.0\times10^6\text{ V} \end{aligned} \]

The work done by the electric field is:

\[ W_{\mathrm{electric}}=-q\Delta V \]

\[ \begin{aligned} W_{\mathrm{electric}} &=-(1.60\times10^{-19})(-2.0\times10^6)\\ &=+3.2\times10^{-13}\text{ J} \end{aligned} \]

The electric field does positive work, transferring electric potential energy into kinetic energy.

(c) Calculate the final speed

The proton starts its return journey from rest.

Therefore, its initial kinetic energy is zero, and its final kinetic energy equals the work done by the electric field:

\[ \Delta K=3.2\times10^{-13}\text{ J} \]

Using:

\[ K=\frac12mv^2 \]

Rearrange for speed:

\[ v=\sqrt{\frac{2K}{m}} \]

Substitute:

\[ \begin{aligned} v&=\sqrt{\frac{2(3.2\times10^{-13})}{1.67\times10^{-27}}}\\ &=\sqrt{3.83\times10^{14}}\\ &=1.96\times10^7\text{ m s}^{-1} \end{aligned} \]

(d) Check the approximation

Compare the calculated speed with the speed of light:

\[ \frac{v}{c} = \frac{1.96\times10^7}{3.00\times10^8} \]

\[ \frac{v}{c}\approx0.065 \]

The proton travels at approximately \(6.5\%\) of the speed of light.

This is sufficiently below \(c\) for the non-relativistic kinetic energy approximation to be appropriate here.

The key distinction is that the potential-energy calculation works independently of the speed approximation. The non-relativistic assumption only enters when converting the kinetic energy into a speed.

Question 5

A negatively charged particle moves through a static electrostatic field along a guided path connecting three points, A, B, and C.

Its properties are:

• Charge: \(q=-2.0\times10^{-6}\text{ C}\).
• Mass: \(m=4.0\times10^{-9}\text{ kg}\).
• Potential at A: \(+100\text{ V}\).
• Potential at B: \(-200\text{ V}\).
• Potential at C: \(+400\text{ V}\).

The particle initially rests at A.

The following sequence occurs:

1. An external agent moves the particle slowly from A to B.
2. At B, the agent releases the particle. It travels freely under the electric force from B to C.
3. At C, the agent catches the particle and brings it to rest without changing its position.
4. The agent then moves the particle slowly from C back to A.

Assume the guiding force is perpendicular to the particle’s motion, the field is electrostatic, and no friction or other force does work.

Calculate:

(a) The work done by the external agent from A to B.

(b) The work done by the electric field from B to C.

(c) The speed of the particle when it reaches C.

(d) The work done by the external agent while bringing the particle to rest at C.

(e) The work done by the external agent while moving the particle slowly from C to A.

(f) The total work done by the external agent over the entire journey.

Explain why the external agent’s work on each individual stage does not have to be zero, even though the particle finishes at its starting point with its original speed.

Solution 5

The total work done by the external agent over the complete journey is zero, but the individual stages involve positive and negative work.

The crucial detail is that the particle gains kinetic energy during the free movement from B to C. The agent must remove this energy when catching it at C.

We need to track electric potential energy and kinetic energy separately.

(a) Work done from A to B

The particle moves slowly, so its kinetic energy remains approximately constant.

Calculate the potential difference:

\[ \begin{aligned} \Delta V_{AB} &=V_B-V_A\\ &=-200-100\\ &=-300\text{ V} \end{aligned} \]

The charge is negative:

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

Therefore:

\[ \begin{aligned} W_{\mathrm{ext},AB} &=q\Delta V_{AB}\\ &=(-2.0\times10^{-6})(-300)\\ &=+6.0\times10^{-4}\text{ J} \end{aligned} \]

The external agent supplies energy to move the negative charge towards a lower potential.

(b) Work done by the electric field from B to C

The particle is released at B, so the external agent does no work during this part of the journey.

Calculate the potential difference:

\[ \begin{aligned} \Delta V_{BC} &=V_C-V_B\\ &=400-(-200)\\ &=+600\text{ V} \end{aligned} \]

The change in electric potential energy is:

\[ \begin{aligned} \Delta U_{BC} &=q\Delta V_{BC}\\ &=(-2.0\times10^{-6})(600)\\ &=-1.2\times10^{-3}\text{ J} \end{aligned} \]

The work done by the electric field is:

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

Therefore:

\[ \boxed{W_{\mathrm{electric},BC}=+1.2\times10^{-3}\text{ J}} \]

The electric field transfers \(1.2\times10^{-3}\text{ J}\) into the particle’s kinetic energy.

(c) Calculate the speed at C

The particle starts from rest at B.

Since the external agent does no work between B and C:

\[ \Delta K=W_{\mathrm{electric}} \]

Therefore:

\[ K_C=1.2\times10^{-3}\text{ J} \]

Use the kinetic energy equation:

\[ K=\frac12mv^2 \]

Rearrange:

\[ v=\sqrt{\frac{2K}{m}} \]

Substitute:

\[ \begin{aligned} v&=\sqrt{\frac{2(1.2\times10^{-3})}{4.0\times10^{-9}}}\\ &=\sqrt{6.0\times10^5}\\ &\approx7.75\times10^2\text{ m s}^{-1} \end{aligned} \]

Therefore:

\[ \boxed{v_C\approx775\text{ m s}^{-1}} \]

The particle reaches C with significant kinetic energy, even though it started from rest.

(d) Work done while stopping the particle at C

This is the important part.

The agent catches the particle and brings it to rest without changing its position.

Because the particle does not move, its electric potential energy does not change:

\[ \Delta U=0 \]

However, its kinetic energy decreases from \(1.2\times10^{-3}\text{ J}\) to zero.

Therefore:

\[ \Delta K=0-1.2\times10^{-3} \]

\[ \Delta K=-1.2\times10^{-3}\text{ J} \]

Using the work-energy theorem:

\[ W_{\mathrm{ext}}=\Delta K \]

Hence:

\[ \boxed{W_{\mathrm{ext,stop}}=-1.2\times10^{-3}\text{ J}} \]

The negative sign means the external agent removes energy from the particle.

This is the result students often miss. The agent does not simply apply \(q\Delta V\) when catching the particle because its kinetic energy is changing.

(e) Work done from C to A

After stopping the particle, the external agent moves it slowly back to A.

The kinetic energy remains constant during this movement.

Calculate the potential difference:

\[ \begin{aligned} \Delta V_{CA} &=V_A-V_C\\ &=100-400\\ &=-300\text{ V} \end{aligned} \]

Therefore:

\[ \begin{aligned} W_{\mathrm{ext},CA} &=q\Delta V_{CA}\\ &=(-2.0\times10^{-6})(-300)\\ &=+6.0\times10^{-4}\text{ J} \end{aligned} \]

So:

\[ \boxed{W_{\mathrm{ext},CA}=+6.0\times10^{-4}\text{ J}} \]

The external agent supplies energy to move the negative charge towards a lower potential.

(f) Total external work

Add the work done during all four stages:

\[ \begin{aligned} W_{\mathrm{ext,total}} ={}&W_{\mathrm{ext},AB}\\ &+W_{\mathrm{ext},BC}\\ &+W_{\mathrm{ext,stop}}\\ &+W_{\mathrm{ext},CA} \end{aligned} \]

Remember that the agent is not acting during the free movement from B to C:

\[ W_{\mathrm{ext},BC}=0 \]

Substitute:

\[ \begin{aligned} W_{\mathrm{ext,total}} ={}&6.0\times10^{-4}\\ &+0\\ &-1.2\times10^{-3}\\ &+6.0\times10^{-4}\\ ={}&0\text{ J} \end{aligned} \]

Therefore:

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

The external agent does zero net work over the complete journey.

Why?

The particle finishes at the same position and with the same kinetic energy that it had initially. Its total change in electric potential energy and kinetic energy is zero.

However, this does not mean the external agent does zero work during each stage.

The agent first supplies energy, allows the electric field to transfer energy into the particle’s motion, removes that kinetic energy when catching it, and then supplies energy again during the final slow movement.

This is the key distinction: zero net work over a complete journey does not mean zero work on every part of that journey.

The most useful next step is to become comfortable switching between external work, electric-field work, and changes in kinetic energy. Once you can distinguish those three quantities, even complicated multi-stage charged-particle questions become much easier to organise.