How to Use the Projectile-Motion Analogy for Electric Fields

In 1897, a beam of cathode rays passed between two charged metal plates. If the rays carried electric charge, the field between the plates should bend their path. Yet earlier experiments had failed to produce a lasting deflection. Why did the beam appear to ignore the field?

The answer connects two ideas that might initially seem unrelated: electric fields and projectile motion. A charged particle entering a uniform electric field horizontally behaves like a projectile. Its horizontal velocity remains constant, while the field produces constant vertical acceleration. The crucial substitution is to replace gravitational acceleration, \(g\), with the signed electric acceleration, \(a_y=qE_y/m\).

This analogy explains both the curved path between the plates and the displaced spot on the screen beyond them. It also gives us a reliable way to calculate the deflection without confusing the direction of the electric field with the direction of the particle's acceleration.

## The experiment: why did the beam bend?

J. J. Thomson investigated cathode rays using an evacuated glass discharge tube. His apparatus provided a way to produce a narrow beam, apply an electric field across its path, and measure the resulting deflection.

Follow the apparatus from left to right:

1. A negatively charged cathode produces cathode rays when a high voltage is applied across the tube.
2. The rays pass through a narrow opening in an anode, followed by another opening that produces a more tightly defined beam.
3. The beam enters the gap between two parallel metal plates. A battery connected to the plates establishes a potential difference and creates an approximately uniform electric field.
4. The beam leaves the plates and travels towards a fluorescent screen at the far end of the tube.
5. Where the beam strikes the screen, it produces a glowing patch. A scale attached to the outside of the tube allows the vertical displacement of that patch to be measured.

[IMAGE: Historically informed cross-section of J. J. Thomson's cathode-ray apparatus, viewed from the side. Show the evacuated glass tube, cathode at the left, slitted anode and second aperture, narrow horizontal electron beam, two horizontal parallel deflection plates, and fluorescent screen at the right. Show the upper plate positive and lower plate negative, with downward electric-field arrows, an upward-curving electron trajectory between the plates, and a straight continuation towards the displaced screen spot. Label the accelerating region, deflection region, field-free drift region, and screen displacement. Distinguish the predicted trajectory from the final measured fluorescent spot.]

The original apparatus used aluminium deflection plates approximately \(5\text{ cm}\) long, \(2\text{ cm}\) wide, and \(1.5\text{ cm}\) apart. The important measurement was not a photograph of the entire curved trajectory. It was the position of the glowing patch on the screen. Thomson inferred the beam's behaviour from this measurable displacement. citeturn617296search12turn617296search13

### Why the first experiment failed

Earlier attempts, including Hertz's experiments, had not produced a clear electrostatic deflection.

Thomson investigated why. Cathode rays travelling through the remaining gas in the tube could ionise it, making the gas electrically conducting. This conducting gas could screen the beam from the applied electric field.

In other words, the experiment was supposed to test how the beam responded to an electric field, but the gas was interfering with the field itself.

Thomson improved the vacuum by removing more of the remaining gas. With much less conducting material surrounding the beam, the electric field could act on the cathode rays.

The revised experiment produced a measurable deflection. Reversing the plate polarity reversed the direction of the deflection, and the displacement increased with the potential difference between the plates. The direction of the deflection indicated that the rays carried negative charge. citeturn617296search0turn617296search12

This was important evidence that cathode rays behaved like negatively charged particles.

There is a useful distinction between the historical observation and the model we use today:

- **Observation:** The fluorescent patch moved when the electric field was applied, and its displacement depended on the field's direction and strength.
- **Physical interpretation:** The beam consisted of negatively charged particles experiencing an electric force.
- **Modern motion model:** Within an approximately uniform field, the particles undergo constant acceleration and follow a parabolic trajectory if their initial velocity is perpendicular to the field.

The parabolic path is the prediction of the motion model. Thomson's screen measurement provided evidence for the deflection; it did not directly photograph the parabola.

## How the projectile-motion analogy works

Imagine throwing a ball horizontally from a table. It moves forwards while gravity accelerates it downwards.

Predict what happens to its horizontal velocity. Does gravity gradually slow the ball's forward motion?

No. If air resistance is negligible, gravity acts vertically, so there is no horizontal acceleration. The horizontal velocity stays constant while the downward velocity increases.

A charged particle entering a uniform electric field horizontally behaves in the same mathematical way.

The difference is the cause and direction of the acceleration.

[IMAGE: Side-by-side comparison of a ball launched horizontally under gravity and a negatively charged electron entering a uniform electric field horizontally. Show the ball curving downwards under constant gravitational acceleration and the electron curving upwards when the electric field points downwards. Include horizontal velocity arrows, vertical acceleration arrows, and parabolic trajectories. Make clear that both paths arise from independent horizontal and vertical motion.]

For a projectile near Earth's surface, the vertical acceleration is approximately \(-g\) if upwards is positive.

For a charged particle in a uniform electric field, Newton's second law gives

\[
F_y=qE_y
\]

and therefore

\[
a_y=\frac{qE_y}{m}
\]

Here:

- \(a_y\) is the particle's vertical acceleration, in \(\mathrm{m\,s^{-2}}\).
- \(q\) is its signed electric charge, in coulombs (\(\mathrm C\)).
- \(E_y\) is the vertical component of the electric field, in \(\mathrm{N\,C^{-1}}\).
- \(m\) is the particle's mass, in kilograms (\(\mathrm{kg}\)).

The charge is signed, not just a magnitude. This is essential because a positive and negative particle accelerate in opposite directions in the same electric field.

If the field is uniform, \(E_y\) is constant. For a particle with constant mass and charge, its acceleration is therefore constant too.

### Why the trajectory is parabolic

Choose the horizontal direction as positive \(x\), and upwards as positive \(y\).

Suppose the particle enters the field horizontally, so its initial vertical velocity is zero.

There is no horizontal electric force in this arrangement. Therefore,

\[
a_x=0
\]

and the horizontal displacement after time \(t\) is

\[
x-x_0=v_xt
\]

where \(v_x\) is the constant horizontal velocity.

Vertically, the particle has constant acceleration \(a_y\), so

\[
y-y_0=\frac12a_yt^2
\]

Substitute \(t=(x-x_0)/v_x\):

\[
y-y_0=
\frac{a_y}{2v_x^2}(x-x_0)^2
\]

This is a quadratic relationship between horizontal and vertical position. The trajectory is therefore parabolic.

Notice what has happened: we have obtained the same mathematical structure as horizontal projectile motion. We have simply replaced gravitational acceleration with electric acceleration.

There is one important limitation. The trajectory is parabolic only while the particle experiences the uniform electric field. Once it leaves the plates and enters a region where the electric field is negligible, its acceleration becomes approximately zero. It then travels in a straight line, following the direction of its velocity at the exit.

The sign of the electric field matters

This is where the projectile analogy can catch you out.

Gravity accelerates an ordinary ball downwards regardless of whether it is made of plastic, steel, or rubber. An electric field does not behave like that. The direction of electric acceleration depends on the sign of the particle’s charge.

Set up a consistent coordinate system:

• Positive \(y\) points upwards.
• The upper plate is positive.
• The lower plate is negative.
• The electric field therefore points downwards.

The electric field points from higher electric potential to lower electric potential.

Let the plate separation be \(d\), and define the signed potential difference as

\[ \Delta V_{tb}=V_{\text{top}}-V_{\text{bottom}} \]

For the arrangement above, \(\Delta V_{tb}\) is positive.

The vertical electric field component is

\[ E_y=-\frac{\Delta V_{tb}}{d} \]

The negative sign tells us that the field points downwards, even though the potential difference is positive.

Now use the signed charge in

\[ a_y=\frac{qE_y}{m} \]

This gives two different predictions:

Do not replace \(q\) with \(|q|\) when calculating a signed acceleration. You can use the absolute value when calculating a magnitude, but you must determine the direction separately.

Potential difference and kinetic energy

There is another useful way to check your prediction.

Electric potential energy is related to electric potential by

\[ U=qV \]

For a change in position,

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

If the electric force is the only force doing work, conservation of energy gives

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

Here, \(\Delta V=V_f-V_i\) is the final electric potential minus the initial electric potential.

Consider an electron moving upwards between the plates.

It moves towards the positive upper plate, so its electric potential increases. Therefore, \(\Delta V>0\).

But the electron has negative charge:

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

Consequently,

\[ \Delta K=-q\Delta V>0 \]

Its kinetic energy increases.

This agrees with our force calculation: the electron accelerates upwards, gaining velocity in the direction it is being displaced.

The two approaches should agree:

• The force approach tells you how the particle accelerates.
• The energy approach tells you how its kinetic energy changes as it moves through a potential difference.

If they predict opposite changes in kinetic energy, check your signs.

For a refresher on why the central region between parallel plates has an approximately uniform field, see Uniform Electric Fields Between Parallel Plates: HSC Physics.

01A reliable method for calculating electric deflection

For a particle entering horizontally between parallel plates, use the following sequence.

Here, \(L\) is the plate length and \(v_x\) is the constant horizontal velocity.

The key decision is to calculate the time spent inside the field using horizontal motion. Then use that same time in the vertical motion equation.

02What happens after the particle leaves the plates?

This is an important part of understanding Thomson’s screen measurement.

You might assume that the electron continues following a parabola all the way to the fluorescent screen. But that would mean it continues experiencing a vertical acceleration.

Once it leaves the uniform-field region, the electric force becomes negligible in the idealised model.

Its vertical acceleration is then approximately zero.

However, the electron has already gained an upward vertical velocity. With no further vertical force, that velocity remains constant.

The path therefore changes from a parabola to a straight line.

03When the projectile analogy gives a different prediction

The analogy is useful, but it has limits. Three differences are particularly important.

1. Positive and negative charges bend in opposite directions

Suppose you replace the electron in the worked example with a positive ion, keeping the same initial horizontal velocity and electric field.

The electric field still points downwards. However, the positive ion now accelerates downwards rather than upwards.

Its trajectory curves downwards.

The field has not changed. The particle’s charge has.

2. A curved path is not guaranteed

The parabolic trajectory requires the particle to have a horizontal velocity component while accelerating vertically.

What if the particle enters parallel to the electric field?

It accelerates along a straight line. Its speed changes, but its direction of travel does not.

The field still produces acceleration. It simply does not produce sideways deflection.

This is why you should check the initial velocity direction before automatically applying the projectile-motion analogy.

3. The electric field only acts in the region where it exists

For a projectile near Earth’s surface, gravity continues acting throughout its flight.

In the parallel-plate experiment, the electric field is concentrated mainly between the plates. Once the particle leaves that region, its path becomes approximately straight.

This means you must treat the motion in two stages:

• Inside the plates: constant horizontal velocity and constant vertical acceleration.
• Outside the plates: constant horizontal and vertical velocities.

The overall path is not one continuous parabola.

04Two potential differences, two different jobs

An electron beam can be accelerated before it reaches the deflection plates.

This introduces two separate potential differences, and confusing them is a common source of errors.

The accelerating potential difference determines the electron’s initial kinetic energy.

Suppose an electron starts from rest at a cathode and moves towards an anode at a higher electric potential.

Define

\[ V_a=V_{\text{anode}}-V_{\text{cathode}}>0 \]

Because the electron has charge \(q=-e\),

\[ \Delta K=-qV_a=eV_a \]

where \(e=1.60\times10^{-19}\\,\mathrm C\) is the magnitude of the electron’s charge.

Using non-relativistic kinetic energy,

\[ \frac12m_ev_0^2=eV_a \]

Therefore,

\[ \boxed{v_0=\sqrt{\frac{2eV_a}{m_e}}} \]

A larger accelerating voltage gives the electron a greater initial horizontal speed.

The deflection potential difference determines the strength of the electric field between the plates.

For an upper positive plate and a lower negative plate, define

\[ V_d=V_{\text{top}}-V_{\text{bottom}}>0 \]

Then the downward field component is

\[ E_y=-\frac{V_d}{d} \]

A larger deflection voltage produces a stronger field and greater vertical acceleration of the electron.

A useful relationship for screen deflection

We can combine these two effects to predict the screen position.

Let:

• \(L\) be the length of the deflection plates, in metres.
• \(s\) be the distance from the end of the plates to the screen, in metres.
• \(d\) be the plate separation, in metres.
• \(V_a\) be the positive accelerating potential difference.
• \(V_d\) be the positive potential difference from the lower plate to the upper plate.

For an electron entering horizontally from rest after acceleration through \(V_a\), its vertical displacement inside the plates is

\[ \Delta y_1=\frac{V_dL^2}{4dV_a} \]

After leaving the plates, its vertical velocity remains constant. Its additional displacement before reaching the screen is

\[ \Delta y_2=\frac{V_dLs}{2dV_a} \]

Adding the two contributions gives

\[ \boxed{ \Delta y_{\text{screen}} =\frac{V_dL(L+2s)}{4dV_a} } \]

This result assumes a uniform deflection field, negligible fringing, negligible gravity, and non-relativistic motion.

It reveals something interesting:

For fixed apparatus dimensions, screen deflection depends on the ratio of the deflection voltage to the accelerating voltage.

Consider the predictions:

The last result is less obvious than it looks.

Doubling the deflection voltage doubles the vertical acceleration. But doubling the accelerating voltage increases the initial horizontal speed by a factor of \(\sqrt{2}\), reducing the time spent in both the deflection and drift regions.

The two changes balance exactly in the non-relativistic model.

This is a useful example of why you should consider the entire motion, rather than looking at the electric force in isolation.

05Assumptions and speed limits

The calculations in this article use the non-relativistic equations of motion.

They assume:

• The particle’s speed is much smaller than the speed of light, \(c=3.00\times10^8\\,\mathrm{m\\,s^{-1}}\).
• The electric field is approximately uniform between the plates.
• Edge effects and fringing fields are negligible.
• The particle does not collide with either plate.
• Gravity and other external forces are negligible.
• The particle’s charge and mass remain constant.

For a non-relativistic particle, kinetic energy is

\[ K=\frac12mv^2 \]

This is not an exact relationship at all speeds.

If the calculated speed approaches a significant fraction of the speed of light, relativistic corrections are needed. The kinetic energy must instead be calculated using

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

where

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

The simple relationship \(a=qE/m\) and the resulting constant-acceleration projectile model also cease to be generally valid at relativistic speeds.

For the illustrative electron in our worked examples, the final speed is approximately \(0.068c\). The non-relativistic approximation is appropriate at the precision used here.

Do not assume that every electron-deflection calculation automatically permits the use of classical mechanics. Check the speed implied by the accelerating voltage.

For a broader comparison of projectile, orbital, electric, and magnetic motion, see How to Predict Projectile, Orbital, Electric, and Magnetic Motion.

06Questions and solutions

The following questions progress from a standard electric-deflection calculation to problems involving potential differences, competing particle models, and changes in the field.

Question 1

A proton enters horizontally into a uniform electric field directed vertically downwards.

The field strength is \(2.0\times10^3\\,\mathrm{N\\,C^{-1}}\), and the proton enters at \(4.0\times10^5\\,\mathrm{m\\,s^{-1}}\).

It experiences the field for \(4.0\times10^{-7}\\,\mathrm s\).

Calculate:

(a) The proton’s vertical acceleration.

(b) Its vertical displacement during this time.

Assume its initial vertical velocity is zero and ignore other forces.

Use \(q_p=+1.60\times10^{-19}\\,\mathrm C\) and \(m_p=1.67\times10^{-27}\\,\mathrm{kg}\).

Solution 1

(a) The proton accelerates downwards at \(1.92\times10^{11}\\,\mathrm{m\\,s^{-2}}\).

(b) Its vertical displacement is \(15.3\\,\mathrm{mm}\) downwards.

Part (a): Calculate acceleration

Take upwards as positive. The electric field therefore has a negative vertical component:

\[ E_y=-2.0\times10^3\\,\mathrm{N\\,C^{-1}} \]

Using Newton’s second law,

\[ a_y=\frac{q_pE_y}{m_p} \]

Substitute:

\[ \begin{aligned} a_y&=\frac{(1.60\times10^{-19})(-2.0\times10^3)} {1.67\times10^{-27}}\\\ &=-1.92\times10^{11}\\,\mathrm{m\\,s^{-2}} \end{aligned} \]

The negative sign indicates downward acceleration.

Part (b): Calculate displacement

Use the constant-acceleration equation:

\[ \Delta y=v_{0y}t+\frac12a_yt^2 \]

Since the initial vertical velocity is zero,

\[ \begin{aligned} \Delta y&=\frac12(-1.92\times10^{11}) (4.0\times10^{-7})^2\\\ &=-1.53\times10^{-2}\\,\mathrm m\\\ &=-15.3\\,\mathrm{mm} \end{aligned} \]

The proton moves \(15.3\\,\mathrm{mm}\) downwards.

Notice that the positive proton accelerates in the direction of the electric field. An electron in the same field would accelerate upwards instead.

Question 2

Two horizontal parallel plates are separated by \(16\\,\mathrm{mm}\).

The upper plate is maintained at \(-120\\,\mathrm V\), and the lower plate is maintained at \(+80\\,\mathrm V\).

An electron enters halfway between the plates with a horizontal velocity of \(1.2\times10^7\\,\mathrm{m\\,s^{-1}}\). The plate length is \(30\\,\mathrm{mm}\).

Calculate the electron’s vertical displacement when it leaves the plates.

Determine whether it strikes either plate before exiting.

Assume a uniform field, negligible edge effects, and zero initial vertical velocity.

Use \(q_e=-1.60\times10^{-19}\\,\mathrm C\) and \(m_e=9.11\times10^{-31}\\,\mathrm{kg}\).

Solution 2

The electron moves \(6.86\\,\mathrm{mm}\) downwards and does not strike either plate before exiting.

Step 1: Calculate the signed potential difference and electric field.

Take upwards as positive.

\[ \Delta V_{tb}=V_{\text{top}}-V_{\text{bottom}} \]

\[ \Delta V_{tb}=-120-(+80)=-200\\,\mathrm V \]

The plate separation is

\[ d=0.016\\,\mathrm m \]

Therefore,

\[ \begin{aligned} E_y&=-\frac{\Delta V_{tb}}{d}\\\ &=-\frac{-200}{0.016}\\\ &=+1.25\times10^4\\,\mathrm{V\\,m^{-1}} \end{aligned} \]

The electric field points upwards.

Step 2: Calculate the electron’s acceleration.

The electron has negative charge, so it accelerates opposite to the field.

\[ a_y=\frac{q_eE_y}{m_e} \]

\[ \begin{aligned} a_y&=\frac{(-1.60\times10^{-19})(1.25\times10^4)} {9.11\times10^{-31}}\\\ &=-2.20\times10^{15}\\,\mathrm{m\\,s^{-2}} \end{aligned} \]

The acceleration is downwards.

Step 3: Calculate the time inside the plates.

\[ \begin{aligned} t&=\frac{L}{v_x}\\\ &=\frac{0.030}{1.2\times10^7}\\\ &=2.5\times10^{-9}\\,\mathrm s \end{aligned} \]

Step 4: Calculate vertical displacement.

\[ \begin{aligned} \Delta y&=\frac12a_yt^2\\\ &=\frac12(-2.20\times10^{15})(2.5\times10^{-9})^2\\\ &=-6.86\times10^{-3}\\,\mathrm m\\\ &=-6.86\\,\mathrm{mm} \end{aligned} \]

The negative sign indicates downward displacement.

Step 5: Check whether the electron hits a plate.

The electron begins halfway across the \(16\\,\mathrm{mm}\) gap.

Its initial clearance to either plate is therefore

\[ \frac{16}{2}=8.0\\,\mathrm{mm} \]

It moves downwards by \(6.86\\,\mathrm{mm}\).

Its remaining clearance above the lower plate is

\[ 8.0-6.86=1.14\\,\mathrm{mm} \]

Therefore, it exits without striking either plate.

This final check matters. A mathematically correct displacement is not necessarily a physically possible trajectory if the particle hits a plate before reaching the end.

Question 3

A proton and an electron enter the same uniform electric field horizontally. Both particles have an initial kinetic energy of \(1.0\\,\mathrm{keV}\).

The electric field has a magnitude of \(4.0\times10^3\\,\mathrm{N\\,C^{-1}}\) and points vertically downwards.

Both particles travel through the field for a horizontal distance of \(40\\,\mathrm{mm}\).

Assume each particle enters with zero vertical velocity and that the field is uniform throughout this distance.

(a) Calculate the vertical displacement of the electron.

(b) Calculate the vertical displacement of the proton.

(c) Explain why the two displacements have equal magnitudes despite the very different particle masses.

Use \(1\\,\mathrm{keV}=1.60\times10^{-16}\\,\mathrm J\).

Solution 3

(a) The electron moves \(1.60\\,\mathrm{mm}\) upwards.

(b) The proton moves \(1.60\\,\mathrm{mm}\) downwards.

(c) Their displacement magnitudes are equal because the slower proton spends longer in the field, exactly compensating for its smaller acceleration in this idealised non-relativistic model.

Step 1: Convert the kinetic energy and identify the field direction.

Both particles have

\[ K=1.60\times10^{-16}\\,\mathrm J \]

The field points downwards, so

\[ E_y=-4.0\times10^3\\,\mathrm{N\\,C^{-1}} \]

The electron accelerates upwards, while the proton accelerates downwards.

Step 2: Derive a displacement expression using kinetic energy.

For either particle, the horizontal speed is determined by

\[ K=\frac12mv_x^2 \]

Rearranging,

\[ v_x^2=\frac{2K}{m} \]

The time spent in the field is

\[ t=\frac{L}{v_x} \]

Since the initial vertical velocity is zero,

\[ \Delta y=\frac12\frac{qE_y}{m} \left(\frac{L}{v_x}\right)^2 \]

Substitute \(v_x^2=2K/m\):

\[ \begin{aligned} \Delta y &=\frac{qE_yL^2}{2mv_x^2}\\\ &=\frac{qE_yL^2}{4K} \end{aligned} \]

Notice that the mass has cancelled.

This is the key to the question. For particles with equal kinetic energies and equal charge magnitudes, the displacement magnitudes are equal in the same field.

Step 3: Calculate the electron’s displacement.

For the electron,

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

and

\[ L=0.040\\,\mathrm m \]

Therefore,

\[ \begin{aligned} \Delta y_e &=\frac{(-1.60\times10^{-19})(-4.0\times10^3)(0.040)^2} {4(1.60\times10^{-16})}\\\ &=+1.60\times10^{-3}\\,\mathrm m\\\ &=+1.60\\,\mathrm{mm} \end{aligned} \]

The electron moves upwards.

Step 4: Calculate the proton’s displacement.

For the proton,

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

Using the same expression,

\[ \begin{aligned} \Delta y_p &=\frac{(+1.60\times10^{-19})(-4.0\times10^3)(0.040)^2} {4(1.60\times10^{-16})}\\\ &=-1.60\times10^{-3}\\,\mathrm m\\\ &=-1.60\\,\mathrm{mm} \end{aligned} \]

The proton moves downwards.

The two particles have equal displacement magnitudes but opposite directions.

The tempting prediction is that the electron must deflect much more because it has a much smaller mass. That prediction considers acceleration but ignores time.

The proton accelerates less, but it also moves horizontally much more slowly at the same kinetic energy. It therefore experiences the field for longer.

The mass dependence cancels when the initial kinetic energy, rather than the initial speed, is held constant.

Question 4

An electron travels horizontally through two consecutive uniform-field regions.

Each region is \(20\\,\mathrm{mm}\) long. The electron enters the first region at \(2.0\times10^7\\,\mathrm{m\\,s^{-1}}\), with zero initial vertical velocity.

The electric field in the first region is \(2.0\times10^3\\,\mathrm{N\\,C^{-1}}\) downwards.

In the second region, the field has the same magnitude but points upwards.

Assume the electron experiences each field for the same time, and that there is no significant field in the small gap between the regions.

Calculate:

(a) The electron’s total vertical displacement after both regions.

(b) Its final vertical velocity.

Explain why the electron does not return to its original vertical position, even though the two electric accelerations have equal magnitudes and opposite directions.

Use \(q_e=-1.60\times10^{-19}\\,\mathrm C\) and \(m_e=9.11\times10^{-31}\\,\mathrm{kg}\).

Solution 4

(a) The electron finishes \(0.351\\,\mathrm{mm}\) above its initial position.

(b) Its final vertical velocity is zero.

The two accelerations cancel the change in vertical velocity, but they do not cancel the displacement that has already occurred.

Step 1: Calculate the acceleration in each region.

In the first region, the electric field points downwards:

\[ E_{y1}=-2.0\times10^3\\,\mathrm{N\\,C^{-1}} \]

For the electron,

\[ \begin{aligned} a_{y1}&=\frac{q_eE_{y1}}{m_e}\\\ &=\frac{(-1.60\times10^{-19})(-2.0\times10^3)} {9.11\times10^{-31}}\\\ &=+3.51\times10^{14}\\,\mathrm{m\\,s^{-2}} \end{aligned} \]

The acceleration is upwards.

In the second region, the electric field reverses direction. Therefore,

\[ a_{y2}=-3.51\times10^{14}\\,\mathrm{m\\,s^{-2}} \]

The electron now accelerates downwards.

Step 2: Calculate the time in each region.

The horizontal velocity remains constant because there is no horizontal electric force.

\[ \begin{aligned} t&=\frac{0.020}{2.0\times10^7}\\\ &=1.0\times10^{-9}\\,\mathrm s \end{aligned} \]

Step 3: Calculate the displacement in the first region.

The electron begins with zero vertical velocity.

\[ \begin{aligned} \Delta y_1&=\frac12a_{y1}t^2\\\ &=\frac12(3.51\times10^{14})(1.0\times10^{-9})^2\\\ &=1.755\times10^{-4}\\,\mathrm m \end{aligned} \]

Its vertical velocity at the end of the first region is

\[ \begin{aligned} v_{y1}&=a_{y1}t\\\ &=(3.51\times10^{14})(1.0\times10^{-9})\\\ &=3.51\times10^5\\,\mathrm{m\\,s^{-1}} \end{aligned} \]

The electron is now moving upwards.

Step 4: Calculate the displacement in the second region.

The electron enters the second region with an upward velocity, even though its acceleration is now downwards.

Therefore,

\[ \Delta y_2=v_{y1}t+\frac12a_{y2}t^2 \]

Substituting,

\[ \begin{aligned} \Delta y_2 &=(3.51\times10^5)(1.0\times10^{-9})\\\ &\quad-\frac12(3.51\times10^{14})(1.0\times10^{-9})^2\\\ &=1.755\times10^{-4}\\,\mathrm m \end{aligned} \]

The electron continues moving upwards throughout the second region, but it slows down vertically.

Step 5: Calculate the final displacement and velocity.

The total displacement is

\[ \begin{aligned} \Delta y_{\text{total}} &=\Delta y_1+\Delta y_2\\\ &=3.51\times10^{-4}\\,\mathrm m\\\ &=0.351\\,\mathrm{mm} \end{aligned} \]

The final vertical velocity is

\[ \begin{aligned} v_{yf}&=v_{y1}+a_{y2}t\\\ &=3.51\times10^5-3.51\times10^5\\\ &=0 \end{aligned} \]

The electron therefore exits travelling horizontally, but it is still \(0.351\\,\mathrm{mm}\) above its original position.

The tempting approach is to say that equal and opposite accelerations cancel, so the electron returns to where it started. That confuses velocity with position.

The second field reverses the vertical velocity change. It does not reverse the displacement already accumulated. During the second region, the electron continues travelling upwards before its vertical velocity reaches zero.

Question 5

An electron starts from rest at a cathode and is accelerated towards an anode.

The anode is maintained at a potential \(1.0\\,\mathrm{kV}\) higher than the cathode.

The electron then enters horizontally between two deflection plates. The upper plate is at \(+50\\,\mathrm V\), and the lower plate is at \(-50\\,\mathrm V\).

The apparatus has the following dimensions:

• Plate separation: \(20\\,\mathrm{mm}\).
• Plate length: \(40\\,\mathrm{mm}\).
• Distance from the plate exit to the fluorescent screen: \(60\\,\mathrm{mm}\).

Assume the electron enters the deflection plates halfway between them, with zero vertical velocity.

The electric field is uniform between the plates, and fringe fields, gravity, and other external forces are negligible.

(a) Calculate the electron’s initial horizontal speed as it enters the deflection plates.

(b) Calculate its vertical displacement when it reaches the screen.

(c) Both the accelerating potential difference and the magnitude of the deflection-plate potential difference are now doubled. The apparatus dimensions remain unchanged.

Predict what happens to:

• the position of the fluorescent spot;
• the electron’s speed at the screen.

Justify both predictions using the projectile-motion model and conservation of energy.

Use \(q_e=-1.60\times10^{-19}\\,\mathrm C\), \(m_e=9.11\times10^{-31}\\,\mathrm{kg}\), and \(c=3.00\times10^8\\,\mathrm{m\\,s^{-1}}\).

Solution 5

(a) The electron enters the deflection plates at approximately \(1.88\times10^7\\,\mathrm{m\\,s^{-1}}\).

(b) It reaches the screen \(8.0\\,\mathrm{mm}\) above its original horizontal entry level.

(c) Doubling both potential differences leaves the fluorescent spot in the same position, but increases the electron’s speed at the screen by a factor of \(\sqrt{2}\).

This result follows from the interaction between the initial horizontal speed and the vertical electric acceleration.

Part (a): Calculate the initial horizontal speed

The accelerating potential difference is

\[ \Delta V_a=+1000\\,\mathrm V \]

The electron starts from rest, so its initial kinetic energy after acceleration is

\[ \Delta K=-q_e\Delta V_a \]

Since the electron has negative charge,

\[ \begin{aligned} K_i&=-(-1.60\times10^{-19})(1000)\\\ &=1.60\times10^{-16}\\,\mathrm J \end{aligned} \]

Using non-relativistic kinetic energy,

\[ \frac12m_ev_0^2=K_i \]

Rearrange:

\[ v_0=\sqrt{\frac{2K_i}{m_e}} \]

Substituting,

\[ \begin{aligned} v_0&=\sqrt{\frac{2(1.60\times10^{-16})} {9.11\times10^{-31}}}\\\ &=1.88\times10^7\\,\mathrm{m\\,s^{-1}} \end{aligned} \]

Part (b): Calculate the screen displacement

Step 1: Calculate the electric field.

The potential difference between the upper and lower plates is

\[ V_d=50-(-50)=100\\,\mathrm V \]

The field points downwards:

\[ \begin{aligned} E_y&=-\frac{100}{0.020}\\\ &=-5.0\times10^3\\,\mathrm{V\\,m^{-1}} \end{aligned} \]

Step 2: Calculate the electron’s vertical acceleration.

\[ \begin{aligned} a_y&=\frac{q_eE_y}{m_e}\\\ &=\frac{(-1.60\times10^{-19})(-5.0\times10^3)} {9.11\times10^{-31}}\\\ &=8.78\times10^{14}\\,\mathrm{m\\,s^{-2}} \end{aligned} \]

The acceleration is upwards.

Step 3: Calculate the displacement inside the plates.

The time spent in the field is

\[ \begin{aligned} t_1&=\frac{0.040}{1.88\times10^7}\\\ &=2.13\times10^{-9}\\,\mathrm s \end{aligned} \]

The vertical displacement is

\[ \begin{aligned} \Delta y_1&=\frac12a_yt_1^2\\\ &=\frac12(8.78\times10^{14})(2.13\times10^{-9})^2\\\ &=2.0\times10^{-3}\\,\mathrm m\\\ &=2.0\\,\mathrm{mm} \end{aligned} \]

Step 4: Calculate the vertical velocity at the plate exit.

\[ \begin{aligned} v_y&=a_yt_1\\\ &=(8.78\times10^{14})(2.13\times10^{-9})\\\ &=1.87\times10^6\\,\mathrm{m\\,s^{-1}} \end{aligned} \]

Step 5: Calculate the additional displacement towards the screen.

The electron now travels through a field-free region.

Its horizontal velocity remains approximately \(1.88\times10^7\\,\mathrm{m\\,s^{-1}}\), and its vertical velocity remains \(1.87\times10^6\\,\mathrm{m\\,s^{-1}}\).

The drift time is

\[ \begin{aligned} t_2&=\frac{0.060}{1.88\times10^7}\\\ &=3.20\times10^{-9}\\,\mathrm s \end{aligned} \]

The additional vertical displacement is

\[ \begin{aligned} \Delta y_2&=v_yt_2\\\ &=(1.87\times10^6)(3.20\times10^{-9})\\\ &=6.0\times10^{-3}\\,\mathrm m\\\ &=6.0\\,\mathrm{mm} \end{aligned} \]

Step 6: Calculate the total screen displacement.

\[ \begin{aligned} \Delta y_{\text{screen}} &=\Delta y_1+\Delta y_2\\\ &=2.0+6.0\\\ &=8.0\\,\mathrm{mm} \end{aligned} \]

The electron hits the screen \(8.0\\,\mathrm{mm}\) above its original horizontal entry level.

Part (c): Double both potential differences

This is the most interesting part of the question.

The accelerating potential difference increases from \(1000\\,\mathrm V\) to \(2000\\,\mathrm V\).

The initial kinetic energy therefore doubles:

\[ K_i'=2K_i \]

Since \(K=\frac12mv^2\),

\[ v_0'=\sqrt{2}\\,v_0 \]

The electron enters the plates faster by a factor of \(\sqrt{2}\).

Now consider the deflection plates.

Their potential difference also doubles, from \(100\\,\mathrm V\) to \(200\\,\mathrm V\).

Therefore, the electric field doubles, and so does the electron’s vertical acceleration.

However, the time spent between the plates decreases by a factor of \(\sqrt{2}\), because the horizontal speed has increased.

The vertical displacement inside the plates depends on

\[ \Delta y_1=\frac12a_yt_1^2 \]

Doubling the acceleration while reducing the time by \(\sqrt{2}\) gives

\[ \Delta y_1' =\frac12(2a_y) \left(\frac{t_1}{\sqrt2}\right)^2 =\Delta y_1 \]

The electron therefore leaves the plates at the same vertical position.

Its vertical exit velocity is

\[ v_y'=2a_y\frac{t_1}{\sqrt2} =\sqrt2\\,v_y \]

The drift time to the screen also decreases by \(\sqrt2\):

\[ t_2'=\frac{t_2}{\sqrt2} \]

Consequently, the additional vertical displacement is unchanged:

\[ \Delta y_2' =(\sqrt2\\,v_y) \left(\frac{t_2}{\sqrt2}\right) =\Delta y_2 \]

The total screen displacement remains

\[ \boxed{\Delta y_{\text{screen}}'=8.0\\,\mathrm{mm}} \]

The fluorescent spot does not move.

Now consider the electron’s speed.

Both the initial horizontal velocity and the vertical velocity at the screen increase by \(\sqrt2\).

Therefore, the magnitude of the final velocity also increases by \(\sqrt2\):

\[ v_f'=\sqrt2\\,v_f \]

Using the original screen speed of approximately \(1.88\times10^7\\,\mathrm{m\\,s^{-1}}\),

\[ \begin{aligned} v_f'&=\sqrt2(1.88\times10^7)\\\ &\approx2.66\times10^7\\,\mathrm{m\\,s^{-1}} \end{aligned} \]

The electron reaches the same screen position, but travels faster.

This also agrees with conservation of energy. Doubling the accelerating voltage doubles the initial kinetic energy. Doubling the deflection voltage also doubles the kinetic energy gained from the electric field, because the vertical displacement within the plates remains unchanged while the field strength doubles.

The total final kinetic energy therefore doubles, giving a final speed increase of \(\sqrt2\).

The tempting prediction is that doubling the deflection voltage must double the screen displacement. That would be correct if the initial horizontal speed remained fixed. Here, however, the accelerating voltage changes too. The shorter time spent in both regions exactly compensates for the stronger electric acceleration.

The result is a useful distinction: the screen position depends on the ratio of the two potential differences, while the particle’s speed depends on their overall energy contribution.