How to Calculate Electron Speed After Acceleration Through Voltage

An electron starts from rest at a cathode at \(-80\ \mathrm V\) and accelerates towards an anode at \(+170\ \mathrm V\). How fast is it moving when it reaches the anode?

The potential difference is \(250\ \mathrm V\). This gives the electron \(250\ \mathrm{eV}\) of kinetic energy, producing a final speed of approximately \(9.37\times10^6\ \mathrm{m\,s^{-1}}\).

The calculation uses conservation of energy, not the equations of motion. The important part is handling the signs correctly. An electron has negative charge, so it gains kinetic energy when it moves towards a higher electric potential.

01Understand how voltage changes an electron’s kinetic energy

Before using an equation, make a prediction. If an electron moves from a lower electric potential to a higher one, does it speed up or slow down?

It speeds up. This might seem backwards because electric field direction is often associated with the direction of acceleration. However, an electron is negatively charged, so the electric force acts opposite to the electric field.

The relationship between charge and potential difference

Electric potential energy is given by

\[ U=qV \]

where:

• \(U\) is electric potential energy in joules (J).
• \(q\) is the particle’s charge in coulombs (C).
• \(V\) is electric potential in volts (V).

The change in electric potential energy is therefore

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

Conservation of energy tells us that, if no energy is transferred to other forms,

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

Combining these equations gives the key relationship:

\[ \boxed{\Delta K=-q\Delta V} \]

Here, the potential difference is defined as

\[ \Delta V=V_f-V_i \]

where \(V_i\) is the initial potential and \(V_f\) is the final potential.

For an electron,

\[ q=-e=-1.602\times10^{-19}\ \mathrm C \]

The symbol \(e\) represents the positive magnitude of the elementary charge.

Substituting the electron’s charge gives

\[ \boxed{\Delta K=e\Delta V} \]

This is the relationship to remember when calculating the kinetic energy gained or lost by an electron.

Notice that the potential difference is still signed. We have not replaced it with its absolute value.

For an electron, moving to a higher potential is like moving downhill in its own potential-energy landscape. The analogy is useful because it helps explain why increasing electric potential can reduce potential energy.

The limitation is that electric potential itself is not physical height. The relationship between potential and potential energy depends on the sign of the particle’s charge. A positive charge experiences the opposite energy change.

02Convert kinetic energy into electron speed

Once you know the electron’s kinetic energy, use the familiar kinetic-energy equation:

\[ K=\frac12 mv^2 \]

For an electron, its mass is

\[ m_e=9.11\times10^{-31}\ \mathrm{kg} \]

Rearrange the equation to make speed the subject:

\[ \boxed{v=\sqrt{\frac{2K}{m_e}}} \]

This gives the speed from the electron’s kinetic energy.

If the electron starts from rest, its initial kinetic energy is zero. If it accelerates through a positive potential difference, its final kinetic energy is

\[ K_f=e\Delta V \]

Substituting this into the speed equation gives

\[ \boxed{v=\sqrt{\frac{2e\Delta V}{m_e}}} \]

This is the standard shortcut for calculating the final speed of an electron starting from rest.

It assumes that:

• The electron starts with negligible kinetic energy.
• The electric field is the only significant source of energy change.
• The electron does not lose energy through collisions.
• Its speed is sufficiently low for classical kinetic energy to be accurate.

If the electron already has an initial speed, you must include its initial kinetic energy. We will work through that case shortly.

A useful note about electronvolts

An electronvolt (eV) is a unit of energy, not voltage.

One electronvolt is the energy gained by an electron moving through a potential difference of one volt:

\[ 1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm J \]

For an electron, a potential increase of \(250\ \mathrm V\) produces a kinetic-energy increase of \(250\ \mathrm{eV}\).

You can work in electronvolts for energy calculations, provided you convert the final energy into joules before substituting into \(K=\frac12 mv^2\).

03How could an experiment check the calculated speed?

There is an interesting measurement problem here. We can set the accelerating voltage accurately, but how do we check that an electron actually reaches the predicted speed?

A bright spot on a fluorescent screen tells us that electrons have arrived. It does not directly tell us how quickly they travelled.

An idealised time-of-flight experiment provides a better test.

The apparatus and its prediction

The apparatus works in the following order:

1. A cathode releases a short pulse of electrons inside an evacuated tube.
2. A voltage supply establishes a potential difference between the cathode and anode.
3. The electrons accelerate across the gap and gain kinetic energy.
4. After leaving the accelerating region, the electrons enter a section where the electric field is negligible.
5. They pass two detector grids separated by a known distance. The detectors record the time at which the electron pulse passes each grid.

The distance and time allow us to calculate the average speed:

\[ v=\frac{\text{distance}}{\text{time}} \]

For the previous example, the predicted speed was \(9.37\times10^6\ \mathrm{m\,s^{-1}}\).

Suppose the detector grids are separated by \(0.100\ \mathrm m\). The predicted transit time is

\[ \begin{aligned} t&=\frac{d}{v}\\ &=\frac{0.100}{9.37\times10^6}\\ &=1.07\times10^{-8}\ \mathrm s \end{aligned} \]

That is approximately \(10.7\ \mathrm{ns}\), where ns means nanoseconds.

What should the experiment show?

If the energy calculation is accurate, the electron pulse should take approximately \(10.7\ \mathrm{ns}\) to travel between the detectors.

Consider illustrative measurement data in which the measured interval is \(10.8\ \mathrm{ns}\).

The measured speed would be

\[ \begin{aligned} v&=\frac{0.100}{10.8\times10^{-9}}\\ &=9.26\times10^6\ \mathrm{m\,s^{-1}} \end{aligned} \]

This is close to the predicted \(9.37\times10^6\ \mathrm{m\,s^{-1}}\).

The agreement supports the energy model. A small difference could result from timing uncertainty, the electrons having a small initial kinetic energy, or residual electric fields in the drift region.

Why the measurement design matters

A fluorescent screen alone cannot distinguish between a fast, low-current beam and a slower, brighter beam. Brightness depends on the number of electrons arriving and other properties of the screen, not just their speed.

The two-detector arrangement addresses this problem by measuring the time taken to travel a known distance.

This is an idealised experimental design, not a claim about a particular historical apparatus. It also relies on several assumptions: the electron pulse is sufficiently short, the detector separation is known, and the electrons travel at approximately constant speed between the detectors.

The experiment illustrates an important distinction: conservation of energy predicts the speed, while time-of-flight measurements provide an independent way to estimate it.

04What if the electron already has an initial speed?

The previous example assumed the electron started from rest. That is a useful starting point, but not every question will make that assumption.

If an electron already has kinetic energy, the voltage adds to or subtracts from that energy.

Start with

\[ \Delta K=-q\Delta V \]

Since \(\Delta K=K_f-K_i\),

\[ K_f=K_i-q\Delta V \]

Using the kinetic-energy equation for the initial and final states,

\[ \frac12m_ev_f^2 = \frac12m_ev_i^2-q\Delta V \]

Rearranging gives

\[ \boxed{v_f=\sqrt{v_i^2-\frac{2q\Delta V}{m_e}}} \]

This is the more general equation for calculating an electron’s final speed.

Notice that \(q\) is still negative. Do not substitute the positive magnitude of the electron’s charge without accounting for its sign.

05What happens when an electron moves to a lower potential?

This is where students often make a sign error.

Suppose an electron is already moving towards a region of lower electric potential. It might be tempting to use the magnitude of the potential difference and assume the electron gains energy.

That is incorrect.

Consider an electron with an initial kinetic energy of \(400\ \mathrm{eV}\). It moves from a region at \(+200\ \mathrm V\) towards another region at \(-100\ \mathrm V\).

Step 1: Calculate the potential difference.

\[ \begin{aligned} \Delta V&=V_f-V_i\\ &=-100-(+200)\\ &=-300\ \mathrm V \end{aligned} \]

Step 2: Calculate the change in kinetic energy.

\[ \begin{aligned} \Delta K&=e\Delta V\\ &=-300\ \mathrm{eV} \end{aligned} \]

The electron loses \(300\ \mathrm{eV}\) of kinetic energy.

Step 3: Calculate the final kinetic energy.

\[ \begin{aligned} K_f&=K_i+\Delta K\\ &=400-300\\ &=100\ \mathrm{eV} \end{aligned} \]

Step 4: Calculate the final speed.

Convert the energy into joules:

\[ K_f=100(1.602\times10^{-19}) =1.602\times10^{-17}\ \mathrm J \]

Then,

\[ \begin{aligned} v_f&=\sqrt{\frac{2K_f}{m_e}}\\ &=\sqrt{\frac{2(1.602\times10^{-17})}{9.11\times10^{-31}}}\\ &=5.93\times10^6\ \mathrm{m\,s^{-1}} \end{aligned} \]

The final speed is

\[ \boxed{5.93\times10^6\ \mathrm{m\,s^{-1}}} \]

The electron has slowed down to half its initial speed.

Why half? Its kinetic energy has fallen from \(400\ \mathrm{eV}\) to \(100\ \mathrm{eV}\), which is one quarter of its original value. Since kinetic energy is proportional to the square of speed, its speed becomes half as large.

There is another important consequence.

If this electron had started from rest instead, it could not have travelled towards the lower potential under these conditions. The calculation would give a negative final kinetic energy, which is physically impossible.

A negative calculated kinetic energy is a warning that the particle cannot reach the specified final position under the stated assumptions.

06When is the speed equation accurate?

The equation

\[ K=\frac12m_ev^2 \]

is the classical, or non-relativistic, kinetic-energy equation.

It assumes that the electron travels much slower than the speed of light.

The speed of light is approximately

\[ c=3.00\times10^8\ \mathrm{m\,s^{-1}} \]

For the first worked example, the electron reached

\[ v=9.37\times10^6\ \mathrm{m\,s^{-1}} \]

This is only about \(3.1\%\) of the speed of light. The non-relativistic calculation is therefore appropriate.

However, the equation cannot be used at arbitrarily high speeds.

An electron’s rest energy is approximately \(511\ \mathrm{keV}\). When its kinetic energy becomes a significant fraction of this value, relativistic effects become important.

For example, if you incorrectly used the classical equation at a kinetic energy of \(511\ \mathrm{keV}\), it would predict a speed of \(\sqrt{2}c\), which is greater than the speed of light.

That result is physically impossible.

There is no single sharp voltage at which the classical equation suddenly stops working. Its accuracy gradually decreases as the electron’s speed approaches the speed of light.

For the few-hundred-electronvolt examples in this article, the non-relativistic equation is a suitable approximation.

07Common mistakes to avoid

One final distinction is worth remembering: the energy calculation gives you speed, not direction.

To determine direction, consider the electron’s initial motion and the direction of the electric force. The force is opposite to the electric field because the electron has negative charge.

If you want to connect this method to other HSC questions, Acceleration-Time Graphs: Area and Velocity Change explains how acceleration over time determines velocity change. For the earlier stages of electron emission and beam formation, see How an Electron Gun Uses Voltage to Produce a Beam.

08Questions and solutions

Question 1

An electron starts from rest at a cathode with an electric potential of \(-40\ \mathrm V\). It accelerates towards an anode at \(+140\ \mathrm V\).

Calculate the electron’s final speed. Give your answer in \(\mathrm{m\,s^{-1}}\).

Solution 1

The electron reaches a final speed of approximately \(7.96\times10^6\ \mathrm{m\,s^{-1}}\).

Step 1: Calculate the potential difference.

\[ \begin{aligned} \Delta V&=V_f-V_i\\ &=140-(-40)\\ &=+180\ \mathrm V \end{aligned} \]

Step 2: Calculate the kinetic energy gained.

Since the electron starts from rest,

\[ K_f=e\Delta V \]

\[ \begin{aligned} K_f&=(1.602\times10^{-19})(180)\\ &=2.884\times10^{-17}\ \mathrm J \end{aligned} \]

Step 3: Calculate the final speed.

\[ \begin{aligned} v&=\sqrt{\frac{2K_f}{m_e}}\\ &=\sqrt{\frac{2(2.884\times10^{-17})}{9.11\times10^{-31}}}\\ &=7.96\times10^6\ \mathrm{m\,s^{-1}} \end{aligned} \]

Therefore,

\[ \boxed{v=7.96\times10^6\ \mathrm{m\,s^{-1}}} \]

The electron gains kinetic energy because it moves through a positive potential difference.

Question 2

An electron has an initial kinetic energy of \(450\ \mathrm{eV}\) and is travelling towards the right.

It moves from position A, at an electric potential of \(+240\ \mathrm V\), towards position B, at \(-60\ \mathrm V\).

Assume that the potential decreases continuously between A and B and that no collisions occur.

Calculate:

(a) The electron’s kinetic energy at B, in electronvolts.

(b) Its speed at B, in \(\mathrm{m\,s^{-1}}\).

(c) Whether the electron speeds up or slows down as it travels from A to B.

Solution 2

The electron reaches B with \(150\ \mathrm{eV}\) of kinetic energy and a speed of approximately \(7.26\times10^6\ \mathrm{m\,s^{-1}}\). It slows down, but it has enough initial kinetic energy to reach B.

Step 1: Calculate the signed potential difference.

\[ \begin{aligned} \Delta V&=V_f-V_i\\ &=-60-240\\ &=-300\ \mathrm V \end{aligned} \]

The negative potential difference means the electron loses kinetic energy.

Step 2: Calculate the kinetic energy change.

For an electron, the numerical value of the kinetic-energy change in eV equals the signed potential difference in volts.

\[ \Delta K=-300\ \mathrm{eV} \]

Step 3: Calculate the final kinetic energy.

\[ \begin{aligned} K_f&=K_i+\Delta K\\ &=450-300\\ &=150\ \mathrm{eV} \end{aligned} \]

Therefore, the answer to part (a) is

\[ \boxed{K_f=150\ \mathrm{eV}} \]

Step 4: Convert the kinetic energy into joules.

\[ \begin{aligned} K_f&=150(1.602\times10^{-19})\\ &=2.403\times10^{-17}\ \mathrm J \end{aligned} \]

Step 5: Calculate the final speed.

\[ \begin{aligned} v_f&=\sqrt{\frac{2K_f}{m_e}}\\ &=\sqrt{\frac{2(2.403\times10^{-17})}{9.11\times10^{-31}}}\\ &=7.26\times10^6\ \mathrm{m\,s^{-1}} \end{aligned} \]

Therefore, part (b) gives

\[ \boxed{v_f=7.26\times10^6\ \mathrm{m\,s^{-1}}} \]

Step 6: Explain the change in speed.

The electron loses \(300\ \mathrm{eV}\) of kinetic energy as it moves towards a lower potential.

However, it starts with \(450\ \mathrm{eV}\), so it still has \(150\ \mathrm{eV}\) remaining when it reaches B.

It therefore reaches B while travelling more slowly than it was initially.

Question 3

An electron starts from rest at position A and travels through two successive accelerating regions.

The electric potentials are:

• Position A: \(-200\ \mathrm V\)
• Position B: \(+150\ \mathrm V\)
• Position C: \(-50\ \mathrm V\)

The electron travels from A to B, then from B to C. Assume that the potential changes continuously between the positions and that no collisions occur.

Calculate:

(a) The electron’s speed at B.

(b) The electron’s speed at C.

(c) Explain why the electron slows down between B and C, even though C is at a higher potential than A.

Solution 3

The electron reaches B at approximately \(1.11\times10^7\ \mathrm{m\,s^{-1}}\) and reaches C at approximately \(7.26\times10^6\ \mathrm{m\,s^{-1}}\). It speeds up in the first region and slows down in the second.

Step 1: Calculate the potential difference between A and B.

\[ \begin{aligned} \Delta V_{AB}&=V_B-V_A\\ &=150-(-200)\\ &=+350\ \mathrm V \end{aligned} \]

Step 2: Calculate the kinetic energy at B.

The electron starts from rest, so

\[ K_B=e\Delta V_{AB} \]

In electronvolts,

\[ \boxed{K_B=350\ \mathrm{eV}} \]

Convert to joules:

\[ K_B=350(1.602\times10^{-19}) =5.607\times10^{-17}\ \mathrm J \]

Step 3: Calculate the speed at B.

\[ \begin{aligned} v_B&=\sqrt{\frac{2K_B}{m_e}}\\ &=\sqrt{\frac{2(5.607\times10^{-17})}{9.11\times10^{-31}}}\\ &=1.11\times10^7\ \mathrm{m\,s^{-1}} \end{aligned} \]

Therefore,

\[ \boxed{v_B=1.11\times10^7\ \mathrm{m\,s^{-1}}} \]

Step 4: Calculate the potential difference between B and C.

\[ \begin{aligned} \Delta V_{BC}&=V_C-V_B\\ &=-50-150\\ &=-200\ \mathrm V \end{aligned} \]

The electron now moves towards a lower potential, so it loses kinetic energy.

Step 5: Calculate the kinetic energy at C.

\[ \begin{aligned} K_C&=K_B+e\Delta V_{BC}\\ &=350-200\\ &=150\ \mathrm{eV} \end{aligned} \]

Step 6: Calculate the speed at C.

Convert the kinetic energy into joules:

\[ K_C=150(1.602\times10^{-19}) =2.403\times10^{-17}\ \mathrm J \]

Then,

\[ \begin{aligned} v_C&=\sqrt{\frac{2K_C}{m_e}}\\ &=\sqrt{\frac{2(2.403\times10^{-17})}{9.11\times10^{-31}}}\\ &=7.26\times10^6\ \mathrm{m\,s^{-1}} \end{aligned} \]

Therefore,

\[ \boxed{v_C=7.26\times10^6\ \mathrm{m\,s^{-1}}} \]

Step 7: Explain why the electron slows down.

The potential rises from A to B, so the electron gains \(350\ \mathrm{eV}\).

It then falls by \(200\ \mathrm V\) between B and C, causing the electron to lose \(200\ \mathrm{eV}\).

Although C is at a higher potential than A, it is at a lower potential than B. The electron therefore leaves C with less kinetic energy than it had at B.

This demonstrates why you must consider each potential change rather than simply assuming that an electron always speeds up throughout an accelerator.

Question 4

Two electron accelerators have different plate separations but identical accelerating potential differences.

Both electrons start from rest and travel through a positive potential difference of \(400\ \mathrm V\).

• Accelerator P has a plate separation of \(1.0\ \mathrm{cm}\).
• Accelerator Q has a plate separation of \(4.0\ \mathrm{cm}\).

Assume uniform electric fields, negligible fringe effects, and no collisions.

(a) Calculate the final speed of the electron in each accelerator.

(b) Calculate the electric field strength and acceleration magnitude in each accelerator.

(c) Compare the times taken for the electrons to cross their respective gaps. Assume constant acceleration.

Explain why the final speeds are identical even though the accelerations are different.

Solution 4

Both electrons leave their accelerators at approximately \(1.19\times10^7\ \mathrm{m\,s^{-1}}\). However, accelerator P produces four times the electric field strength and acceleration of accelerator Q. Its electron also takes one quarter of the time to cross the gap.

Part (a): Calculate the final speeds.

Step 1: Identify the potential difference.

Both electrons start from rest and move through

\[ \Delta V=+400\ \mathrm V \]

Step 2: Calculate the kinetic energy gained.

\[ K_f=e\Delta V \]

\[ \begin{aligned} K_f&=(1.602\times10^{-19})(400)\\ &=6.408\times10^{-17}\ \mathrm J \end{aligned} \]

Step 3: Calculate the final speed.

\[ \begin{aligned} v_f&=\sqrt{\frac{2K_f}{m_e}}\\ &=\sqrt{\frac{2(6.408\times10^{-17})}{9.11\times10^{-31}}}\\ &=1.19\times10^7\ \mathrm{m\,s^{-1}} \end{aligned} \]

The potential difference is identical in both accelerators, so

\[ \boxed{v_P=v_Q=1.19\times10^7\ \mathrm{m\,s^{-1}}} \]

Part (b): Compare the electric fields and accelerations.

Step 4: Calculate the electric field strength.

For a uniform electric field,

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

For accelerator P, the separation is

\[ d_P=1.0\ \mathrm{cm}=0.010\ \mathrm m \]

Therefore,

\[ \begin{aligned} E_P&=\frac{400}{0.010}\\ &=4.0\times10^4\ \mathrm{N\,C^{-1}} \end{aligned} \]

For accelerator Q,

\[ d_Q=4.0\ \mathrm{cm}=0.040\ \mathrm m \]

Therefore,

\[ \begin{aligned} E_Q&=\frac{400}{0.040}\\ &=1.0\times10^4\ \mathrm{N\,C^{-1}} \end{aligned} \]

Accelerator P has four times the electric field strength.

Step 5: Calculate the acceleration magnitudes.

The electric force magnitude is

\[ F=eE \]

Using Newton’s second law,

\[ a=\frac{F}{m_e}=\frac{eE}{m_e} \]

For accelerator P,

\[ \begin{aligned} a_P&=\frac{(1.602\times10^{-19})(4.0\times10^4)} {9.11\times10^{-31}}\\ &=7.03\times10^{15}\ \mathrm{m\,s^{-2}} \end{aligned} \]

For accelerator Q,

\[ \begin{aligned} a_Q&=\frac{(1.602\times10^{-19})(1.0\times10^4)} {9.11\times10^{-31}}\\ &=1.76\times10^{15}\ \mathrm{m\,s^{-2}} \end{aligned} \]

Therefore,

\[ \boxed{a_P=4a_Q} \]

The shorter gap produces the stronger field and greater acceleration.

Part (c): Compare the crossing times.

Step 6: Calculate the time taken in each accelerator.

Both electrons start from rest and experience constant acceleration, so

\[ v_f=v_i+at \]

Since \(v_i=0\),

\[ t=\frac{v_f}{a} \]

For accelerator P,

\[ \begin{aligned} t_P&=\frac{1.19\times10^7}{7.03\times10^{15}}\\ &=1.69\times10^{-9}\ \mathrm s \end{aligned} \]

For accelerator Q,

\[ \begin{aligned} t_Q&=\frac{1.19\times10^7}{1.76\times10^{15}}\\ &=6.76\times10^{-9}\ \mathrm s \end{aligned} \]

Thus,

\[ \boxed{t_Q=4t_P} \]

The electron in accelerator Q takes four times as long to cross the gap.

The key idea is that electric field strength determines the acceleration, while potential difference determines the total energy transferred.

A weaker field acting over a longer distance can produce exactly the same final kinetic energy as a stronger field acting over a shorter distance.

Question 5

This final problem tests an important distinction between calculating an electron’s energy at a particular position and determining whether it can actually reach that position.

An electron starts at position A with an initial kinetic energy of \(120\ \mathrm{eV}\). It travels to the right along a straight evacuated channel.

The electric potential varies with position as follows:

The potential changes linearly between A and B, and between B and C.

Assume a static electric field and no collisions or other energy transfers.

A student calculates the potential difference between A and C:

\[ \Delta V=+300\ \mathrm V \]

They then conclude that the electron reaches C with \(420\ \mathrm{eV}\) of kinetic energy.

Determine whether the student’s conclusion is physically correct.

Your response must:

(a) Determine whether the electron can reach B.

(b) If it cannot reach B, calculate the position and electric potential at which it stops momentarily.

(c) Describe what happens when it reverses direction and returns towards A.

(d) Explain why calculating the energy difference between A and C gives a mathematically correct result but an incorrect prediction of the electron’s actual motion.

Solution 5

The electron does not reach B or C. It stops momentarily \(4.0\ \mathrm{cm}\) from A, where the potential is \(-120\ \mathrm V\), then reverses direction and returns towards A.

The student’s calculation correctly predicts the kinetic energy the electron would have at C if it could reach C. The mistake is assuming that the electron can travel through every intermediate position.

Part (a): Determine whether the electron reaches B.

Step 1: Calculate the potential difference between A and B.

\[ \begin{aligned} \Delta V_{AB}&=V_B-V_A\\ &=-180-0\\ &=-180\ \mathrm V \end{aligned} \]

Step 2: Calculate the kinetic energy the electron would have at B.

For an electron,

\[ K_B=K_A+e\Delta V_{AB} \]

Using electronvolts,

\[ \begin{aligned} K_B&=120-180\\ &=-60\ \mathrm{eV} \end{aligned} \]

This result is physically impossible.

Kinetic energy cannot be negative, so the electron cannot reach B.

It must stop somewhere between A and B, before its kinetic energy reaches zero.

Part (b): Find the turning position.

Step 3: Determine the potential at which the electron stops.

At the turning point, its instantaneous speed is zero.

Therefore,

\[ K=0 \]

Using conservation of energy,

\[ K=K_A+e(V-V_A) \]

Substitute the initial kinetic energy and potential:

\[ 0=120+(V-0) \]

Therefore,

\[ \boxed{V=-120\ \mathrm V} \]

The electron stops momentarily when the electric potential reaches \(-120\ \mathrm V\).

Step 4: Calculate the position of the turning point.

The potential changes linearly from \(0\ \mathrm V\) at A to \(-180\ \mathrm V\) at B.

The potential drop per centimetre is

\[ \frac{-180}{6.0}=-30\ \mathrm{V\,cm^{-1}} \]

To reach \(-120\ \mathrm V\),

\[ \begin{aligned} x&=\frac{-120}{-30}\\ &=4.0\ \mathrm{cm} \end{aligned} \]

Therefore, the turning point is

\[ \boxed{x=4.0\ \mathrm{cm}} \]

The electron stops before reaching B.

Part (c): Describe the electron’s return towards A.

At the turning point, the electron has zero instantaneous velocity, but it still experiences an electric force.

Between A and B, the potential decreases as position increases. The electric field therefore points towards the right.

Since the electron is negatively charged, its electric force points towards the left.

Consequently, the electron accelerates back towards A.

At A, it has returned to its original electric potential. Since no energy has been lost, conservation of energy tells us that it must also regain its original kinetic energy:

\[ \boxed{K_A=120\ \mathrm{eV}} \]

Its speed at A is therefore equal to its original speed, although its direction of travel has reversed.

Part (d): Explain the student’s mistake.

The student calculated the potential difference between A and C:

\[ \Delta V_{AC}=300-0=+300\ \mathrm V \]

If the electron could reach C, its kinetic energy would be

\[ \begin{aligned} K_C&=K_A+e\Delta V_{AC}\\ &=120+300\\ &=420\ \mathrm{eV} \end{aligned} \]

The energy calculation is correct.

However, it describes the electron’s energy at C under the condition that it reaches C. It does not prove that the electron can travel there.

The electron first encounters a region of decreasing electric potential. For a negative charge, this increases electric potential energy and reduces kinetic energy.

It does not have enough initial kinetic energy to pass through this region. It therefore turns around before reaching the potential minimum at B.

This is the crucial distinction: the potential difference between two endpoints determines the net change in kinetic energy, but the electron must still have sufficient energy to pass through every intermediate point.

The problem is not with conservation of energy. It is with assuming that a mathematically calculated final state is necessarily physically accessible.

Understanding this distinction will help with more complex electrostatic potential diagrams, electron barriers, and questions involving the motion of charged particles through non-uniform electric fields.