How to Calculate a Charged Particle’s Exit Angle in an Electric Field

An electron enters horizontally between two parallel plates. The upper plate is at a higher electric potential than the lower plate, so the electron accelerates upwards. By the time it reaches the end of the plates, its velocity is no longer horizontal.

How do you calculate the angle at which it leaves?

The key is to find the particle’s horizontal and vertical velocity components at the instant it exits the electric field. The exit angle comes from their ratio, not from the particle’s total sideways displacement.

For a particle entering horizontally at speed \(u\), travelling through a uniform vertical electric field over a plate length \(L\), the exit angle is

\[
\boxed{\theta=\tan^{-1}\left(
-\frac{q\Delta V_p L}{mdu^2}
\right)}
\]

Here:

• \(q\) is the signed particle charge, in coulombs (C).
• \(\Delta V_p=V_{\text{top}}-V_{\text{bottom}}\) is the signed potential difference between the plates, in volts (V).
• \(d\) is the plate separation, in metres (m).
• \(L\) is the horizontal length of the electric field, in metres (m).
• \(m\) is the particle mass, in kilograms (kg).
• \(u\) is its initial horizontal speed, in metres per second (\(\mathrm{m\,s^{-1}}\)).
• \(\theta\) is measured from the positive horizontal direction. An upward exit angle is positive, and a downward exit angle is negative.

This expression assumes a uniform vertical electric field, horizontal entry, and non-relativistic motion.

Let’s build the reasoning behind it, including how to handle negative charges and potential differences without losing track of signs.

01The exit angle is determined by velocity, not displacement

Imagine a charged particle entering the space between two horizontal plates.

The electric field acts vertically. If there is no horizontal electric force, the particle keeps its horizontal velocity while gaining vertical velocity.

Its path therefore curves as it travels through the plates.

Notice something important about the exit angle.

It describes the direction in which the particle is moving at the instant it leaves the field. It does not describe the angle of the straight line joining its entry and exit positions.

That distinction catches students out.

The velocity triangle

At the exit, the particle has two velocity components:

• \(v_x\): horizontal velocity.
• \(v_y\): vertical velocity.

These combine to produce the resultant velocity, \(v\).

From basic trigonometry,

\[
\tan\theta=\frac{v_y}{v_x}
\]

Therefore,

\[
\boxed{\theta=\tan^{-1}\left(\frac{v_y}{v_x}\right)}
\]

This is the relationship to remember.

You need the two velocity components at the exit. Once you have them, finding the angle is straightforward.

Why you cannot simply use the displacement

Suppose the particle enters horizontally and moves upwards by a vertical displacement \(\Delta y\) over a horizontal distance \(L\).

You might try:

\[
\tan\theta=\frac{\Delta y}{L}
\]

That gives the angle of the straight chord joining the entry and exit positions. It does not give the exit angle.

Why?

The particle starts with zero vertical velocity and accelerates vertically throughout its journey. Its vertical velocity is therefore smaller at the beginning than at the end.

For constant vertical acceleration,

\[
\Delta y=\frac12 a_y t^2
\]

and

\[
v_y=a_y t
\]

Since the horizontal velocity remains constant,

\[
L=v_x t
\]

Combining these expressions gives

\[
\boxed{\tan\theta=\frac{2\Delta y}{L}}
\]

This shortcut works only when the particle enters with zero vertical velocity and experiences constant vertical acceleration.

It is another useful way to check an answer, but the velocity-component method is more general.

02Get the electric field direction right

Before calculating anything, establish which way the electric field points.

The electric field points from higher electric potential towards lower electric potential. It also points in the direction of the force on a positive test charge.

A negative charge experiences a force in the opposite direction.

This is where the sign of the charge matters.

The electric force is

\[
\vec F=q\vec E
\]

Newton’s second law gives

\[
\vec a=\frac{q\vec E}{m}
\]

Here, \(q\) is the actual signed charge, not simply its magnitude.

For example:

A positive particle accelerates towards lower electric potential. A negative particle accelerates towards higher electric potential.

Use a consistent sign convention

For the calculations in this article, choose:

• Positive \(x\): horizontally to the right.
• Positive \(y\): vertically upwards.
• Positive angle: above the horizontal.
• Negative angle: below the horizontal.

Define the signed potential difference between the plates as

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

The vertical electric field is then

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

The negative sign is important.

If the upper plate is at a higher potential, \(\Delta V_p\) is positive, and the electric field points downwards.

If the upper plate is at a lower potential, \(\Delta V_p\) is negative, and the electric field points upwards.

For a positive particle, the force follows this direction. For a negative particle, it points the other way.

You can summarise the possibilities:

Do not automatically make the electric field strength positive by replacing every voltage with its magnitude. You can use magnitudes when calculating the size of a quantity, but you must keep its direction separately.

For the signed method, retain the negative sign in \(E_y=-\Delta V_p/d\).

03Calculate the velocity components at the exit

Now that the directions are clear, the calculation is mostly projectile-motion physics.

The electric force acts vertically, so:

\[
a_x=0
\]

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

The horizontal velocity therefore stays constant.

If the particle enters horizontally with speed \(u\),

\[
v_x=u
\]

The time spent between the plates is

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

The vertical velocity changes because of the electric acceleration:

\[
v_y=u_y+a_y t
\]

For horizontal entry, the initial vertical velocity \(u_y\) is zero. Therefore,

\[
v_y=a_y t
\]

Substituting the time gives

\[
v_y=\frac{qE_yL}{mu}
\]

We now have both velocity components.

The exit angle is

\[
\tan\theta=\frac{v_y}{v_x}
\]

Substituting the expressions for the components,

\[
\tan\theta=
\frac{qE_yL}{mu^2}
\]

Finally, using \(E_y=-\Delta V_p/d\),

\[
\boxed{
\theta=\tan^{-1}\left(
-\frac{q\Delta V_pL}{mdu^2}
\right)
}
\]

That’s the complete method for horizontal entry.

What if the particle enters at an angle?

The same method still works.

Suppose the particle enters with speed \(u\) at an angle \(\alpha\) above the horizontal.

Resolve its initial velocity:

\[
u_x=u\cos\alpha
\]

\[
u_y=u\sin\alpha
\]

The time inside the field is now

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

The exit components become

\[
v_x=u_x
\]

\[
v_y=u_y+\frac{qE_yL}{mu_x}
\]

The exit angle is therefore

\[
\boxed{
\theta=\tan^{-1}\left(\frac{v_y}{v_x}\right)
}
\]

For a more general velocity direction, use the two-component angle function \(\operatorname{atan2}(v_y,v_x)\) to identify the correct quadrant.

The important point is that you must include the particle’s initial vertical velocity. The electric field changes it; it does not automatically start from zero.

05What if the entry speed comes from an accelerating voltage?

Many HSC questions do not give you the particle’s initial speed directly.

Instead, they tell you that the particle starts from rest and accelerates through a potential difference before entering the deflecting plates.

You need to calculate its speed first.

This is where conservation of energy comes in.

The change in electric potential energy is

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

where \(\Delta V=V_f-V_i\) is the potential difference experienced by the particle.

If electric potential energy decreases, kinetic energy increases.

Therefore,

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

This is the safest way to handle accelerating voltages because it keeps the charge sign consistent.

For a particle starting from rest,

\[
\frac12mu^2=-q\Delta V_{\rm acc}
\]

where \(\Delta V_{\rm acc}\) is the signed potential change between the starting position and the particle’s entry point into the deflecting field.

Rearranging,

\[
\boxed{
u=\sqrt{\frac{-2q\Delta V_{\rm acc}}{m}}
}
\]

The expression inside the square root must be positive if the particle starts from rest and gains kinetic energy.

Consider the signs:

• A proton gains kinetic energy when it moves to a lower potential, so \(\Delta V_{\rm acc}\) is negative.
• An electron gains kinetic energy when it moves to a higher potential, so \(\Delta V_{\rm acc}\) is positive.

Both situations produce a positive kinetic energy change.

If you are unsure how to calculate a particle’s speed from an accelerating voltage, see How to Calculate Electron Speed After Acceleration Through Voltage.

A useful shortcut when the particle starts from rest

There is an interesting simplification when you combine the accelerating voltage equation with the exit-angle equation.

Recall:

\[
\tan\theta=
-\frac{q\Delta V_pL}{mdu^2}
\]

From the accelerating voltage,

\[
u^2=-\frac{2q\Delta V_{\rm acc}}{m}
\]

Substituting,

\[
\tan\theta=
-\frac{q\Delta V_pL}
{md\left(-\frac{2q\Delta V_{\rm acc}}{m}\right)}
\]

The charge and mass cancel:

\[
\boxed{
\tan\theta=
\frac{\Delta V_pL}
{2d\Delta V_{\rm acc}}
}
\]

This shortcut applies when:

• The particle starts from rest.
• It is accelerated through the specified potential difference.
• It enters the deflecting field horizontally.
• The deflecting field is uniform and vertical.
• Its motion is non-relativistic.

Be careful: the voltages are still signed. The cancellation does not mean you can ignore the direction of acceleration.

07An important distinction: Plate voltage versus the particle’s potential change

There is one more voltage mistake worth avoiding.

The potential difference between the plates is not necessarily the potential difference experienced by the particle.

The plate voltage describes the potential difference across the entire gap.

But a particle entering halfway between the plates might only travel a few millimetres vertically before exiting.

It therefore experiences only part of the potential difference.

For a uniform vertical electric field,

\[
\Delta V_{\rm particle}=-E_y\Delta y
\]

Since

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

we can also write

\[
\boxed{
\Delta V_{\rm particle}
=\frac{\Delta V_p}{d}\Delta y
}
\]

Here:

• \(\Delta V_{\rm particle}\) is the signed potential change experienced along the particle’s actual path.
• \(\Delta y\) is its signed vertical displacement.
• \(\Delta V_p\) is the signed potential difference between the plates.

The kinetic energy change is then

\[
\boxed{\Delta K=-q\Delta V_{\rm particle}}
\]

This matters because using the full plate voltage in the energy equation would assume that the particle travels through the entire plate separation.

It usually doesn’t.

For a particle entering horizontally, the vertical kinetic energy at the exit can also be checked using

\[
\frac12mv_y^2=qE_y\Delta y
\]

This is a useful independent check on a longer calculation.

For a fuller explanation of the relationship between charge, electric potential energy, and movement through an electric field, see Can a Charged Particle Move Without Changing Electric Potential Energy?.

08Questions and solutions

Try each question before reading its solution. The first uses the standard horizontal-entry method. The later questions introduce signed accelerating voltages, initial vertical velocity, and a more subtle energy constraint.

Question 1

An electron enters horizontally between two parallel plates at a speed of \(2.5\times10^7\,\mathrm{m\,s^{-1}}\).

The upper plate is at \(+150\,\mathrm V\), and the lower plate is at \(0\,\mathrm V\). The plate separation is \(2.5\times10^{-2}\,\mathrm m\), and the plate length is \(4.0\times10^{-2}\,\mathrm m\).

Calculate the electron’s exit angle, including its direction.

Use:

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

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

Solution 1

The electron exits at approximately \(3.9^\circ\) above the horizontal.

The upper plate is at a higher potential, so the electric field points downwards. The electron experiences an upward force because its charge is negative.

First, calculate the electric field:

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

\[
E_y=-\frac{150}{0.025}
\]

\[
E_y=-6.0\times10^3\,\mathrm{N\,C^{-1}}
\]

Next, calculate the electron’s acceleration:

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

\[
a_y=
\frac{(-1.60\times10^{-19})(-6.0\times10^3)}
{9.11\times10^{-31}}
\]

\[
a_y=1.05\times10^{15}\,\mathrm{m\,s^{-2}}
\]

The time inside the field is

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

\[
t=\frac{0.040}{2.5\times10^7}
\]

\[
t=1.60\times10^{-9}\,\mathrm s
\]

The vertical exit velocity is

\[
v_y=a_yt
\]

\[
v_y=(1.05\times10^{15})(1.60\times10^{-9})
\]

\[
v_y=1.69\times10^6\,\mathrm{m\,s^{-1}}
\]

The horizontal component remains

\[
v_x=2.5\times10^7\,\mathrm{m\,s^{-1}}
\]

Therefore,

\[
\theta=\tan^{-1}\left(
\frac{1.69\times10^6}{2.5\times10^7}
\right)
\]

\[
\boxed{\theta=3.9^\circ\text{ above the horizontal}}
\]

The important sign decision was recognising that the electron accelerates opposite to the electric field.

Question 2

A proton enters a uniform electric field at a speed of \(4.0\times10^5\,\mathrm{m\,s^{-1}}\), travelling \(5.0^\circ\) above the horizontal.

The field exists between two horizontal parallel plates:

• Upper plate potential: \(-90\,\mathrm V\).
• Lower plate potential: \(+90\,\mathrm V\).
• Plate separation: \(3.0\times10^{-2}\,\mathrm m\).
• Plate length: \(4.0\times10^{-2}\,\mathrm m\).

Calculate:

a. The proton’s horizontal and vertical velocity components at the exit.

b. Its exit angle.

c. Whether its vertical velocity has reversed direction.

Use:

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

\[
m_p=1.67\times10^{-27}\,\mathrm{kg}
\]

Solution 2

The proton exits at approximately \(13.1^\circ\) above the horizontal, with its vertical velocity still pointing upwards.

The important difference from Question 1 is that the proton already has an initial vertical velocity. We must include it.

Step 1: Determine the electric field direction.

The signed plate potential difference is

\[
\Delta V_p=-90-(+90)=-180\,\mathrm V
\]

Therefore,

\[
E_y=-\frac{-180}{0.030}
\]

\[
E_y=+6.0\times10^3\,\mathrm{N\,C^{-1}}
\]

The field points upwards.

Since the proton is positive, its force and acceleration also point upwards.

Step 2: Resolve the initial velocity.

The initial horizontal component is

\[
u_x=u\cos5^\circ
\]

\[
u_x=(4.0\times10^5)\cos5^\circ
\]

\[
u_x=3.98\times10^5\,\mathrm{m\,s^{-1}}
\]

The initial vertical component is

\[
u_y=u\sin5^\circ
\]

\[
u_y=(4.0\times10^5)\sin5^\circ
\]

\[
u_y=3.49\times10^4\,\mathrm{m\,s^{-1}}
\]

Step 3: Calculate the vertical acceleration.

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

\[
a_y=
\frac{(1.60\times10^{-19})(6.0\times10^3)}
{1.67\times10^{-27}}
\]

\[
a_y=5.75\times10^{11}\,\mathrm{m\,s^{-2}}
\]

Step 4: Calculate the time inside the plates.

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

\[
t=\frac{0.040}{3.98\times10^5}
\]

\[
t=1.00\times10^{-7}\,\mathrm s
\]

Step 5: Calculate the exit velocity components.

The horizontal component is unchanged:

\[
v_x=3.98\times10^5\,\mathrm{m\,s^{-1}}
\]

The vertical component is

\[
v_y=u_y+a_yt
\]

\[
v_y=3.49\times10^4+
(5.75\times10^{11})(1.00\times10^{-7})
\]

\[
v_y\approx9.26\times10^4\,\mathrm{m\,s^{-1}}
\]

The vertical velocity is still positive.

Step 6: Calculate the exit angle.

\[
\theta=\tan^{-1}\left(
\frac{9.26\times10^4}{3.98\times10^5}
\right)
\]

\[
\boxed{\theta\approx13.1^\circ}
\]

The proton exits at approximately \(13.1^\circ\) above the horizontal.

It has not reversed its vertical direction. The upward electric force has increased its already upward vertical velocity.

This is an important distinction: the direction of acceleration does not necessarily match the direction in which a particle is travelling.

Question 3

An electron starts from rest and is accelerated through a potential change of \(+1.50\times10^3\,\mathrm V\).

It then enters horizontally between two parallel deflecting plates.

The upper plate is at \(+150\,\mathrm V\), and the lower plate is at \(0\,\mathrm V\). The plate separation is \(2.5\times10^{-2}\,\mathrm m\), and the plate length is \(5.0\times10^{-2}\,\mathrm m\).

Calculate:

a. The electron’s speed as it enters the deflecting field.

b. Its exit angle.

Use:

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

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

Solution 3

The electron enters the deflecting field at approximately \(2.30\times10^7\,\mathrm{m\,s^{-1}}\) and exits at \(5.7^\circ\) above the horizontal.

We need to use energy conservation before calculating the deflection.

Step 1: Calculate the kinetic energy gained during acceleration.

The electron has negative charge, and its potential change is positive.

Using

\[
\Delta K=-q\Delta V_{\rm acc}
\]

\[
\Delta K=
-(-1.60\times10^{-19})(1.50\times10^3)
\]

\[
\Delta K=2.40\times10^{-16}\,\mathrm J
\]

Step 2: Calculate the entry speed.

The electron starts from rest, so

\[
\frac12mu^2=\Delta K
\]

Rearranging,

\[
u=\sqrt{\frac{2\Delta K}{m}}
\]

\[
u=\sqrt{
\frac{2(2.40\times10^{-16})}
{9.11\times10^{-31}}
}
\]

\[
\boxed{u=2.30\times10^7\,\mathrm{m\,s^{-1}}}
\]

Step 3: Calculate the electric field in the deflecting region.

The signed plate voltage is

\[
\Delta V_p=150-0=+150\,\mathrm V
\]

Therefore,

\[
E_y=-\frac{150}{0.025}
\]

\[
E_y=-6.0\times10^3\,\mathrm{N\,C^{-1}}
\]

The electric field points downwards.

The electron accelerates upwards.

Step 4: Calculate the time inside the field.

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

\[
t=\frac{0.050}{2.30\times10^7}
\]

\[
t=2.18\times10^{-9}\,\mathrm s
\]

Step 5: Calculate the vertical exit velocity.

The vertical acceleration is

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

\[
a_y=
\frac{(-1.60\times10^{-19})(-6.0\times10^3)}
{9.11\times10^{-31}}
\]

\[
a_y=1.05\times10^{15}\,\mathrm{m\,s^{-2}}
\]

Since the electron enters horizontally,

\[
v_y=a_yt
\]

\[
v_y=(1.05\times10^{15})(2.18\times10^{-9})
\]

\[
v_y=2.29\times10^6\,\mathrm{m\,s^{-1}}
\]

Step 6: Calculate the exit angle.

The horizontal velocity remains

\[
v_x=2.30\times10^7\,\mathrm{m\,s^{-1}}
\]

Therefore,

\[
\theta=\tan^{-1}\left(
\frac{2.29\times10^6}{2.30\times10^7}
\right)
\]

\[
\boxed{\theta=5.7^\circ\text{ above the horizontal}}
\]

A useful shortcut would have been to substitute directly into

\[
\tan\theta=
\frac{\Delta V_pL}{2d\Delta V_{\rm acc}}
\]

The result is the same. However, understanding the energy calculation is more important than memorising the shortcut.

Question 4

A proton enters horizontally oriented parallel plates at a speed of \(3.0\times10^5\,\mathrm{m\,s^{-1}}\), travelling \(4.0^\circ\) above the horizontal.

The upper plate is at \(+12.4\,\mathrm V\), and the lower plate is at \(0\,\mathrm V\).

The plates are separated by \(2.0\times10^{-2}\,\mathrm m\), and their length is \(6.0\times10^{-2}\,\mathrm m\).

Determine:

a. The proton’s exit angle.

b. Its vertical displacement while between the plates.

c. Whether its vertical velocity has reversed direction.

Explain why using \(\tan\theta=\Delta y/L\) would give an incorrect exit angle.

Use:

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

\[
m_p=1.67\times10^{-27}\,\mathrm{kg}
\]

Solution 4

The proton exits at approximately \(1.7^\circ\) above the horizontal, after moving approximately \(3.0\,\mathrm{mm}\) upwards. Its vertical velocity has not reversed.

The interesting part is that the electric force acts downwards, even though the proton is still travelling upwards when it exits.

Step 1: Determine the electric field.

The signed potential difference is

\[
\Delta V_p=12.4-0=+12.4\,\mathrm V
\]

Therefore,

\[
E_y=-\frac{12.4}{0.020}
\]

\[
E_y=-620\,\mathrm{N\,C^{-1}}
\]

The electric field points downwards.

Because the proton is positive, its acceleration also points downwards.

Step 2: Resolve the initial velocity.

\[
u_x=u\cos4^\circ
\]

\[
u_x=(3.0\times10^5)\cos4^\circ
\]

\[
u_x=2.99\times10^5\,\mathrm{m\,s^{-1}}
\]

The initial vertical velocity is

\[
u_y=u\sin4^\circ
\]

\[
u_y=(3.0\times10^5)\sin4^\circ
\]

\[
u_y=2.09\times10^4\,\mathrm{m\,s^{-1}}
\]

Step 3: Calculate the vertical acceleration.

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

\[
a_y=
\frac{(1.60\times10^{-19})(-620)}
{1.67\times10^{-27}}
\]

\[
a_y=-5.94\times10^{10}\,\mathrm{m\,s^{-2}}
\]

The negative sign confirms that the acceleration points downwards.

Step 4: Calculate the time inside the plates.

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

\[
t=\frac{0.060}{2.99\times10^5}
\]

\[
t=2.01\times10^{-7}\,\mathrm s
\]

Step 5: Calculate the vertical exit velocity.

Using

\[
v_y=u_y+a_yt
\]

\[
v_y=
2.09\times10^4+
(-5.94\times10^{10})(2.01\times10^{-7})
\]

\[
v_y\approx9.02\times10^3\,\mathrm{m\,s^{-1}}
\]

The result is still positive.

This means the proton is still travelling upwards, but its upward vertical velocity has decreased.

Step 6: Calculate the exit angle.

The horizontal velocity remains unchanged:

\[
v_x=2.99\times10^5\,\mathrm{m\,s^{-1}}
\]

Therefore,

\[
\theta=\tan^{-1}\left(
\frac{9.02\times10^3}{2.99\times10^5}
\right)
\]

\[
\boxed{\theta\approx1.7^\circ\text{ above the horizontal}}
\]

Step 7: Calculate the vertical displacement.

Use the vertical displacement equation:

\[
\Delta y=u_yt+\frac12a_yt^2
\]

Substitute:

\[
\Delta y=
(2.09\times10^4)(2.01\times10^{-7})
\]

\[
\qquad+
\frac12(-5.94\times10^{10})
(2.01\times10^{-7})^2
\]

\[
\Delta y\approx3.00\times10^{-3}\,\mathrm m
\]

Therefore,

\[
\boxed{\Delta y=3.0\,\mathrm{mm}\text{ upwards}}
\]

Why is this not enough to calculate the exit angle using \(\tan\theta=\Delta y/L\)?

Because that expression gives the angle of the chord, not the exit velocity.

The proton starts with an upward vertical velocity, and the downward acceleration reduces that velocity as it travels.

Its average vertical velocity is therefore different from its final vertical velocity.

In fact,

\[
\tan\theta_{\rm out}
=
\frac{2\Delta y}{L}-\frac{u_y}{u_x}
\]

Substituting,

\[
\tan\theta_{\rm out}
=
\frac{2(0.0030)}{0.060}-\tan4^\circ
\]

\[
\tan\theta_{\rm out}\approx0.030
\]

This gives the same exit angle of approximately \(1.7^\circ\).

The key lesson is that a downward force does not immediately make an upward-moving particle travel downwards. It first reduces its upward velocity.

Question 5

An electron starts from rest and is accelerated through a potential change of \(+1.00\times10^3\,\mathrm V\).

It then enters horizontally, halfway between two parallel deflecting plates.

The upper plate is at \(+200\,\mathrm V\) relative to the lower plate. The plates are separated by \(2.0\times10^{-2}\,\mathrm m\), and the electric field extends for \(6.0\times10^{-2}\,\mathrm m\) horizontally.

A student makes the following claim:

The electron gains \(200\,\mathrm{eV}\) of kinetic energy inside the deflecting plates because the potential difference between the plates is \(200\,\mathrm V\).

Determine:

a. The electron’s entry speed.

b. Its exit angle.

c. Whether it reaches the end of the plates without hitting either plate.

d. The actual kinetic energy gained inside the deflecting region.

Assess the student’s claim, explaining the difference between the plate-to-plate potential difference and the potential change experienced by the electron.

Use:

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

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

Solution 5

The electron exits at approximately \(16.7^\circ\) above the horizontal, clears the upper plate by approximately \(1.0\,\mathrm{mm}\), and gains approximately \(90\,\mathrm{eV}\) of kinetic energy inside the deflecting region.

The student’s claim is incorrect.

The \(200\,\mathrm V\) is the potential difference across the entire gap. The electron does not travel through the entire gap, so it does not experience the full \(200\,\mathrm V\) potential change.

This question requires us to combine energy conservation, electric acceleration, exit velocity, and the geometry of the plates.

Step 1: Calculate the electron’s entry speed.

The electron starts from rest and experiences a positive potential change of \(1000\,\mathrm V\).

Its kinetic energy increases because its charge is negative.

Using

\[
\Delta K=-q\Delta V_{\rm acc}
\]

\[
\Delta K=
-(-1.60\times10^{-19})(1000)
\]

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

Now use

\[
\frac12mu^2=\Delta K
\]

Rearranging,

\[
u=\sqrt{\frac{2\Delta K}{m}}
\]

\[
u=
\sqrt{
\frac{2(1.60\times10^{-16})}
{9.11\times10^{-31}}
}
\]

\[
\boxed{u=1.87\times10^7\,\mathrm{m\,s^{-1}}}
\]

Step 2: Determine the electric field direction and acceleration.

The signed plate voltage is

\[
\Delta V_p=200-0=+200\,\mathrm V
\]

Therefore,

\[
E_y=-\frac{200}{0.020}
\]

\[
E_y=-1.00\times10^4\,\mathrm{N\,C^{-1}}
\]

The electric field points downwards.

The electron is negative, so its acceleration points upwards.

Calculate the acceleration:

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

\[
a_y=
\frac{(-1.60\times10^{-19})(-1.00\times10^4)}
{9.11\times10^{-31}}
\]

\[
\boxed{a_y=1.76\times10^{15}\,\mathrm{m\,s^{-2}}}
\]

Step 3: Calculate the time spent between the plates.

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

\[
t=\frac{0.060}{1.87\times10^7}
\]

\[
t=3.20\times10^{-9}\,\mathrm s
\]

Step 4: Calculate the exit velocity components.

The horizontal velocity remains constant:

\[
v_x=1.87\times10^7\,\mathrm{m\,s^{-1}}
\]

The vertical velocity is

\[
v_y=a_yt
\]

\[
v_y=(1.76\times10^{15})(3.20\times10^{-9})
\]

\[
v_y=5.62\times10^6\,\mathrm{m\,s^{-1}}
\]

Step 5: Calculate the exit angle.

\[
\tan\theta=\frac{v_y}{v_x}
\]

\[
\tan\theta=
\frac{5.62\times10^6}{1.87\times10^7}
\]

\[
\theta=\tan^{-1}(0.300)
\]

\[
\boxed{\theta=16.7^\circ\text{ above the horizontal}}
\]

The electron’s exit direction is now established.

Step 6: Check whether the electron hits a plate.

We need to calculate its vertical displacement while between the plates.

Since the electron enters horizontally,

\[
\Delta y=\frac12a_yt^2
\]

Substitute:

\[
\Delta y=
\frac12(1.76\times10^{15})
(3.20\times10^{-9})^2
\]

\[
\Delta y=9.0\times10^{-3}\,\mathrm m
\]

Therefore, the electron rises by

\[
\boxed{\Delta y=9.0\,\mathrm{mm}}
\]

The electron enters halfway between the plates.

The distance from its starting position to the upper plate is

\[
\frac d2=\frac{0.020}{2}
\]

\[
\frac d2=10\,\mathrm{mm}
\]

The remaining clearance is

\[
10-9.0=1.0\,\mathrm{mm}
\]

The electron reaches the end of the plates without hitting either plate.

This check matters. If the calculated vertical displacement had exceeded \(10\,\mathrm{mm}\), the electron would have struck the upper plate before reaching the end. In that situation, calculating an exit angle at the right-hand edge would not describe its actual motion.

Step 7: Calculate the actual potential change experienced by the electron.

This is the crucial part of the question.

The plate-to-plate potential difference is \(200\,\mathrm V\), but the electron only moves upwards by \(9.0\,\mathrm{mm}\) through a gap of \(20\,\mathrm{mm}\).

The actual potential change is

\[
\Delta V_{\rm particle}=-E_y\Delta y
\]

Substitute:

\[
\Delta V_{\rm particle}
=-(-1.00\times10^4)(9.0\times10^{-3})
\]

\[
\boxed{\Delta V_{\rm particle}=+90\,\mathrm V}
\]

The electron moves upwards towards the positive plate, so its electric potential increases.

Step 8: Calculate the kinetic energy gained.

Use

\[
\Delta K=-q\Delta V_{\rm particle}
\]

\[
\Delta K=
-(-1.60\times10^{-19})(90)
\]

\[
\Delta K=1.44\times10^{-17}\,\mathrm J
\]

Converting to electronvolts,

\[
\Delta K=
\frac{1.44\times10^{-17}}
{1.60\times10^{-19}}
\]

\[
\boxed{\Delta K=90\,\mathrm{eV}}
\]

The electron gains \(90\,\mathrm{eV}\), not \(200\,\mathrm{eV}\).

Its total kinetic energy at the exit is therefore approximately

\[
K_{\rm exit}=1000+90
\]

\[
\boxed{K_{\rm exit}=1090\,\mathrm{eV}}
\]

The original \(1000\,\mathrm{eV}\) came from the upstream accelerating region. The additional \(90\,\mathrm{eV}\) came from the vertical motion through the deflecting field.

The student’s mistake was treating the entire plate-to-plate voltage as the potential change experienced by the electron. That would only be correct if the electron travelled through the full vertical separation between the plates.

08Choosing the right method in an exam

The quickest way to approach an exit-angle problem is to identify what information the question gives you.

Before submitting your answer, check these five things:

• Have you used the signed charge \(q\), rather than automatically using its magnitude?
• Have you established the electric field direction from the plate potentials?
• Have you included any initial vertical velocity?
• Have you calculated the angle from the exit velocity components rather than the total displacement?
• Have you checked that the particle actually reaches the end of the plates?

One final limitation: Relativistic speeds

The calculations in this article use the non-relativistic equations

\[
K=\frac12mv^2
\]

and

\[
a=\frac{qE}{m}
\]

These are appropriate when the particle’s speed is much smaller than the speed of light,

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

If your calculated speed becomes a substantial fraction of \(c\), relativistic energy and momentum must be considered instead. Do not assume the non-relativistic equations remain accurate at arbitrarily high speeds.

For the usual non-relativistic HSC-style problems, the method developed here is the one to use.

09Where this connects next

Finding the exit angle is one part of understanding the complete trajectory of a charged particle.

Once the particle leaves the plates, the electric force disappears in the idealised model. Its horizontal and vertical velocity components remain constant, so it continues in a straight line along its exit direction.

That lets you tackle questions involving the position where a particle strikes a screen, the distance it travels after leaving the plates, and the relationship between electric deflection and particle speed.

For the broader treatment of the parabolic path inside the field and the straight-line motion afterwards, continue with Charged-Particle Trajectories in Electric Fields: HSC Physics.