How to Find the Stopping Voltage for a Charged Particle

An electron is moving through an evacuated tube at \(6.0\times10^6\ \mathrm{m\,s^{-1}}\). You want it to reach a point where its speed is zero. How much potential difference is needed to stop it?

You could calculate its acceleration and the distance over which it slows down. There is a much quicker method: use conservation of energy to determine the potential difference that converts all its kinetic energy into electric potential energy.

First, predict the direction. Must the electron move towards a higher or lower electric potential to slow down?

It must move towards a lower potential. An electron has negative charge, so moving to a lower potential increases its electric potential energy. That increase comes from its kinetic energy.

The key result is:

\[
\boxed{\Delta V_{\mathrm{stop}}=\frac{K_i}{q}}
\]

Here, \(K_i\) is the particle's initial kinetic energy in joules, and \(q\) is its signed charge in coulombs. The sign of the answer tells you whether the particle must move towards a higher or lower electric potential.

## 1. The stopping voltage equation

A charged particle can gain or lose kinetic energy as it moves through an electric potential difference.

The relationship between its change in kinetic energy and the potential difference is:

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

The symbols mean:

- \(\Delta K=K_f-K_i\): change in kinetic energy, measured in joules (J).
- \(q\): signed electric charge, measured in coulombs (C).
- \(\Delta V=V_f-V_i\): change in electric potential, measured in volts (V).
- \(K_i\): initial kinetic energy.
- \(K_f\): final kinetic energy.

The negative sign is important. When a particle gains electric potential energy, it loses kinetic energy, assuming no energy is transferred to other forms.

### What changes when the particle stops?

At the stopping point, the particle's instantaneous speed is zero.

Therefore:

\[
K_f=0
\]

Its change in kinetic energy is:

\[
\Delta K=0-K_i=-K_i
\]

Substitute this into the energy equation:

\[
-K_i=-q\Delta V_{\mathrm{stop}}
\]

Rearranging gives:

\[
\boxed{\Delta V_{\mathrm{stop}}=\frac{K_i}{q}}
\]

This is the signed stopping potential difference.

If the particle's initial speed is known, substitute the classical kinetic energy equation:

\[
K_i=\frac12 mv_i^2
\]

Therefore:

\[
\boxed{\Delta V_{\mathrm{stop}}
=\frac{mv_i^2}{2q}}
\]

Here, \(m\) is the particle's mass in kilograms (kg), and \(v_i\) is its initial speed in metres per second (\(\mathrm{m\,s^{-1}}\)).

[IMAGE: Two-panel annotated scientific diagram showing positive and negative charged particles, each travelling horizontally to the right towards a stopping point. In the positive-charge panel, show an electric field pointing left, an electric force pointing left, and electric potential increasing from the starting point to the stopping point. In the negative-charge panel, show an electric field pointing right, an electric force pointing left, and electric potential decreasing towards the stopping point. Label initial velocity, initial potential, stopping point, and the signed potential change. Make clear that the electric force opposes the motion in both cases.]

### Why the sign of charge matters

This is where students often lose marks.

A proton and an electron have charges of equal magnitude but opposite signs. Consequently, the same potential difference produces opposite changes in their kinetic energies.

Consider a potential increase of \(+120\ \mathrm V\).

| Particle | Charge | Change in kinetic energy |
|---|---:|---:|
| Proton | \(+e\) | \(-120\ \mathrm{eV}\) |
| Electron | \(-e\) | \(+120\ \mathrm{eV}\) |

Here, \(e=1.60\times10^{-19}\ \mathrm C\) is the positive magnitude of the elementary charge.

The proton loses kinetic energy because it moves towards a higher potential. The electron gains kinetic energy because its negative charge makes its electric potential energy decrease.

If both particles initially have \(120\ \mathrm{eV}\) of kinetic energy, the proton stops, while the electron speeds up to \(240\ \mathrm{eV}\).

The important distinction is that you must use the signed charge in the stopping voltage equation.

### The sign decision

For a particle to lose all its initial kinetic energy, its electric potential energy must increase by exactly \(K_i\).

Since:

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

the stopping condition is:

\[
q\Delta V_{\mathrm{stop}}=K_i
\]

This gives the following decision rule:

| Particle | Sign of \(q\) | Required sign of \(\Delta V_{\mathrm{stop}}\) |
|---|---:|---:|
| Positive ion or proton | Positive | Positive |
| Electron or negative ion | Negative | Negative |

Therefore:

- A positive particle must reach a higher electric potential.
- A negative particle must reach a lower electric potential.

The magnitude of the stopping voltage is:

\[
\boxed{|\Delta V_{\mathrm{stop}}|
=\frac{K_i}{|q|}}
\]

Use this equation only when the question asks for the magnitude. Keep the signed version when determining the actual stopping potential.

## 2. A reliable method for calculating stopping voltage

Use the following steps in HSC Physics questions.

**Step 1: Identify the particle's charge.**

Write its charge with the correct sign. For example:

\[
q_e=-1.60\times10^{-19}\ \mathrm C
\]

\[
q_p=+1.60\times10^{-19}\ \mathrm C
\]

For a doubly charged positive ion:

\[
q=+3.20\times10^{-19}\ \mathrm C
\]

**Step 2: Calculate its initial kinetic energy.**

If the initial speed is provided:

\[
K_i=\frac12 mv_i^2
\]

If kinetic energy is already provided, use that value directly.

**Step 3: Apply the stopping condition.**

At the stopping point:

\[
K_f=0
\]

Therefore:

\[
\Delta V_{\mathrm{stop}}=\frac{K_i}{q}
\]

**Step 4: Calculate the final electric potential if required.**

Remember that a potential difference is not the same as an absolute potential.

Use:

\[
\boxed{V_s=V_i+\Delta V_{\mathrm{stop}}}
\]

where \(V_s\) is the stopping potential and \(V_i\) is the initial potential.

**Step 5: Check the sign.**

A positive particle should finish at a higher potential. A negative particle should finish at a lower potential.

If your answer contradicts this, check the sign of \(q\) and your subtraction order.

### A useful shortcut when kinetic energy is in electronvolts

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

For a particle with charge \(q=ze\), where \(z\) is its signed charge number, you can use:

\[
\boxed{\Delta V_{\mathrm{stop}}
=\frac{K_i\text{ (in eV)}}{z}
\text{ volts}}
\]

For example:

- An electron with \(250\ \mathrm{eV}\) of kinetic energy needs a stopping potential difference of \(-250\ \mathrm V\).
- An alpha particle with \(250\ \mathrm{eV}\) of kinetic energy needs a stopping potential difference of \(+125\ \mathrm V\), because its charge is \(+2e\).

This shortcut saves time, but remember that the charge number \(z\) is signed.

## 3. Worked examples

### Worked example: Finding the stopping voltage for an electron

An electron travels through an evacuated tube at \(6.0\times10^6\ \mathrm{m\,s^{-1}}\). Its initial electric potential is \(+25.0\ \mathrm V\).

Calculate:

a. The potential difference required to stop the electron.

b. The electric potential at which it stops.

Use:

- Electron mass: \(m_e=9.11\times10^{-31}\ \mathrm{kg}\).
- Electron charge: \(q=-1.60\times10^{-19}\ \mathrm C\).

[IMAGE: Side-view scientific diagram of an electron travelling horizontally to the right through a region between two parallel plates. Show the initial position labelled A, its velocity arrow pointing right, and its initial potential of +25.0 V. Show an unknown stopping position B further along the tube. Draw the electric field pointing right, so the electron's electric force opposes its motion. Do not label the stopping potential or stopping distance.]

**Step 1**: Calculate the electron's initial kinetic energy.

Use:

\[
K_i=\frac12 mv_i^2
\]

Substitute the values:

\[
\begin{aligned}
K_i&=\frac12(9.11\times10^{-31})
(6.0\times10^6)^2\\
&=1.64\times10^{-17}\ \mathrm J
\end{aligned}
\]

The electron initially has \(1.64\times10^{-17}\ \mathrm J\) of kinetic energy.

**Step 2**: Calculate the signed stopping potential difference.

Use:

\[
\Delta V_{\mathrm{stop}}=\frac{K_i}{q}
\]

Substitute:

\[
\begin{aligned}
\Delta V_{\mathrm{stop}}
&=\frac{1.64\times10^{-17}}
{-1.60\times10^{-19}}\\
&=-102.5\ \mathrm V
\end{aligned}
\]

The negative sign tells us that the electron must move towards a lower electric potential.

**Step 3**: Calculate the stopping potential.

The initial potential is \(+25.0\ \mathrm V\).

\[
V_s=V_i+\Delta V_{\mathrm{stop}}
\]

\[
\begin{aligned}
V_s&=25.0-102.5\\
&=-77.5\ \mathrm V
\end{aligned}
\]

Therefore:

\[
\boxed{\Delta V_{\mathrm{stop}}=-102.5\ \mathrm V}
\]

\[
\boxed{V_s=-77.5\ \mathrm V}
\]

The electron must experience a potential decrease of \(102.5\ \mathrm V\) to reach its stopping point.

Notice that the stopping potential is negative, but the initial potential was positive. There is nothing wrong with this: electric potential can be positive or negative depending on the chosen reference.

### Worked example: Stopping an alpha particle

An alpha particle travels through an evacuated region at \(4.00\times10^5\ \mathrm{m\,s^{-1}}\). Its initial electric potential is \(-200.0\ \mathrm V\).

Calculate the stopping potential.

Use:

- Alpha particle mass: \(m_\alpha=6.64\times10^{-27}\ \mathrm{kg}\).
- Alpha particle charge: \(q=+3.20\times10^{-19}\ \mathrm C\).

Assume the electric force acts opposite to the particle's motion and no energy is lost through collisions.

[IMAGE: Side-view diagram of a positively charged alpha particle moving right through a retarding electric field. Show the electric field and electric force arrows pointing left, the initial position labelled A at -200.0 V, and a possible stopping position B further right with its potential unknown. Keep the particle, field direction, and initial velocity spatially clear.]

**Step 1**: Calculate the initial kinetic energy.

\[
K_i=\frac12 mv_i^2
\]

Substitute:

\[
\begin{aligned}
K_i&=\frac12(6.64\times10^{-27})
(4.00\times10^5)^2\\
&=5.312\times10^{-16}\ \mathrm J
\end{aligned}
\]

**Step 2**: Calculate the signed stopping potential difference.

The alpha particle has a positive charge:

\[
q=+3.20\times10^{-19}\ \mathrm C
\]

Therefore:

\[
\begin{aligned}
\Delta V_{\mathrm{stop}}
&=\frac{5.312\times10^{-16}}
{3.20\times10^{-19}}\\
&=+1660\ \mathrm V
\end{aligned}
\]

The positive result tells us that the alpha particle must move towards a higher electric potential.

**Step 3**: Calculate the stopping potential.

\[
V_s=V_i+\Delta V_{\mathrm{stop}}
\]

\[
\begin{aligned}
V_s&=-200.0+1660\\
&=+1460\ \mathrm V
\end{aligned}
\]

Therefore:

\[
\boxed{\Delta V_{\mathrm{stop}}=+1660\ \mathrm V}
\]

\[
\boxed{V_s=+1460\ \mathrm V}
\]

The alpha particle requires a potential increase of \(1660\ \mathrm V\) to convert its initial kinetic energy into electric potential energy.

Compare this with the electron example. The electron needed a negative potential difference, while the alpha particle needed a positive one. The method was identical; only the sign of the charge changed.

## 4. Important assumptions and common mistakes

The stopping voltage equation is powerful, but you need to understand what it does and does not tell you.

### A particle stops momentarily, not necessarily permanently

When a particle reaches its stopping point, its instantaneous velocity is zero.

However, the electric force does not automatically disappear.

If the electric field remains present, the particle accelerates in the opposite direction and starts moving back. The stopping point is therefore a turning point, not necessarily a permanent resting position.

[IMAGE: Scientific side-view illustration of a positive particle moving right in a uniform electric field directed left. Show its initial motion, its turning point where the instantaneous velocity is zero, and its subsequent motion back to the left. Keep the electric field direction constant throughout and distinguish the momentary stopping point from permanent rest.]

### The stopping voltage depends on kinetic energy and charge

Look at:

\[
\Delta V_{\mathrm{stop}}=\frac{K_i}{q}
\]

Notice what is missing from this equation if kinetic energy is already known: mass.

Two particles with different masses can require the same stopping voltage if their initial kinetic energies and signed charges are equal.

However, if their initial speeds are equal, their kinetic energies will generally differ because:

\[
K_i=\frac12 mv_i^2
\]

This distinction becomes important when comparing particles such as protons and alpha particles.

### The classical speed equation has a limit

The equation:

\[
K=\frac12 mv^2
\]

is the non-relativistic kinetic energy equation. It assumes that the particle's speed is much smaller than the speed of light.

Use:

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

as a reference when checking your calculated speed.

There is no single sharp speed at which the classical equation suddenly fails. However, at \(0.1c\), its kinetic energy estimate is already approximately 0.8% below the relativistic value.

For speeds approaching the speed of light, use the relativistic expression:

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

where:

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

The stopping voltage relationship remains:

\[
\Delta V_{\mathrm{stop}}=\frac{K_i}{q}
\]

You simply need the correct expression for kinetic energy.

The worked examples and practice questions below use non-relativistic speeds.

### Four mistakes worth avoiding

| Common mistake | Correct approach |
|---|---|
| Using the magnitude of charge in the main equation | Use signed \(q\), including its positive or negative sign. |
| Confusing potential with potential difference | Calculate \(\Delta V=V_f-V_i\) first. |
| Substituting zero for the initial kinetic energy | At the stopping point, it is the final kinetic energy that equals zero. |
| Assuming zero instantaneous velocity means the particle stays still | Check whether an electric force remains and causes it to reverse direction. |

One final qualification: the simple stopping-voltage method assumes the electric field can remove the particle's kinetic energy along its direction of motion. If the particle has velocity components perpendicular to the electric force, you must consider those components separately. The final practice question explores this situation.

## Questions and solutions

Try each question before reading its solution. The questions progress from a straightforward stopping-voltage calculation to more demanding applications of energy conservation.

### Question 1

A proton travels horizontally through an evacuated tube at a speed of \(3.0\times10^5\ \mathrm{m\,s^{-1}}\).

Its initial electric potential is \(-50.0\ \mathrm V\).

Calculate:

a. The signed potential difference required to bring the proton momentarily to rest.

b. The electric potential at its stopping point.

Use:

- Proton mass: \(m_p=1.67\times10^{-27}\ \mathrm{kg}\).
- Proton charge: \(q_p=+1.60\times10^{-19}\ \mathrm C\).

[IMAGE: Side-view diagram of a proton travelling horizontally to the right along an evacuated tube. Label its initial position A, initial potential -50.0 V, and initial velocity pointing right. Show a second position B further along the tube with an unknown stopping potential. Do not include electric force arrows or the calculated potential.]

### Solution 1

The signed stopping potential difference is approximately \(+470\ \mathrm V\), giving a stopping potential of approximately \(+420\ \mathrm V\).

The proton has a positive charge, so it must move towards a higher electric potential to lose kinetic energy.

**Step 1**: Calculate the proton's initial kinetic energy.

\[
K_i=\frac12 mv_i^2
\]

Substitute:

\[
\begin{aligned}
K_i&=\frac12(1.67\times10^{-27})
(3.0\times10^5)^2\\
&=7.515\times10^{-17}\ \mathrm J
\end{aligned}
\]

**Step 2**: Calculate the stopping potential difference.

Use:

\[
\Delta V_{\mathrm{stop}}=\frac{K_i}{q}
\]

Substitute:

\[
\begin{aligned}
\Delta V_{\mathrm{stop}}
&=\frac{7.515\times10^{-17}}
{1.60\times10^{-19}}\\
&=+469.7\ \mathrm V
\end{aligned}
\]

Therefore:

\[
\boxed{\Delta V_{\mathrm{stop}}
\approx+4.7\times10^2\ \mathrm V}
\]

**Step 3**: Calculate the stopping potential.

\[
V_s=V_i+\Delta V_{\mathrm{stop}}
\]

\[
\begin{aligned}
V_s&=-50.0+469.7\\
&=419.7\ \mathrm V
\end{aligned}
\]

Therefore:

\[
\boxed{V_s\approx+420\ \mathrm V}
\]

The proton must move from \(-50.0\ \mathrm V\) to approximately \(+420\ \mathrm V\) to lose all its initial kinetic energy.

The tempting mistake is to report \(470\ \mathrm V\) as the stopping potential. It is actually the potential difference. The stopping potential is the final potential relative to the chosen zero reference.

### Question 2

A negatively charged ion has an initial kinetic energy of \(480\ \mathrm{eV}\). Its charge is:

\[
q=-3.20\times10^{-19}\ \mathrm C
\]

It begins at an electric potential of \(-40.0\ \mathrm V\) and travels towards a region of lower electric potential.

Calculate:

a. The signed stopping potential difference.

b. The absolute electric potential at which the ion stops.

c. The change in kinetic energy if the ion instead moves through a potential increase of \(+100\ \mathrm V\). Explain whether this would slow it down or speed it up.

[IMAGE: Side-view diagram of a negatively charged ion travelling horizontally to the right along an evacuated tube. Label its initial position A at -40.0 V, its initial velocity to the right, and a possible stopping position B further along the tube with unknown potential. Do not show force arrows or numerical stopping results.]

### Solution 2

The ion requires a stopping potential difference of \(-240\ \mathrm V\), giving an absolute stopping potential of \(-280\ \mathrm V\). A potential increase of \(+100\ \mathrm V\), however, would increase its kinetic energy by \(200\ \mathrm{eV}\).

**Part (a): Calculate the signed stopping potential difference.**

**Step 1**: Convert the initial kinetic energy into joules.

We are given:

\[
K_i=480\ \mathrm{eV}
\]

Using:

\[
1\ \mathrm{eV}=1.60\times10^{-19}\ \mathrm J
\]

gives:

\[
K_i=480(1.60\times10^{-19})
\]

\[
K_i=7.68\times10^{-17}\ \mathrm J
\]

**Step 2**: Apply the stopping voltage equation.

\[
\Delta V_{\mathrm{stop}}=\frac{K_i}{q}
\]

Substitute:

\[
\begin{aligned}
\Delta V_{\mathrm{stop}}
&=\frac{7.68\times10^{-17}}
{-3.20\times10^{-19}}\\
&=-240\ \mathrm V
\end{aligned}
\]

Therefore:

\[
\boxed{\Delta V_{\mathrm{stop}}=-240\ \mathrm V}
\]

The negative charge requires a negative potential difference.

**Part (b): Calculate the absolute stopping potential.**

\[
V_s=V_i+\Delta V_{\mathrm{stop}}
\]

\[
\begin{aligned}
V_s&=-40.0-240\\
&=-280\ \mathrm V
\end{aligned}
\]

Therefore:

\[
\boxed{V_s=-280\ \mathrm V}
\]

**Part (c): Determine what happens during a potential increase of \(+100\ \mathrm V\).**

Use the general energy relationship:

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

Substitute:

\[
\begin{aligned}
\Delta K
&=-(-3.20\times10^{-19})(100)\\
&=+3.20\times10^{-17}\ \mathrm J
\end{aligned}
\]

Convert to electronvolts:

\[
\Delta K=
\frac{3.20\times10^{-17}}
{1.60\times10^{-19}}
\]

\[
\boxed{\Delta K=+200\ \mathrm{eV}}
\]

The ion gains \(200\ \mathrm{eV}\) of kinetic energy, so it speeds up rather than slows down.

The trap is using the magnitude of the charge and ignoring its sign. A potential increase does not automatically slow a charged particle. Its effect depends on whether the particle is positive or negative.

### Question 3

An unknown charged particle has a mass of:

\[
m=3.34\times10^{-27}\ \mathrm{kg}
\]

It travels through an evacuated region at \(1.20\times10^5\ \mathrm{m\,s^{-1}}\).

At position A, its electric potential is \(+20.0\ \mathrm V\). It travels towards position B, where it momentarily comes to rest at a potential of \(+170\ \mathrm V\).

Assume the electric field is responsible for the entire change in kinetic energy.

Determine:

a. The signed charge of the particle.

b. Its charge in terms of the elementary charge \(e\).

c. Whether the particle is positively or negatively charged.

Use \(e=1.60\times10^{-19}\ \mathrm C\).

[IMAGE: Simple side-view scientific diagram of an unknown charged particle travelling horizontally from A to B. Show an initial velocity arrow pointing right, potential +20.0 V at A, and potential +170 V at B. Do not identify the particle's charge or draw electric force arrows.]

### Solution 3

The particle has a charge of approximately \(+1.60\times10^{-19}\ \mathrm C\), or \(+e\). It is therefore positively charged.

We can determine the sign before doing any calculations. The potential increases by \(150\ \mathrm V\), and the particle loses all its kinetic energy. Its electric potential energy must therefore increase, meaning its charge must be positive.

**Step 1**: Calculate the initial kinetic energy.

\[
K_i=\frac12 mv_i^2
\]

Substitute:

\[
\begin{aligned}
K_i&=\frac12(3.34\times10^{-27})
(1.20\times10^5)^2\\
&=2.4048\times10^{-17}\ \mathrm J
\end{aligned}
\]

**Step 2**: Calculate the potential difference.

\[
\Delta V=V_f-V_i
\]

\[
\begin{aligned}
\Delta V&=170-20\\
&=+150\ \mathrm V
\end{aligned}
\]

**Step 3**: Rearrange the stopping voltage equation to find charge.

Starting with:

\[
\Delta V_{\mathrm{stop}}=\frac{K_i}{q}
\]

Rearrange:

\[
q=\frac{K_i}{\Delta V_{\mathrm{stop}}}
\]

Substitute:

\[
\begin{aligned}
q&=\frac{2.4048\times10^{-17}}{150}\\
&=1.6032\times10^{-19}\ \mathrm C
\end{aligned}
\]

Therefore:

\[
\boxed{q\approx+1.60\times10^{-19}\ \mathrm C}
\]

**Step 4**: Express the charge in terms of \(e\).

\[
q=\frac{1.60\times10^{-19}}
{1.60\times10^{-19}}e
\]

\[
\boxed{q=+e}
\]

The particle has a single positive elementary charge.

Notice that the charge could not have been negative. A negative charge moving towards a higher potential would gain kinetic energy, not lose it.

### Question 4

A proton and an alpha particle travel through separate evacuated tubes.

Both particles have the same initial speed of:

\[
v_i=2.0\times10^5\ \mathrm{m\,s^{-1}}
\]

Both begin at an electric potential of \(-100.0\ \mathrm V\). Their respective electric fields are arranged to oppose their motion.

Use the following data:

| Property | Proton | Alpha particle |
|---|---:|---:|
| Mass | \(1.67\times10^{-27}\ \mathrm{kg}\) | \(6.68\times10^{-27}\ \mathrm{kg}\) |
| Charge | \(+1.60\times10^{-19}\ \mathrm C\) | \(+3.20\times10^{-19}\ \mathrm C\) |

Calculate:

a. The stopping potential difference for each particle.

b. The absolute stopping potential for each particle.

c. The ratio of their stopping voltage magnitudes.

Explain why the alpha particle requires a greater stopping voltage despite having twice the charge.

[IMAGE: Two-panel comparison diagram. In the upper panel, show a proton moving right with initial speed 2.0 x 10^5 m/s. In the lower panel, show an alpha particle moving right at the same initial speed. Both travel through separate one-dimensional retarding electric-field regions, starting at -100.0 V. Label the proton charge +e and the alpha particle charge +2e. Show their different masses as supplied data. Do not include calculated stopping voltages.]

### Solution 4

The proton requires a stopping potential difference of approximately \(+210\ \mathrm V\), while the alpha particle requires approximately \(+420\ \mathrm V\).

The alpha particle needs twice the stopping voltage because its mass, and therefore its initial kinetic energy at the same speed, is four times greater. Its charge is only twice as large.

**Part (a): Calculate each stopping potential difference.**

**Step 1**: Calculate the proton's initial kinetic energy.

\[
K_p=\frac12m_pv_i^2
\]

\[
\begin{aligned}
K_p&=\frac12(1.67\times10^{-27})
(2.0\times10^5)^2\\
&=3.34\times10^{-17}\ \mathrm J
\end{aligned}
\]

**Step 2**: Calculate the proton's stopping voltage.

\[
\Delta V_p=\frac{K_p}{q_p}
\]

\[
\begin{aligned}
\Delta V_p
&=\frac{3.34\times10^{-17}}
{1.60\times10^{-19}}\\
&=+208.75\ \mathrm V
\end{aligned}
\]

Therefore:

\[
\boxed{\Delta V_p\approx+210\ \mathrm V}
\]

**Step 3**: Calculate the alpha particle's initial kinetic energy.

\[
K_\alpha=\frac12m_\alpha v_i^2
\]

\[
\begin{aligned}
K_\alpha&=\frac12(6.68\times10^{-27})
(2.0\times10^5)^2\\
&=1.336\times10^{-16}\ \mathrm J
\end{aligned}
\]

**Step 4**: Calculate the alpha particle's stopping voltage.

\[
\Delta V_\alpha=\frac{K_\alpha}{q_\alpha}
\]

\[
\begin{aligned}
\Delta V_\alpha
&=\frac{1.336\times10^{-16}}
{3.20\times10^{-19}}\\
&=+417.5\ \mathrm V
\end{aligned}
\]

Therefore:

\[
\boxed{\Delta V_\alpha\approx+420\ \mathrm V}
\]

**Part (b): Calculate the absolute stopping potentials.**

Both particles start at \(-100.0\ \mathrm V\).

For the proton:

\[
\begin{aligned}
V_{s,p}&=-100.0+208.75\\
&=+108.75\ \mathrm V
\end{aligned}
\]

\[
\boxed{V_{s,p}\approx+110\ \mathrm V}
\]

For the alpha particle:

\[
\begin{aligned}
V_{s,\alpha}&=-100.0+417.5\\
&=+317.5\ \mathrm V
\end{aligned}
\]

\[
\boxed{V_{s,\alpha}\approx+320\ \mathrm V}
\]

**Part (c): Compare the stopping voltages.**

Calculate the ratio:

\[
\begin{aligned}
\frac{|\Delta V_\alpha|}
{|\Delta V_p|}
&=\frac{417.5}{208.75}\\
&=2.0
\end{aligned}
\]

Therefore:

\[
\boxed{\frac{|\Delta V_\alpha|}
{|\Delta V_p|}=2.0}
\]

The alpha particle requires twice the stopping voltage.

Why?

Both particles have the same initial speed, but the alpha particle has four times the mass.

Since:

\[
K_i=\frac12mv_i^2
\]

its initial kinetic energy is four times greater.

Its charge is only twice as large, so:

\[
|\Delta V_{\mathrm{stop}}|
=\frac{K_i}{|q|}
\]

gives twice the stopping voltage.

The tempting prediction is that twice the charge should mean half the stopping voltage. That would only follow if the initial kinetic energies were equal. Here, they are not.

This is why you should compare kinetic energy and charge together, rather than considering charge alone.

### Question 5

An electron enters a uniform electric field with two components of velocity:

- Horizontal velocity: \(v_x=3.0\times10^6\ \mathrm{m\,s^{-1}}\).
- Vertical velocity: \(v_y=4.0\times10^6\ \mathrm{m\,s^{-1}}\).

The electric field points horizontally to the right.

The electron initially moves diagonally upwards and to the right. Its mass is \(9.11\times10^{-31}\ \mathrm{kg}\), and its charge is \(-1.60\times10^{-19}\ \mathrm C\).

Assume:

- The electric field is uniform and remains horizontal.
- There is no vertical electric force.
- There are no collisions or other significant forces.

A student calculates the electron's total speed and uses its total kinetic energy to determine a stopping voltage of approximately \(-71\ \mathrm V\).

The student claims that applying this potential difference will bring the electron completely to rest.

Determine whether this claim is correct.

Your response must:

a. Calculate the electron's initial total kinetic energy.

b. Calculate the signed potential difference at which its horizontal velocity first becomes zero.

c. Determine the electron's speed and remaining kinetic energy at this instant.

d. Explain why the electron cannot be brought completely to rest by this horizontal electric field.

[IMAGE: Scientific vector diagram showing an electron entering a horizontal electric-field region from the lower-left. Its initial velocity vector points diagonally upwards and to the right. Draw horizontal and vertical coordinate axes, label the supplied velocity components vx = 3.0 x 10^6 m/s and vy = 4.0 x 10^6 m/s, and show the uniform electric field pointing right. Show the electron at the entrance but do not draw its electric force, subsequent trajectory, turning point, or any calculated results.]

### Solution 5

The student's claim is incorrect. The electron's horizontal velocity becomes zero after a potential decrease of approximately \(25.6\ \mathrm V\), but it continues moving upwards at \(4.0\times10^6\ \mathrm{m\,s^{-1}}\).

The key issue is that the electric field only acts horizontally. It can remove the horizontal component of kinetic energy, but it cannot remove the vertical component.

**Part (a): Calculate the initial total kinetic energy.**

First, determine the total initial speed.

The horizontal and vertical velocity components are perpendicular, so use Pythagoras:

\[
v_i^2=v_x^2+v_y^2
\]

\[
\begin{aligned}
v_i^2&=(3.0\times10^6)^2\\
&\quad +(4.0\times10^6)^2\\
&=25\times10^{12}
\end{aligned}
\]

Therefore:

\[
v_i=5.0\times10^6\ \mathrm{m\,s^{-1}}
\]

Now calculate the total kinetic energy:

\[
K_i=\frac12mv_i^2
\]

\[
\begin{aligned}
K_i&=\frac12(9.11\times10^{-31})
(5.0\times10^6)^2\\
&=1.13875\times10^{-17}\ \mathrm J
\end{aligned}
\]

Therefore:

\[
\boxed{K_i\approx1.14\times10^{-17}\ \mathrm J}
\]

The student used this total kinetic energy to calculate a potential difference of:

\[
\begin{aligned}
\Delta V_{\mathrm{claimed}}
&=\frac{1.13875\times10^{-17}}
{-1.60\times10^{-19}}\\
&=-71.2\ \mathrm V
\end{aligned}
\]

The arithmetic is correct, but the physical interpretation is not.

This voltage would only stop the electron if the electric force could remove its entire velocity.

**Part (b): Calculate the potential difference when the horizontal velocity becomes zero.**

The electric field is horizontal, so the electron experiences a horizontal force only.

Its horizontal kinetic energy is:

\[
K_x=\frac12mv_x^2
\]

Substitute:

\[
\begin{aligned}
K_x&=\frac12(9.11\times10^{-31})
(3.0\times10^6)^2\\
&=4.0995\times10^{-18}\ \mathrm J
\end{aligned}
\]

When the horizontal velocity becomes zero, this horizontal kinetic energy has been converted into electric potential energy.

Therefore:

\[
\Delta V_x=\frac{K_x}{q}
\]

Substitute:

\[
\begin{aligned}
\Delta V_x
&=\frac{4.0995\times10^{-18}}
{-1.60\times10^{-19}}\\
&=-25.62\ \mathrm V
\end{aligned}
\]

Therefore:

\[
\boxed{\Delta V_x\approx-25.6\ \mathrm V}
\]

The electron moves towards a potential that is \(25.6\ \mathrm V\) lower than its starting potential before its horizontal velocity reaches zero.

**Part (c): Calculate the remaining speed and kinetic energy.**

The electric field has no vertical component, so there is no vertical electric force.

Consequently, the vertical velocity remains constant:

\[
v_y=4.0\times10^6\ \mathrm{m\,s^{-1}}
\]

At the instant the horizontal velocity reaches zero:

\[
v_x=0
\]

The total speed is therefore:

\[
v=\sqrt{v_x^2+v_y^2}
\]

\[
\begin{aligned}
v&=\sqrt{0^2+(4.0\times10^6)^2}\\
&=4.0\times10^6\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Therefore:

\[
\boxed{v=4.0\times10^6\ \mathrm{m\,s^{-1}}}
\]

Now calculate the remaining kinetic energy.

Since only the vertical velocity remains:

\[
K_f=\frac12mv_y^2
\]

\[
\begin{aligned}
K_f&=\frac12(9.11\times10^{-31})
(4.0\times10^6)^2\\
&=7.288\times10^{-18}\ \mathrm J
\end{aligned}
\]

Therefore:

\[
\boxed{K_f\approx7.29\times10^{-18}\ \mathrm J}
\]

The electron still has approximately \(45.5\ \mathrm{eV}\) of kinetic energy.

**Part (d): Explain why the electron cannot be completely stopped.**

The electron cannot be brought completely to rest by this horizontal electric field because its vertical velocity is unaffected.

The electric force acts horizontally, opposing the electron's horizontal motion. As the electron slows horizontally, it continues travelling upwards.

When its horizontal velocity reaches zero, the electric force is still present. The electron therefore begins moving horizontally in the opposite direction while continuing upwards.

[IMAGE: Explanatory side-view vector diagram showing the electron's trajectory in a uniform horizontal electric field directed right. Show its initial diagonal upwards-right velocity, the horizontal electric force pointing left, and its curved path. Mark the point where horizontal velocity becomes zero, showing the remaining upward velocity. Continue the trajectory as it curves back towards the left while still moving upwards. Clearly distinguish the zero horizontal velocity from zero total velocity.]

The important distinction is:

- Zero horizontal velocity does not mean zero total velocity.
- The electric field can remove kinetic energy associated with horizontal motion.
- Without a vertical force, the vertical kinetic energy remains unchanged.

The student's total-energy calculation would be appropriate if the electric force acted directly opposite to the entire initial velocity. It does not do so here.

This is a useful boundary case for the stopping voltage equation. Before using the total kinetic energy, check that the electric force can actually remove every component of the particle's velocity.

## 5. What to use next

The stopping voltage method is an application of conservation of energy.

If a particle's initial speed is given, calculate its kinetic energy and divide by its signed charge. If its kinetic energy is already known, you can go straight to the stopping voltage equation.

If the particle starts from rest and is first accelerated through a potential difference, you can calculate its kinetic energy using:

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

Then use that kinetic energy to determine the required stopping voltage.

For a particle that starts from rest and is accelerated through a potential difference before being brought back to rest, the required stopping potential difference is equal in magnitude and opposite in sign to the original accelerating potential difference, provided no other energy transfers occur.

For the next step, see [How to Calculate Electron Speed After Acceleration Through Voltage](https://tutorgum.com/learn/resources/hsc-physics-how-to-calculate-electron-speed-after-acceleration-through-a-voltage/).

If a question asks you to predict the actual path of a charged particle rather than simply calculate its energy change, you'll also need to consider the electric force, acceleration, and velocity components. See [Charged-Particle Trajectories in Electric Fields: HSC Physics](https://tutorgum.com/learn/resources/hsc-physics-charged-particle-trajectories-in-electric-fields/).