How to Calculate Electron Deflection Between Parallel Plates

An electron beam passes between two parallel plates and lands on a fluorescent screen. The glowing spot appears above the beam’s original horizontal path. How far did the electron actually deflect while it was between the plates?

It might be tempting to use the screen’s displacement as the answer. However, the electron continues moving vertically after it leaves the electric field, so the screen displacement can be considerably larger than the deflection inside the plates.

To calculate the deflection between the plates, we need the electron’s vertical acceleration and the time it spends in the electric field.

For an electron entering horizontally with speed \(u_x\), the time spent between plates of length \(L\) is

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

The vertical displacement at the exit is therefore

\[
y_{\mathrm{exit}}=\frac12 a_y t^2
\]

Using the signed potential difference between the plates, we can write the complete result as

\[
\boxed{
y_{\mathrm{exit}}
=-\frac{q\Delta V_p L^2}
{2m_e d u_x^2}
}
\]

Here, \(q\) is the electron’s charge, \(m_e\) is its mass, \(d\) is the plate separation, and \(\Delta V_p\) is the potential difference between the upper and lower plates.

The important detail is that this equation gives a signed displacement. You can use its sign to determine which direction the electron bends, rather than calculating a magnitude and guessing the direction afterwards.

01The experimental setup: where does the deflection come from?

Consider a typical electron-beam tube.

A heated cathode at the left emits electrons. An anode at a higher electric potential accelerates them into a narrow beam. The beam then travels horizontally between two parallel metal plates.

A voltage supply maintains a potential difference between the plates. A voltmeter measures this potential difference.

Finally, the beam travels towards a fluorescent screen. The screen glows where the electrons strike it, making the beam’s final position visible.

What should happen when the voltage across the plates is switched on?

If the upper plate is at a higher potential than the lower plate, the electric field points downwards. An electron experiences a force opposite to the electric field, so the beam curves upwards towards the upper plate.

Reversing the plate voltage reverses the electric field and the direction of the electron’s deflection.

This is the characteristic observation in an electron-beam deflection experiment: the position of the glowing spot changes when the potential difference across the plates changes.

There is, however, an important measurement problem.

Why the screen displacement is not necessarily the plate deflection

Once the electron leaves the plates, the electric field is approximately zero. Its vertical velocity does not suddenly disappear.

The electron therefore continues travelling horizontally and vertically, following a straight-line path after leaving the field.

Consequently, a screen placed some distance beyond the plates measures the total displacement, including the additional movement after the electron exits.

If the screen is a horizontal distance \(S\) beyond the plates, its displacement is

\[
y_{\mathrm{screen}}
=
y_{\mathrm{exit}}
+
v_{y,\mathrm{exit}}\frac{S}{u_x}
\]

where \(v_{y,\mathrm{exit}}\) is the electron’s vertical velocity when it leaves the plates.

This distinction matters when an experiment is designed to measure the deflection specifically inside the electric field.

A single screen at a fixed distance cannot independently distinguish the two contributions. A useful refinement is to repeat the measurement with the fluorescent screen at several known distances beyond the plates.

Because the electron follows a straight path after leaving the field, plotting screen displacement against downstream distance gives a straight line. Extrapolating that line back to zero downstream distance gives the displacement at the plate exit.

We can now calculate that displacement directly.

02Step 1: Establish the direction of the electric field

Before substituting numbers, choose a consistent coordinate system.

Take:

• The positive \(x\)-direction as the electron’s initial direction of travel.
• The positive \(y\)-direction as vertically upwards.
• The upper plate as the plate at position \(y=d\).
• The lower plate as the plate at position \(y=0\).

Define the signed potential difference as

\[
\Delta V_p=V_{\mathrm{top}}-V_{\mathrm{bottom}}
\]

This definition is important. We are subtracting the lower plate’s potential from the upper plate’s potential.

The electric field points from higher potential towards lower potential.

For a uniform electric field between parallel plates, its signed vertical component is

\[
E_y=-\frac{\Delta V_p}{d}
\]

The negative sign appears because the electric field points in the direction of decreasing potential.

Consider the two possible arrangements:

Notice that the electric field direction and the electron’s force direction are opposite.

This is because the electron has a negative charge.

The electron’s charge is

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

Its mass is

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

Do not substitute a positive charge for an electron. The negative sign in \(q\) is what automatically gives the correct force direction.

03Step 2: Calculate the electron’s vertical acceleration

The electric force acting on a charged particle is

\[
F_y=qE_y
\]

Newton’s second law gives

\[
F_y=m_ea_y
\]

Combining these equations:

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

Substituting the signed electric field:

\[
\boxed{
a_y=-\frac{q\Delta V_p}{m_e d}
}
\]

For an electron, \(q=-e\), where \(e\) is the positive magnitude of the elementary charge.

Therefore,

\[
a_y=\frac{e\Delta V_p}{m_e d}
\]

This tells us that an electron accelerates upwards when the upper plate is at a higher potential.

It also shows that increasing the potential difference or decreasing the plate separation increases the electron’s vertical acceleration.

04Step 3: Calculate the time spent between the plates

The electron enters horizontally, and the electric force acts vertically.

There is therefore no horizontal acceleration in the idealised model.

Its horizontal velocity remains constant:

\[
v_x=u_x
\]

Using the horizontal motion equation,

\[
x=u_xt
\]

At the end of the plates, the horizontal displacement is \(L\).

Therefore,

\[
\boxed{
t=\frac{L}{u_x}
}
\]

This is one of the most important steps in the calculation.

A faster electron spends less time between the plates. It therefore has less time to develop vertical velocity.

05Step 4: Calculate the vertical displacement

The electron enters horizontally, so its initial vertical velocity is zero.

Its vertical displacement follows the constant-acceleration equation:

\[
y=u_{y0}t+\frac12a_yt^2
\]

Since \(u_{y0}=0\),

\[
y=\frac12a_yt^2
\]

Substitute the expressions for acceleration and time:

\[
y=\frac12
\left(-\frac{q\Delta V_p}{m_ed}\right)
\left(\frac{L}{u_x}\right)^2
\]

This gives the final equation:

\[
\boxed{
y_{\mathrm{exit}}
=-\frac{q\Delta V_pL^2}
{2m_ed u_x^2}
}
\]

For an electron, \(q=-e\), so we can also write

\[
\boxed{
y_{\mathrm{exit}}
=\frac{e\Delta V_pL^2}
{2m_ed u_x^2}
}
\]

The second version is particularly convenient for electron calculations.

Remember that \(\Delta V_p\) is still signed. A positive result means upwards, and a negative result means downwards.

Why does doubling the electron’s speed reduce the deflection to one-quarter?

Look at the final equation:

\[
y_{\mathrm{exit}}\propto\frac{1}{u_x^2}
\]

If the electron’s horizontal speed doubles, its time between the plates halves.

However, vertical displacement depends on the square of time.

Therefore,

\[
y_{\mathrm{new}}
=\frac{y_{\mathrm{original}}}{2^2}
\]

\[
\boxed{
y_{\mathrm{new}}=\frac14y_{\mathrm{original}}
}
\]

The faster electron does not experience a weaker electric force. It simply spends less time experiencing that force.

This distinction is useful when comparing electron beams travelling through identical deflection systems.

07What if the electron’s initial speed comes from an accelerating voltage?

Sometimes a question does not provide the electron’s horizontal entry speed.

Instead, it tells you that the electron starts from rest and is accelerated through a potential difference before entering the deflection plates.

You then need to calculate its initial speed using conservation of energy.

There are two different potential differences to keep separate:

Do not confuse the accelerating voltage with the voltage across the deflection plates.

Calculate the entry speed using energy

Define the accelerating potential difference as

\[
\Delta V_a=V_{\mathrm{anode}}-V_{\mathrm{cathode}}
\]

For an electron accelerated from rest towards an anode at a higher potential, \(\Delta V_a\) is positive.

The change in kinetic energy is equal to the negative of the change in electric potential energy:

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

Since the electron starts from rest,

\[
\frac12m_eu_x^2=-q\Delta V_a
\]

Rearranging:

\[
\boxed{
u_x=\sqrt{\frac{-2q\Delta V_a}{m_e}}
}
\]

Because \(q=-e\), this becomes

\[
\boxed{
u_x=\sqrt{\frac{2e\Delta V_a}{m_e}}
}
\]

You can now substitute this speed into the electron deflection equation.

Interestingly, the electron’s charge and mass cancel.

Starting with

\[
y_{\mathrm{exit}}
=-\frac{q\Delta V_pL^2}
{2m_ed u_x^2}
\]

and substituting

\[
u_x^2=-\frac{2q\Delta V_a}{m_e}
\]

gives

\[
\boxed{
y_{\mathrm{exit}}
=\frac{\Delta V_pL^2}
{4d\Delta V_a}
}
\]

This useful shortcut applies when the electron starts from rest and is accelerated through \(\Delta V_a\) before entering the deflection plates.

The plate voltage remains signed, so the equation still gives the correct deflection direction.

08Check the assumptions and speed limits

The equations above are powerful, but they rely on several assumptions.

1. The electric field must be approximately uniform

The equation

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

assumes a uniform electric field.

This is a good approximation in the central region between large, closely spaced parallel plates.

Near the plate edges, the field begins to curve and its magnitude changes. These effects are called fringing.

For a detailed explanation of when the uniform-field approximation works, see Uniform Electric Fields Between Parallel Plates: HSC Physics.

2. The electron must not hit a plate before reaching the exit

If the electron enters halfway between the plates with zero initial vertical velocity, compare its exit displacement with half the plate separation:

\[
|y_{\mathrm{exit}}|<\frac d2
\]

If this condition is satisfied, the electron clears both plates.

If it enters away from the centreline, you must calculate its distance to the nearer plate instead.

If the electron reaches a plate before the end of the field region, solve the vertical displacement equation for the time of contact. Then use

\[
x=u_xt
\]

to determine how far along the plates it travels before striking the surface.

You cannot report a hypothetical exit displacement for an electron that has already collided with a plate.

3. The electron must be moving sufficiently slowly for classical mechanics

The calculations in this article use Newtonian mechanics and the kinetic energy equation

\[
K=\frac12m_ev^2
\]

These are non-relativistic equations.

They require the electron’s total speed to be much smaller than the speed of light:

\[
v\ll c
\]

For an electron accelerated from rest, the classical speed estimate is

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

For example, an accelerating potential of \(2.0\text{ kV}\) gives approximately \(0.088c\).

At \(50\text{ kV}\), however, the classical estimate is approximately \(0.44c\). This is no longer a speed that can safely be treated as much smaller than \(c\).

Relativistic energy and momentum must be considered when greater accuracy is required.

There is no single sharp voltage at which the classical approximation suddenly stops working. The important question is whether the expected error is acceptable.

Also remember that the electron gains vertical speed inside the plates. If checking the speed limit, consider its total speed, not just its horizontal component.

09Questions and solutions

Unless stated otherwise, assume:

• The electric field is uniform between the plates.
• Fringing effects are negligible.
• The electron enters horizontally with zero initial vertical velocity.
• Gravity, magnetic forces, and collisions can be ignored.
• Non-relativistic mechanics applies.

Use the following constants:

\[
\begin{aligned}
e&=1.60\times10^{-19}\text{ C}\\
m_e&=9.11\times10^{-31}\text{ kg}\\
m_p&=1.67\times10^{-27}\text{ kg}
\end{aligned}
\]

Question 1

An electron enters horizontally halfway between two parallel plates.

The plates are \(40\text{ mm}\) long and separated by \(10\text{ mm}\). The upper plate is at \(+80\text{ V}\) relative to the lower plate.

The electron enters with a horizontal speed of \(2.0\times10^7\text{ m s}^{-1}\).

Calculate:

a) The vertical acceleration of the electron.

b) Its vertical displacement at the exit.

c) Whether it strikes either plate.

Solution 1

The electron is deflected \(+2.81\text{ mm}\) upwards and clears both plates.

Part a: Calculate the vertical acceleration.

First, calculate the electric field:

\[
E_y=-\frac{\Delta V_p}{d}
\]

\[
E_y=-\frac{80}{0.010}
\]

\[
E_y=-8.0\times10^3\text{ V m}^{-1}
\]

The electric field points downwards.

The electron’s acceleration is

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

\[
a_y=
\frac{(-1.60\times10^{-19})(-8.0\times10^3)}
{9.11\times10^{-31}}
\]

\[
\boxed{
a_y=+1.40\times10^{15}\text{ m s}^{-2}
}
\]

The electron accelerates upwards.

Part b: Calculate the vertical displacement.

First, find the time spent between the plates:

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

\[
t=\frac{0.040}{2.0\times10^7}
\]

\[
t=2.0\times10^{-9}\text{ s}
\]

The initial vertical velocity is zero, so

\[
y=\frac12a_yt^2
\]

\[
\begin{aligned}
y&=\frac12(1.40\times10^{15})
(2.0\times10^{-9})^2\\
&=2.81\times10^{-3}\text{ m}
\end{aligned}
\]

Therefore,

\[
\boxed{y=+2.81\text{ mm}}
\]

Part c: Check for a collision.

The electron starts halfway between the plates, so the upper plate is \(5.0\text{ mm}\) away.

Its displacement is \(2.81\text{ mm}\), which is less than \(5.0\text{ mm}\).

The electron therefore exits the plates without striking either surface.

Question 2

An electron enters horizontally between two parallel plates.

The plates are \(50\text{ mm}\) long and separated by \(12\text{ mm}\).

The upper plate is maintained at \(0\text{ V}\), while the lower plate is at \(+240\text{ V}\).

The electron enters \(2.0\text{ mm}\) above the lower plate with a horizontal speed of \(3.0\times10^7\text{ m s}^{-1}\).

Determine whether the electron reaches the end of the plates. If it strikes a plate, calculate how far it travels horizontally before the collision.

Solution 2

The electron strikes the lower plate after travelling approximately \(32.0\text{ mm}\) horizontally. It does not reach the end of the plates.

The important detail is that we must calculate when the electron first reaches the plate, rather than simply calculating its hypothetical displacement at the exit.

Step 1: Calculate the signed potential difference.

\[
\Delta V_p=V_{\mathrm{top}}-V_{\mathrm{bottom}}
\]

\[
\Delta V_p=0-240
\]

\[
\Delta V_p=-240\text{ V}
\]

Step 2: Calculate the electric field.

\[
E_y=-\frac{\Delta V_p}{d}
\]

\[
E_y=-\frac{-240}{0.012}
\]

\[
E_y=+2.0\times10^4\text{ V m}^{-1}
\]

The electric field points upwards.

Since the electron is negatively charged, its force and acceleration point downwards.

Step 3: Calculate the vertical acceleration.

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

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

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

Step 4: Calculate the time until the electron reaches the lower plate.

The electron begins \(2.0\text{ mm}\) above the lower plate.

Therefore, its vertical displacement before contact is

\[
y=-0.0020\text{ m}
\]

Using

\[
y=\frac12a_yt^2
\]

we obtain

\[
-0.0020
=\frac12(-3.51\times10^{15})t^2
\]

Rearranging:

\[
t=\sqrt{\frac{2(0.0020)}{3.51\times10^{15}}}
\]

\[
t=1.07\times10^{-9}\text{ s}
\]

Step 5: Calculate the horizontal distance travelled.

The horizontal velocity remains constant.

\[
x=u_xt
\]

\[
x=(3.0\times10^7)(1.07\times10^{-9})
\]

\[
\boxed{x=0.0320\text{ m}=32.0\text{ mm}}
\]

The collision occurs after \(32.0\text{ mm}\), which is less than the plate length of \(50\text{ mm}\).

Therefore, the electron strikes the lower plate before reaching the exit.

A common mistake would be to calculate the displacement after the full \(50\text{ mm}\) and then report that as the exit position. That calculation assumes the electron continues travelling after it has already hit the plate.

Question 3

Electrons start from rest and are accelerated through a potential difference of \(+1.5\text{ kV}\) before entering a parallel-plate deflector.

The upper plate is \(120\text{ V}\) above the lower plate. The plates are \(40\text{ mm}\) long and separated by \(10\text{ mm}\).

Each electron enters horizontally halfway between the plates.

a) Calculate the vertical displacement at the plate exit.

b) Calculate the accelerating potential difference at which an electron just reaches the upper plate at the exit.

c) Determine whether an accelerating potential difference of \(1.0\text{ kV}\) allows the electron to clear the plates.

Solution 3

At an accelerating potential difference of \(1.5\text{ kV}\), the electron deflects \(3.20\text{ mm}\) upwards and clears the plates. The grazing threshold is \(960\text{ V}\), so an accelerating voltage of \(1.0\text{ kV}\) also clears the plates, but with very little margin.

Part a: Calculate the deflection at \(1.5\text{ kV}\).

Since the electron starts from rest, we can use

\[
y_{\mathrm{exit}}
=\frac{\Delta V_pL^2}
{4d\Delta V_a}
\]

Substitute:

\[
y_{\mathrm{exit}}
=
\frac{(120)(0.040)^2}
{4(0.010)(1500)}
\]

\[
y_{\mathrm{exit}}
=\frac{0.192}{60}
\]

\[
\boxed{y_{\mathrm{exit}}=+3.20\text{ mm}}
\]

The positive result means the electron deflects upwards.

Part b: Find the grazing threshold.

The electron enters halfway between the plates.

The upper plate is therefore \(5.0\text{ mm}\) above its starting position.

At the grazing threshold,

\[
y_{\mathrm{exit}}=\frac d2
\]

Substituting into the voltage-ratio equation:

\[
\frac d2=
\frac{\Delta V_pL^2}
{4d\Delta V_a}
\]

Rearranging:

\[
\Delta V_a
=\frac{\Delta V_pL^2}{2d^2}
\]

Substitute the values:

\[
\Delta V_a
=
\frac{(120)(0.040)^2}
{2(0.010)^2}
\]

\[
\boxed{\Delta V_a=960\text{ V}}
\]

At this voltage, the electron just reaches the upper plate at the exit.

Part c: Check an accelerating voltage of \(1.0\text{ kV}\).

Using the same equation:

\[
y_{\mathrm{exit}}
=
\frac{(120)(0.040)^2}
{4(0.010)(1000)}
\]

\[
y_{\mathrm{exit}}=4.80\text{ mm}
\]

The available clearance is \(5.0\text{ mm}\).

Therefore, the electron clears the upper plate by

\[
5.0-4.8=0.20\text{ mm}
\]

It reaches the exit without colliding, although the clearance is very small.

This example demonstrates an important relationship: increasing the accelerating voltage increases the electron’s entry speed and reduces its time between the plates. Consequently, the deflection decreases.

Question 4

An electron and a proton enter separately into the same parallel-plate deflector.

The upper plate is \(120\text{ V}\) above the lower plate. The plates are \(30\text{ mm}\) long and separated by \(10\text{ mm}\).

Both particles enter horizontally halfway between the plates with the same initial speed of \(2.0\times10^7\text{ m s}^{-1}\).

Calculate the signed vertical displacement of each particle at the plate exit.

Explain why the particles experience electric forces of equal magnitude but have very different deflections.

Solution 4

The electron deflects \(+2.37\text{ mm}\) upwards, while the proton deflects approximately \(1.29\text{ μm}\) downwards.

Although their charge magnitudes are equal, their masses are very different.

Step 1: Calculate the electric field.

\[
E_y=-\frac{120}{0.010}
\]

\[
E_y=-1.20\times10^4\text{ V m}^{-1}
\]

The electric field points downwards.

Step 2: Calculate the time spent between the plates.

Both particles have the same horizontal speed.

\[
t=\frac{0.030}{2.0\times10^7}
\]

\[
t=1.50\times10^{-9}\text{ s}
\]

Step 3: Calculate the electron’s displacement.

For the electron,

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

\[
a_e=
\frac{(-1.60\times10^{-19})(-1.20\times10^4)}
{9.11\times10^{-31}}
\]

\[
a_e=+2.11\times10^{15}\text{ m s}^{-2}
\]

Its displacement is

\[
y_e=\frac12a_et^2
\]

\[
y_e=
\frac12(2.11\times10^{15})
(1.50\times10^{-9})^2
\]

\[
\boxed{y_e=+2.37\text{ mm}}
\]

The electron accelerates upwards because it is negatively charged.

Step 4: Calculate the proton’s displacement.

The proton has a positive charge:

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

Its acceleration is

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

\[
a_p=
\frac{(1.60\times10^{-19})(-1.20\times10^4)}
{1.67\times10^{-27}}
\]

\[
a_p=-1.15\times10^{12}\text{ m s}^{-2}
\]

Its displacement is

\[
y_p=\frac12a_pt^2
\]

\[
y_p=
\frac12(-1.15\times10^{12})
(1.50\times10^{-9})^2
\]

\[
\boxed{y_p=-1.29\times10^{-6}\text{ m}}
\]

The proton accelerates downwards, in the same direction as the electric field.

Why are the deflections so different?

The electric force is

\[
F=qE
\]

The electron and proton have equal charge magnitudes, so they experience equal force magnitudes in the same electric field.

However, Newton’s second law gives

\[
a=\frac Fm
\]

The proton is approximately \(1830\) times more massive than the electron.

Its acceleration is therefore approximately \(1830\) times smaller.

Since both particles spend the same amount of time between the plates, their displacement magnitudes also differ by approximately the same factor.

Equal electric forces do not imply equal accelerations or equal deflections.

Question 5

Two parallel-plate deflectors are arranged in series, with a field-free region between them.

Each deflector has a plate length of \(40\text{ mm}\) and a separation of \(20\text{ mm}\).

The first deflector has its upper plate at \(+80\text{ V}\) relative to its lower plate.

The second deflector has its upper plate at \(80\text{ V}\) below its lower plate.

The field-free gap between the two deflectors is \(20\text{ mm}\).

An electron enters the first deflector horizontally, halfway between the plates, with a speed of \(2.0\times10^7\text{ m s}^{-1}\).

Assume the following:

• Both electric fields are uniform.
• The field-free region has negligible electric and magnetic fields.
• There are no significant edge effects.
• The electron maintains constant horizontal velocity throughout.

Determine:

a) The electron’s vertical displacement and vertical velocity at the exit of the first deflector.

b) Its additional vertical displacement while travelling through the field-free gap.

c) Its final vertical displacement and vertical velocity at the exit of the second deflector.

Does the second deflector return the electron to its original horizontal path? Explain your reasoning.

Solution 5

The electron exits the second deflector \(4.22\text{ mm}\) above its original horizontal path, with zero vertical velocity.

The second deflector restores the electron’s original horizontal direction, but it does not restore its original vertical position.

This is the crucial distinction between cancelling a velocity change and cancelling a displacement.

Step 1: Calculate the acceleration in the first deflector.

The signed potential difference is

\[
\Delta V_{p1}=+80\text{ V}
\]

The electric field is

\[
E_{y1}=-\frac{80}{0.020}
\]

\[
E_{y1}=-4.0\times10^3\text{ V m}^{-1}
\]

For an electron,

\[
a_{y1}=\frac{qE_{y1}}{m_e}
\]

\[
a_{y1}=
\frac{(-1.60\times10^{-19})(-4.0\times10^3)}
{9.11\times10^{-31}}
\]

\[
a_{y1}=+7.03\times10^{14}\text{ m s}^{-2}
\]

The electron accelerates upwards.

Step 2: Calculate the time spent in the first deflector.

\[
t_1=\frac{L}{u_x}
\]

\[
t_1=\frac{0.040}{2.0\times10^7}
\]

\[
t_1=2.0\times10^{-9}\text{ s}
\]

Step 3: Calculate the displacement and velocity at the first exit.

The initial vertical velocity is zero.

Therefore,

\[
y_1=\frac12a_{y1}t_1^2
\]

\[
y_1=
\frac12(7.03\times10^{14})
(2.0\times10^{-9})^2
\]

\[
\boxed{y_1=+1.41\text{ mm}}
\]

The vertical exit velocity is

\[
v_{y1}=a_{y1}t_1
\]

\[
v_{y1}=(7.03\times10^{14})(2.0\times10^{-9})
\]

\[
\boxed{v_{y1}=+1.41\times10^6\text{ m s}^{-1}}
\]

The electron therefore leaves the first deflector moving upwards and to the right.

Step 4: Calculate the displacement through the field-free gap.

This is where a common misconception appears.

Some students assume that once the electron leaves the first deflector, its vertical motion stops.

It does not.

There is no vertical acceleration in the gap, but the electron still has an upward vertical velocity of \(1.41\times10^6\text{ m s}^{-1}\).

The time spent travelling through the \(20\text{ mm}\) gap is

\[
t_g=\frac{0.020}{2.0\times10^7}
\]

\[
t_g=1.0\times10^{-9}\text{ s}
\]

Its additional vertical displacement is

\[
\Delta y_g=v_{y1}t_g
\]

\[
\Delta y_g=(1.41\times10^6)(1.0\times10^{-9})
\]

\[
\boxed{\Delta y_g=+1.41\text{ mm}}
\]

Before entering the second deflector, the electron is now

\[
1.41+1.41=2.82\text{ mm}
\]

above its original path.

Step 5: Calculate the acceleration in the second deflector.

The second deflector has the opposite potential difference:

\[
\Delta V_{p2}=-80\text{ V}
\]

Therefore,

\[
E_{y2}=-\frac{-80}{0.020}
\]

\[
E_{y2}=+4.0\times10^3\text{ V m}^{-1}
\]

The electric field points upwards.

The electron’s acceleration is therefore downwards:

\[
a_{y2}=\frac{qE_{y2}}{m_e}
\]

\[
a_{y2}=-7.03\times10^{14}\text{ m s}^{-2}
\]

Notice that this acceleration has exactly the same magnitude as the first acceleration, but the opposite direction.

Step 6: Calculate the additional displacement in the second deflector.

The electron enters the second deflector with an upward vertical velocity.

We must therefore include its initial vertical velocity in the displacement equation:

\[
\Delta y_2
=v_{y1}t_1+\frac12a_{y2}t_1^2
\]

Substitute:

\[
\begin{aligned}
\Delta y_2
&=(1.41\times10^6)(2.0\times10^{-9})\\
&\quad+\frac12(-7.03\times10^{14})
(2.0\times10^{-9})^2\\
&=2.82\times10^{-3}
-1.41\times10^{-3}\\
&=+1.41\times10^{-3}\text{ m}
\end{aligned}
\]

Therefore,

\[
\boxed{\Delta y_2=+1.41\text{ mm}}
\]

The electron continues moving upwards throughout the second deflector, although its upward velocity decreases.

Step 7: Calculate the final vertical displacement.

The total displacement is the sum of the contributions from all three regions:

\[
y_{\mathrm{final}}=y_1+\Delta y_g+\Delta y_2
\]

\[
y_{\mathrm{final}}=1.41+1.41+1.41
\]

\[
\boxed{y_{\mathrm{final}}=+4.22\text{ mm}}
\]

Step 8: Calculate the final vertical velocity.

The second deflector acts for the same amount of time as the first.

Its downward acceleration removes the upward velocity gained in the first deflector:

\[
v_{y,\mathrm{final}}
=v_{y1}+a_{y2}t_1
\]

\[
v_{y,\mathrm{final}}
=1.41\times10^6
-(7.03\times10^{14})(2.0\times10^{-9})
\]

\[
\boxed{v_{y,\mathrm{final}}=0}
\]

The electron leaves the second deflector travelling horizontally, but \(4.22\text{ mm}\) above its original path.

It also remains between the plates throughout both deflectors because its maximum upward displacement is less than the \(10\text{ mm}\) half-gap.

The tempting prediction is that equal and opposite electric fields must cancel the electron’s deflection.

But the fields cancel the change in vertical velocity, not the accumulated displacement. The electron has already moved upwards, and it continues moving upwards during the field-free gap and the second deflector.

This is why you must track both position and velocity when a charged particle passes through more than one region of an electric field.

10What to calculate next

Once you know the electron’s displacement at the plate exit, the next useful quantity is its vertical exit velocity:

\[
v_{y,\mathrm{exit}}
=\frac{qE_yL}{m_eu_x}
\]

Together with its constant horizontal velocity, this determines the angle at which it leaves the plates.

Beyond the plates, the electric acceleration becomes negligible, so the electron travels in a straight line at that new angle.

That is the next step in predicting the final screen position, which is covered in How Charged Particles Deflect Between Parallel Plates.