Why an Electron Can Gain Kinetic Energy While Moving to Higher Potential

An electron moves from a point at \(-120\ \mathrm V\) to a point at \(+80\ \mathrm V\). Its electric potential has increased by \(200\ \mathrm V\). Yet the electron speeds up.

That sounds backwards. Surely moving to a higher potential should increase potential energy and leave less energy available for motion?

Not for an electron. An electron has negative charge, so its electric potential energy actually decreases when it moves to a higher electric potential. If the electric force is the only force doing work, that lost potential energy becomes kinetic energy.

Before going further, predict what would happen to a proton making exactly the same journey. It would lose kinetic energy, because a proton has positive charge. The difference comes down to the sign of the particle’s charge.

01Why higher electric potential means lower potential energy for an electron

The important distinction is between electric potential and electric potential energy. They are related, but they are not the same quantity.

Electric potential, \(V\), describes electric potential energy per unit positive charge. Its unit is the volt, where

\[
1\ \mathrm V=1\ \mathrm{J\,C^{-1}}
\]

Electric potential energy, \(U\), depends on both the electric potential and the charge placed there:

\[
U=qV
\]

Here:

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

The word signed matters. We must use the actual charge, including whether it is positive or negative.

For an electron,

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

where \(e\) is the positive magnitude of the elementary charge.

Substituting the electron’s charge into the potential energy equation gives

\[
U=-eV
\]

Notice the negative sign. When \(V\) increases, \(U\) decreases.

Follow the energy, not just the voltage

Return to our electron moving from \(-120\ \mathrm V\) to \(+80\ \mathrm V\).

Its change in electric potential is

\[
\Delta V=V_f-V_i
\]

\[
\Delta V=80-(-120)=+200\ \mathrm V
\]

The electron’s change in potential energy is therefore

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

\[
\Delta U=(-1.602\times10^{-19})(200)
\]

\[
\Delta U=-3.204\times10^{-17}\ \mathrm J
\]

The negative result tells us that the electron has lost electric potential energy.

If the electric field is the only force doing work, conservation of energy tells us that this energy becomes kinetic energy:

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

Therefore,

\[
\boxed{\Delta K=+3.204\times10^{-17}\ \mathrm J}
\]

The electron gains kinetic energy and speeds up.

Think of electric potential as an inverted landscape

A useful way to picture this is to imagine electric potential as a landscape.

For a positive charge, moving to higher electric potential is like climbing a hill in potential energy. The particle gains potential energy.

For an electron, the landscape is inverted because its charge is negative. A region of higher electric potential corresponds to lower electric potential energy.

The electron is effectively moving downhill in its own potential-energy landscape, even though the electric potential is increasing.

The analogy has one limitation: electric potential is not actual physical height. Its zero is also arbitrary. What matters physically is the change in potential between two positions.

02Why the electric force pushes an electron towards higher potential

There is another way to understand the same result. Instead of tracking energy, look at the direction of the electric force.

An electric field points in the direction a positive test charge would accelerate. In an electrostatic field, it points from higher electric potential towards lower electric potential.

The electric force on a charged particle is

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

where:

• \(\vec F\) is electric force in newtons (N).
• \(q\) is signed charge in coulombs (C).
• \(\vec E\) is electric field strength in newtons per coulomb (N C\(^{-1}\)).

For a positive charge, the force is in the same direction as the electric field.

For an electron, the force is in the opposite direction because \(q\) is negative.

This means the electric force on an electron points towards higher electric potential.

If the electron is released from rest, it accelerates in that direction. Its speed increases, and its kinetic energy increases.

There is an important distinction here: the force points towards higher potential, but an electron with an existing velocity can initially travel towards lower potential. In that situation, the electric force opposes its motion, so it slows down.

Electron versus proton: the sign changes everything

Consider two particles moving through the same potential difference:

\[
\Delta V=+200\ \mathrm V
\]

The electron and proton have equal charge magnitudes but opposite signs.

Assuming the electric field is the only force doing work. The proton must also have enough initial kinetic energy to complete the journey.

The same increase in electric potential produces opposite changes in kinetic energy.

This is why you cannot decide whether a particle speeds up simply by looking at whether its potential increases or decreases. You must consider the sign of its charge.

03The equation that handles every sign correctly

We can combine the relationships into one useful equation.

The change in electric potential energy is

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

When only the electric field does work, the work-energy theorem gives

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

Combining them:

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

The potential difference is always defined as

\[
\Delta V=V_f-V_i
\]

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

For an electron, substitute \(q=-e\):

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

This is the sign-safe equation for an electron moving through an electrostatic potential difference.

Remember that \(e\) is positive. The sign of the kinetic-energy change comes from the signed value of \(\Delta V\).

A note about electronvolts

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

It is the energy transferred to a particle with charge magnitude \(e\) when it moves through a potential difference of magnitude \(1\ \mathrm V\).

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

For an electron, the numerical value of its kinetic-energy change in electronvolts is equal to its signed potential difference in volts.

For example, if an electron moves through a potential increase of \(350\ \mathrm V\),

\[
\Delta K=+350\ \mathrm{eV}
\]

If it moves through a potential decrease of \(350\ \mathrm V\),

\[
\Delta K=-350\ \mathrm{eV}
\]

Notice that we have not taken the absolute value of the potential difference. Keeping the sign is what tells us whether the electron gains or loses energy.

04Worked examples

05Important assumptions and limitations

The equation

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

is extremely useful, but it has a condition: the electric field must be the only force doing work on the particle.

If another force acts, the more general work-energy relationship is

\[
\Delta K=W_E+W_{\text{other}}
\]

where \(W_E\) is the work done by the electric field and \(W_{\text{other}}\) is the net work done by all other forces.

This means an electron can move to a higher potential without gaining the full amount of kinetic energy predicted by the electric field alone. An external retarding force, for example, could remove some of the energy.

There is also a limitation when calculating speed.

The equation

\[
K=\frac12 mv^2
\]

is the classical, non-relativistic kinetic-energy equation. It assumes the particle’s speed is much smaller than the speed of light.

As a practical guide, at \(0.1c\), the classical kinetic energy is already about \(0.8\%\) below the relativistic value. At \(0.2c\), the difference is about \(3\%\). There is no sharp cutoff, but the approximation becomes progressively less accurate as speed increases.

For a relativistic calculation (an extension beyond the non-relativistic model), kinetic energy is given by

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

where

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

and \(c\) is the speed of light.

For the HSC-style calculations in this article, check that your calculated speed is comfortably below \(c\) before relying on the classical equation.

For a broader explanation of how electrical work connects to kinetic energy, see Work and Kinetic Energy of an Accelerated Charge. For a more detailed treatment of converting an electron’s energy gain into speed, see How to Calculate Electron Speed After Acceleration Through Voltage.

06Questions and solutions

Question 1

An electron moves from a point at \(-40\ \mathrm V\) to another point at \(+110\ \mathrm V\).

Assuming the electric field is the only force doing work, calculate:

(a) The potential difference.

(b) The change in the electron’s kinetic energy in joules and electronvolts.

(c) Whether the electron speeds up or slows down.

Solution 1

The electron gains \(150\ \mathrm{eV}\) of kinetic energy, equivalent to \(2.403\times10^{-17}\ \mathrm J\), so it speeds up.

First, calculate the signed potential difference:

\[
\begin{aligned}
\Delta V&=V_f-V_i\\
&=110-(-40)\\
&=+150\ \mathrm V
\end{aligned}
\]

For an electron,

\[
\Delta K=e\Delta V
\]

Therefore,

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

In electronvolts,

\[
\Delta K=+150\ \mathrm{eV}
\]

The positive change means the electron gains kinetic energy.

The important step is using the electron’s negative charge in the original equation, \(\Delta K=-q\Delta V\). Simply assuming that higher potential means higher potential energy would give the wrong result.

Question 2

Two parallel metal plates are separated by \(0.080\ \mathrm m\).

The left plate is at \(0\ \mathrm V\), and the right plate is at \(+480\ \mathrm V\). An electron enters the region near the left plate, travelling horizontally towards the right plate with an initial speed of \(3.0\times10^6\ \mathrm{m\,s^{-1}}\).

Assume the electric field is uniform between the plates, neglect fringing effects, and ignore collisions.

(a) Determine the direction of the electric field.

(b) Determine the direction and magnitude of the electric force on the electron.

(c) Calculate the electron’s acceleration.

(d) Calculate its speed when it reaches the right plate.

Solution 2

The electric field points to the left, but the electron experiences a force to the right. Its final speed is approximately \(1.33\times10^7\ \mathrm{m\,s^{-1}}\).

(a) Electric field direction

The electric field points from higher potential towards lower potential.

The right plate is at \(+480\ \mathrm V\), while the left plate is at \(0\ \mathrm V\).

Therefore, the electric field points from right to left.

(b) Electric force

The force on a charged particle is

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

Because the electron is negatively charged, its force is opposite to the electric field.

Therefore, the electric force points to the right, towards the positive plate.

First, calculate the electric field strength:

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

\[
E=\frac{480}{0.080}
\]

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

The force magnitude is

\[
F=eE
\]

\[
F=(1.602\times10^{-19})(6.0\times10^3)
\]

\[
\boxed{F=9.61\times10^{-16}\ \mathrm N}
\]

The force acts to the right.

(c) Acceleration

Using Newton’s second law,

\[
F=ma
\]

Rearrange:

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

Substitute the electron’s mass:

\[
a=\frac{9.61\times10^{-16}}{9.11\times10^{-31}}
\]

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

The acceleration is to the right, in the direction of the electron’s force.

(d) Final speed

We can use conservation of energy because the electric field is the only force doing work.

The potential difference is

\[
\Delta V=480-0=+480\ \mathrm V
\]

For an electron,

\[
\Delta K=e\Delta V
\]

\[
\Delta K=(1.602\times10^{-19})(480)
\]

\[
\Delta K=7.690\times10^{-17}\ \mathrm J
\]

The initial kinetic energy is

\[
K_i=\frac12 mv_i^2
\]

\[
K_i=\frac12(9.11\times10^{-31})(3.0\times10^6)^2
\]

\[
K_i=4.10\times10^{-18}\ \mathrm J
\]

Therefore,

\[
K_f=K_i+\Delta K
\]

\[
K_f=4.10\times10^{-18}+7.690\times10^{-17}
\]

\[
K_f=8.10\times10^{-17}\ \mathrm J
\]

Now use

\[
v_f=\sqrt{\frac{2K_f}{m}}
\]

\[
v_f=
\sqrt{
\frac{2(8.10\times10^{-17})}
{9.11\times10^{-31}}
}
\]

\[
\boxed{v_f=1.33\times10^7\ \mathrm{m\,s^{-1}}}
\]

The electron speeds up because the electric force acts in the same direction as its motion.

Notice that the electric field points left, but the electron accelerates right. This is exactly what you should expect for a negatively charged particle.

Question 3

Two students describe the same two positions in an electrostatic field, but they have chosen different reference values for electric potential.

Their measurements are:

An electron moves from A to B.

(a) Calculate the potential difference using each student’s measurements.

(b) Determine the change in the electron’s electric potential energy and kinetic energy.

(c) A third student claims that Student B’s electron gains \(450\ \mathrm{eV}\) of kinetic energy because the final potential is \(+450\ \mathrm V\). Explain the error.

Solution 3

Both students obtain a potential difference of \(+200\ \mathrm V\), so the electron gains \(200\ \mathrm{eV}\) of kinetic energy. The choice of zero potential cannot change the physical energy transfer.

(a) Potential difference

For Student A:

\[
\begin{aligned}
\Delta V&=50-(-150)\\
&=+200\ \mathrm V
\end{aligned}
\]

For Student B:

\[
\begin{aligned}
\Delta V&=450-250\\
&=+200\ \mathrm V
\end{aligned}
\]

Both students obtain the same potential difference.

(b) Changes in energy

For an electron,

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

Therefore,

\[
\Delta U=(-e)(200)
\]

\[
\boxed{\Delta U=-200\ \mathrm{eV}}
\]

The electron loses 200 eV of electric potential energy.

If the electric field is the only force doing work,

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

Therefore,

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

The electron gains 200 eV of kinetic energy.

(c) Why the third student’s conclusion is wrong

The third student has confused the final electric potential with the change in electric potential.

Electric potential energy depends on the potential at a position, but the energy transferred depends on the difference between the initial and final values.

For Student B,

\[
\Delta V=450-250=200\ \mathrm V
\]

The electron therefore gains \(200\ \mathrm{eV}\), not \(450\ \mathrm{eV}\).

Changing the reference zero shifts every potential by the same amount. It also shifts the numerical values of electric potential energy, but it cannot change the difference in potential energy between two positions.

This is why you should always calculate \(V_f-V_i\), rather than using the final potential by itself.

Question 4

An electron moves from point A at \(-50\ \mathrm V\) to point B at \(+250\ \mathrm V\).

Its measured kinetic energy increases from \(220\ \mathrm{eV}\) to \(470\ \mathrm{eV}\).

During the journey, an external retarding force also acts on the electron.

(a) Calculate the work done by the electric field.

(b) Calculate the net work done by the external retarding force.

(c) Explain why the electron still gains kinetic energy, even though the measured increase is smaller than the energy transferred by the electric field.

Solution 4

The electric field does \(+300\ \mathrm{eV}\) of work, while the external retarding force does \(-50\ \mathrm{eV}\) of work. The electron still gains kinetic energy because the electric field transfers more energy than the external force removes.

(a) Work done by the electric field

First, calculate the potential difference:

\[
\begin{aligned}
\Delta V&=V_f-V_i\\
&=250-(-50)\\
&=+300\ \mathrm V
\end{aligned}
\]

The work done by the electric field is

\[
W_E=-q\Delta V
\]

For an electron,

\[
W_E=e\Delta V
\]

Therefore,

\[
W_E=300\ \mathrm{eV}
\]

Converting to joules:

\[
W_E=300(1.602\times10^{-19})
\]

\[
\boxed{W_E=+4.806\times10^{-17}\ \mathrm J}
\]

(b) Work done by the external force

The work-energy theorem states

\[
W_{\text{net}}=\Delta K
\]

The measured change in kinetic energy is

\[
\Delta K=K_f-K_i
\]

\[
\Delta K=470-220
\]

\[
\Delta K=+250\ \mathrm{eV}
\]

The net work is the sum of the electric work and the work done by the external force:

\[
\Delta K=W_E+W_{\text{other}}
\]

Rearrange:

\[
W_{\text{other}}=\Delta K-W_E
\]

\[
W_{\text{other}}=250-300
\]

\[
\boxed{W_{\text{other}}=-50\ \mathrm{eV}}
\]

In joules,

\[
W_{\text{other}}=-8.01\times10^{-18}\ \mathrm J
\]

(c) Explaining the result

The electron gains 300 eV from the electric field, but loses 50 eV through the external retarding force.

Its overall kinetic-energy increase is therefore

\[
300-50=250\ \mathrm{eV}
\]

There is no contradiction.

The equation \(\Delta K=-q\Delta V\) gives the kinetic-energy change due to the electric field alone. It does not automatically give the total kinetic-energy change when other forces also do work.

This distinction becomes important in questions involving collisions, friction, or external forces.

Question 5

An electron travels along a one-dimensional evacuated beam tube. Its electric potential varies with position as follows:

• At position A, \(x=0\), the potential is \(0\ \mathrm V\).
• At position B, \(x=3.0\ \mathrm{cm}\), the potential is \(-120\ \mathrm V\).
• At position C, \(x=6.0\ \mathrm{cm}\), the potential is \(+80\ \mathrm V\).

The potential decreases linearly from A to B, then increases linearly from B to C.

Assume the potential is static, the electron’s motion is restricted to the tube axis, and no collisions or non-electric forces affect its motion.

Two electrons are launched from A towards C.

• Electron 1 has an initial kinetic energy of \(100\ \mathrm{eV}\).
• Electron 2 has an initial kinetic energy of \(150\ \mathrm{eV}\).

Answer the following:

(a) Using only the potential difference between A and C, calculate the kinetic energy each electron would have at C if it reached that position.

(b) Determine whether electron 1 actually reaches C. If it does not, calculate its turning position.

(c) Determine the kinetic energy of electron 2 at B and C. Explain why it can reach C while electron 1 cannot.

(d) Explain why calculating the energy change between the starting and finishing positions is not always enough to determine whether a particle can complete its journey.

Solution 5

Electron 1 cannot reach C. It turns at \(x=2.5\ \mathrm{cm}\), while electron 2 passes through B and reaches C with \(230\ \mathrm{eV}\) of kinetic energy.

The important detail is that the potential difference between the endpoints determines the net energy change, but it does not guarantee that the electron has enough energy to cross every intermediate region.

(a) Hypothetical kinetic energy at C

The potential difference from A to C is

\[
\begin{aligned}
\Delta V&=V_C-V_A\\
&=80-0\\
&=+80\ \mathrm V
\end{aligned}
\]

For an electron,

\[
\Delta K=e\Delta V
\]

Therefore, both electrons would gain

\[
\Delta K=+80\ \mathrm{eV}
\]

if they reached C.

For electron 1:

\[
K_C=100+80
\]

\[
\boxed{K_C=180\ \mathrm{eV}}
\]

For electron 2:

\[
K_C=150+80
\]

\[
\boxed{K_C=230\ \mathrm{eV}}
\]

These are the endpoint energy predictions. We still need to check whether each electron can actually reach C.

(b) Does electron 1 reach C?

No. Electron 1 turns around before reaching B.

From A to B, the potential changes by

\[
\begin{aligned}
\Delta V&=-120-0\\
&=-120\ \mathrm V
\end{aligned}
\]

The electron’s change in kinetic energy would be

\[
\Delta K=e\Delta V
\]

\[
\Delta K=-120\ \mathrm{eV}
\]

If electron 1 reached B, its kinetic energy would have to be

\[
K_B=100-120
\]

\[
K_B=-20\ \mathrm{eV}
\]

This is physically impossible. Kinetic energy cannot be negative.

The electron must therefore stop and reverse direction before reaching B.

At the turning point, its kinetic energy is zero:

\[
K_f=0
\]

Its kinetic-energy change is

\[
\Delta K=K_f-K_i
\]

\[
\Delta K=0-100
\]

\[
\Delta K=-100\ \mathrm{eV}
\]

For an electron,

\[
\Delta K=e\Delta V
\]

Therefore,

\[
\Delta V=-100\ \mathrm V
\]

Since the electron starts at \(0\ \mathrm V\), its turning-point potential is

\[
\boxed{V_{\text{turn}}=-100\ \mathrm V}
\]

Now calculate the position.

The potential decreases by \(120\ \mathrm V\) over \(3.0\ \mathrm{cm}\).

The magnitude of the potential gradient is

\[
\frac{120}{3.0}=40\ \mathrm{V\,cm^{-1}}
\]

To reach \(-100\ \mathrm V\), the electron travels

\[
x=\frac{100}{40}
\]

\[
\boxed{x=2.5\ \mathrm{cm}}
\]

Electron 1 turns around \(2.5\ \mathrm{cm}\) from A, before reaching B.

(c) Can electron 2 reach C?

Yes. Electron 2 has enough initial kinetic energy to cross the low-potential region.

At B, the potential difference from A is

\[
\Delta V=-120\ \mathrm V
\]

Therefore,

\[
\Delta K=-120\ \mathrm{eV}
\]

Its kinetic energy at B is

\[
K_B=150-120
\]

\[
\boxed{K_B=30\ \mathrm{eV}}
\]

The electron still has positive kinetic energy at B, so it can continue travelling towards C.

From B to C, the potential increases from \(-120\ \mathrm V\) to \(+80\ \mathrm V\).

\[
\Delta V=80-(-120)
\]

\[
\Delta V=+200\ \mathrm V
\]

The electron therefore gains another \(200\ \mathrm{eV}\) of kinetic energy.

\[
K_C=30+200
\]

\[
\boxed{K_C=230\ \mathrm{eV}}
\]

This agrees with the endpoint calculation in part (a).

(d) Why the endpoint calculation is not enough

This is the crucial distinction between energy change and whether a particle can reach a particular position.

The electric field is conservative, meaning its total work depends only on the starting and finishing positions. If an electron travels from A to C, it gains \(80\ \mathrm{eV}\), regardless of the route taken.

However, the electron must still have enough kinetic energy to get through every intermediate region.

For an electron, lower electric potential means higher electric potential energy. Position B is therefore a potential-energy barrier, even though it is a minimum on the electric-potential graph.

The barrier is \(120\ \mathrm{eV}\) above the electron’s initial potential energy at A.

• Electron 1 starts with only \(100\ \mathrm{eV}\), so it cannot cross the barrier.
• Electron 2 starts with \(150\ \mathrm{eV}\), so it crosses the barrier with \(30\ \mathrm{eV}\) remaining.

To cross B with non-zero kinetic energy, an electron starting at A needs more than \(120\ \mathrm{eV}\) of initial kinetic energy.

The tempting mistake is to look only at the final potential and conclude that both electrons must gain energy and reach C. The final potential does predict the energy change if C is reached. It does not tell us whether the particle can get there.

That is the deeper lesson: a higher final electric potential can mean a gain in kinetic energy for an electron, while an intermediate region of lower potential can still prevent it from reaching that final position.

07Where to take this next

You now have two separate checks to make whenever an electron moves through an electric field:

1. Use its signed charge and the potential difference to determine whether its kinetic energy increases or decreases.
2. If the particle must travel through several regions, check that it has enough kinetic energy to get through any intermediate potential-energy barriers.

The next useful step is to connect this energy approach to electron motion between parallel plates, where the electric field produces a force, changes the electron’s velocity, and can also deflect its trajectory. That is where energy calculations and equations of motion start working together.