Why a Positive Charge Accelerates Towards Lower Electric Potential

A positive charge is released between two parallel plates. The left plate is at \(+12\text{ V}\), and the right plate is at \(+4\text{ V}\). The charge accelerates to the right, towards the plate at lower electric potential.

Why?

The electric field points in the direction of decreasing electric potential. A positive charge experiences an electric force in the direction of the field, so it accelerates towards lower potential, provided the electric force is the only significant force acting.

An electron in the same arrangement would accelerate in the opposite direction, towards higher potential. The difference comes down to the sign of the charge.

To understand why this happens, we need to connect three ideas: electric potential, electric potential energy, and electric force.

01Why lower electric potential means lower energy for a positive charge

Think about a ball rolling down a ramp.

At the top of the ramp, the ball has more gravitational potential energy. As it rolls downhill, gravitational potential energy decreases and kinetic energy increases.

A positive electric charge behaves in a similar way.

For a positive charge, a region of higher electric potential corresponds to higher electric potential energy. Moving towards lower electric potential reduces its electric potential energy.

The relationship is:

\[ U=qV \]

where:

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

Remember that \(1\text{ V}=1\text{ J/C}\).

For a positive charge, \(q\) is positive. Therefore, increasing \(V\) increases \(U\), while decreasing \(V\) decreases \(U\).

The charge loses electric potential energy as it moves towards lower potential. That energy is transferred into kinetic energy if the electric field is the only force doing work.

What happens for a negative charge?

Here is where the ramp analogy needs adjusting.

An electron has negative charge, so its electric potential energy is also negative when the electric potential is positive.

For example, consider an electron moving from \(+10\text{ V}\) to \(+20\text{ V}\).

Its electric potential energy changes according to:

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

Since \(q\) is negative and \(\Delta V\) is positive, its electric potential energy decreases.

In other words, an electron can lose electric potential energy by moving towards higher electric potential.

The relationship works in opposite directions for positive and negative charges.

There is one limitation to the ramp analogy: electric potential is not literally height. Its zero reference is arbitrary, and negative charges have the opposite relationship between potential and potential energy.

The important idea is that a charge responds to changes in its electric potential energy, not simply to whether the potential is positive or negative.

02How electric potential determines the electric field direction

We have established that a positive charge loses electric potential energy when it moves towards lower electric potential.

But how does that produce an acceleration?

The missing link is the electric field.

The electric field points in the direction in which electric potential decreases most rapidly. In a simple one-dimensional situation, this means the field points from higher potential towards lower potential.

For a uniform electric field, we can express this using:

\[ E_x=-\frac{\Delta V}{\Delta x} \]

where:

• \(E_x\) is the signed electric field component in the \(x\)-direction, measured in N/C.
• \(\Delta V=V_f-V_i\) is the signed potential difference, measured in volts.
• \(\Delta x=x_f-x_i\) is the signed displacement, measured in metres.

The negative sign is important. It tells us that the electric field points towards decreasing potential.

For example, suppose we move to the right through a region where the potential decreases from \(+12\text{ V}\) to \(+4\text{ V}\).

Taking right as the positive \(x\)-direction:

\[ \Delta V=4-12=-8\text{ V} \]

Since the displacement is positive and the potential difference is negative:

\[ E_x=-\frac{-8}{\Delta x}>0 \]

The electric field points to the right.

A positive charge experiences an electric force to the right, so its electric acceleration is also to the right if no other significant forces act.

A useful measurement: mapping electric potential

How could we check this relationship experimentally without trying to track a charged particle?

We can measure the electric potential at different positions and use those measurements to determine the electric field direction.

Consider a simple parallel-plate arrangement using conductive paper.

The apparatus works as follows:

1. Connect the two metal electrodes to a DC power supply, with the left electrode at \(+6.0\text{ V}\) relative to the right electrode.
2. Connect the voltmeter’s COM terminal to the right electrode.
3. Touch the movable positive probe to different positions on the conductive paper.
4. Record the potential at each position.

The right electrode provides our \(0\text{ V}\) reference.

What should we predict?

If the electric field points towards lower potential, the voltmeter reading should decrease as we move the probe from left to right.

The following is an illustrative set of readings, not actual experimental measurements.

The potential decreases steadily as we move to the right.

We can estimate the electric field using the change in potential over the \(0.12\text{ m}\) separation:

\[ \begin{aligned} E_x&=-\frac{\Delta V}{\Delta x}\\ &=-\frac{0.0-6.0}{0.12}\\ &=+50\text{ V/m} \end{aligned} \]

The electric field is \(50\text{ V/m}\) to the right.

Since \(1\text{ V/m}=1\text{ N/C}\), this is also \(50\text{ N/C}\).

A positive test charge would therefore experience an electric force to the right.

What does the experiment actually establish?

It maps the potential and allows us to infer the electric field direction. It does not directly measure the acceleration of a freely moving particle.

There is also a limitation to measuring only the voltage across the two electrodes. That gives the overall potential difference, but it cannot reveal whether the potential changes uniformly at every position.

Scanning the probe across the paper provides a more complete potential map. In a real setup, readings near electrode edges may differ from the ideal values because the field is not perfectly uniform there.

This distinction becomes important when we consider non-uniform electric fields.

03Connecting electric potential to force and acceleration

We now have the two ingredients needed to explain the original question.

First, the electric field points towards decreasing electric potential.

Second, the electric force on a charge is:

\[ \vec F_E=q\vec E \]

The sign of \(q\) determines the direction of the electric force.

For a positive charge:

\[ q>0 \]

Therefore, the electric force acts in the same direction as the electric field.

Newton’s second law tells us:

\[ \vec F_{\text{net}}=m\vec a \]

where \(m\) is mass in kilograms and \(\vec a\) is acceleration in metres per second squared.

If the electric force is the only significant force:

\[ m\vec a=q\vec E \]

Rearranging:

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

For a positive charge, acceleration is in the direction of the electric field.

This gives us the complete explanation:

\[ \begin{aligned} &\text{Electric potential decreases}\\ &\Downarrow\\ &\text{Electric field points towards lower potential}\\ &\Downarrow\\ &\text{Positive charge experiences force along the field}\\ &\Downarrow\\ &\text{Positive charge accelerates towards lower potential} \end{aligned} \]

The sign of the charge is essential. A negative charge experiences a force opposite to the electric field and therefore accelerates towards higher potential, assuming the electric force is the only significant force.

For a more detailed explanation of the individual force and acceleration calculations, see tutorgum’s guides to force on a charge in an electric field and acceleration of a charge in an electric field.

The relationship between potential difference and kinetic energy

There is another useful way to understand the same phenomenon.

The change in electric potential energy is:

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

The work done by the electric field is the negative of this change:

\[ W_E=-\Delta U \]

Therefore:

\[ W_E=-q\Delta V \]

If the electric field is the only force doing work, the work-energy theorem gives:

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

Here, \(\Delta K\) is the change in kinetic energy.

For a positive charge moving towards lower potential:

• \(q\) is positive.
• \(\Delta V\) is negative.
• \(\Delta K\) is positive.

The charge gains kinetic energy.

For a negative charge moving towards higher potential:

• \(q\) is negative.
• \(\Delta V\) is positive.
• \(\Delta K\) is also positive.

Both charges gain kinetic energy when moving in the direction of their electric force.

These energy relationships assume that the electric force is the only force doing work.

04Important distinction: acceleration is not the same as velocity

There is a subtle point that often causes mistakes in HSC Physics questions.

A positive charge accelerates towards lower potential, but that does not mean it must always be travelling towards lower potential.

Remember the difference:

• Velocity describes the direction in which an object is moving.
• Acceleration describes how its velocity is changing.

Imagine a positive charge travelling to the right while the electric field points to the left.

Its acceleration is to the left, towards lower potential.

However, its velocity is initially still to the right.

The charge therefore slows down while continuing to travel right. If it loses all its kinetic energy, it momentarily stops and may then reverse direction.

This is exactly like throwing a ball upwards. The ball initially moves upwards even though its acceleration due to gravity points downwards.

The electric field determines the electric force direction, not necessarily the direction of the particle’s instantaneous motion.

Another important limitation: other forces

The statement that a positive charge accelerates towards lower potential assumes that the electric force determines the net force.

If gravity, friction, or another force is significant, you must calculate the resultant force.

For example, a positive charged particle might experience an upward electric force but a stronger downward gravitational force. Its electric force points towards lower potential, but its actual acceleration is downward.

Always distinguish between the direction of the electric force and the direction of the net acceleration.

05Worked examples

05Common mistakes to avoid

Before attempting the questions, make sure you can distinguish these ideas.

A reliable approach to these questions is:

1. Write down the initial and final potentials.
2. Calculate the signed potential difference, \(\Delta V=V_f-V_i\).
3. Use the negative potential gradient to determine the electric field direction.
4. Use the sign of \(q\) to determine the electric force direction.
5. Consider all forces before determining the actual acceleration.
6. Use \(\Delta K=-q\Delta V\) when only the electric force does work.

07Questions and solutions

Question 1

A positive particle with charge \(+2.0\,\mu\text{C}\) moves from point A, where the electric potential is \(-3.0\text{ V}\), to point B, where the potential is \(-11.0\text{ V}\).

Assume the electric force is the only force doing work.

Calculate:

a. The potential difference.

b. The change in electric potential energy.

c. The work done by the electric field.

d. The change in kinetic energy.

Does moving towards lower potential mean moving towards zero volts?

Solution 1

The particle loses \(1.6\times10^{-5}\text{ J}\) of electric potential energy and gains the same amount of kinetic energy.

Part a: Potential difference

\[ \begin{aligned} \Delta V&=V_f-V_i\\ &=-11-(-3)\\ &=-8.0\text{ V} \end{aligned} \]

The potential decreases by \(8.0\text{ V}\).

Part b: Change in electric potential energy

Using:

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

\[ \begin{aligned} \Delta U &=(2.0\times10^{-6})(-8.0)\\ &=-1.6\times10^{-5}\text{ J} \end{aligned} \]

The negative result means the particle loses electric potential energy.

Part c: Work done by the electric field

\[ \begin{aligned} W_E&=-\Delta U\\ &=+1.6\times10^{-5}\text{ J} \end{aligned} \]

The electric field does positive work.

Part d: Change in kinetic energy

Since no other force does work:

\[ \Delta K=W_E \]

Therefore:

\[ \Delta K=+1.6\times10^{-5}\text{ J} \]

The particle gains kinetic energy.

Finally, moving towards lower potential does not mean moving towards zero volts. The potential decreases from \(-3.0\text{ V}\) to \(-11.0\text{ V}\).

The value \(-11.0\text{ V}\) is numerically smaller, even though its magnitude is larger.

The reference chosen for zero potential does not change the potential difference or the physical result.

Question 2

Two parallel plates are separated by \(0.20\text{ m}\).

The left plate is at \(0\text{ V}\), and the right plate is at \(+400\text{ V}\).

An electron has:

• Charge: \(-1.60\times10^{-19}\text{ C}\)
• Mass: \(9.11\times10^{-31}\text{ kg}\)

The electron starts at \(x=0.15\text{ m}\), measured from the left plate, with an initial velocity of \(8.0\times10^6\text{ m/s}\) to the left.

It travels to \(x=0.10\text{ m}\).

Assume a uniform electric field and ignore gravity.

Calculate:

a. The direction of the electric field.

b. The direction and magnitude of the electron’s acceleration.

c. Whether the electron speeds up or slows down.

d. Its speed when it reaches \(x=0.10\text{ m}\).

Solution 2

The electron accelerates to the right, towards higher potential, while initially travelling to the left. It therefore slows down and reaches the destination at approximately \(5.4\times10^6\text{ m/s}\).

Part a: Electric field direction

The potential increases from left to right.

Taking right as positive:

\[ \begin{aligned} E_x&=-\frac{\Delta V}{\Delta x}\\ &=-\frac{400-0}{0.20}\\ &=-2.0\times10^3\text{ N/C} \end{aligned} \]

The negative result means the electric field points to the left.

Part b: Acceleration direction and magnitude

The electron has negative charge, so its electric force acts opposite to the electric field.

Therefore, the electric force points to the right.

Its magnitude is:

\[ F_E=|q|E \]

\[ \begin{aligned} F_E&=(1.60\times10^{-19})(2.0\times10^3)\\ &=3.2\times10^{-16}\text{ N} \end{aligned} \]

Using Newton’s second law:

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

\[ \begin{aligned} a&=\frac{3.2\times10^{-16}}{9.11\times10^{-31}}\\ &=3.51\times10^{14}\text{ m/s}^2 \end{aligned} \]

The electron accelerates to the right.

Part c: Does the electron speed up or slow down?

Its velocity is to the left, but its acceleration is to the right.

Acceleration acts against its current motion, so the electron slows down.

This is an important distinction: the electron is initially moving towards lower potential, even though its acceleration is towards higher potential.

Part d: Final speed

First, determine the potentials at the initial and final positions.

Since the field is uniform, the potential changes linearly with position.

At \(x=0.15\text{ m}\):

\[ V_i=300\text{ V} \]

At \(x=0.10\text{ m}\):

\[ V_f=200\text{ V} \]

Therefore:

\[ \Delta V=200-300=-100\text{ V} \]

Calculate the change in electric potential energy:

\[ \begin{aligned} \Delta U&=q\Delta V\\ &=(-1.60\times10^{-19})(-100)\\ &=+1.60\times10^{-17}\text{ J} \end{aligned} \]

The electron gains electric potential energy, so its kinetic energy decreases:

\[ \Delta K=-1.60\times10^{-17}\text{ J} \]

Calculate its initial kinetic energy:

\[ K_i=\frac12mv_i^2 \]

\[ \begin{aligned} K_i&=\frac12(9.11\times10^{-31})(8.0\times10^6)^2\\ &=2.92\times10^{-17}\text{ J} \end{aligned} \]

Its final kinetic energy is:

\[ \begin{aligned} K_f&=K_i+\Delta K\\ &=2.92\times10^{-17}-1.60\times10^{-17}\\ &=1.32\times10^{-17}\text{ J} \end{aligned} \]

Now calculate its final speed:

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

\[ \begin{aligned} v_f&=\sqrt{\frac{2(1.32\times10^{-17})}{9.11\times10^{-31}}}\\ &\approx5.4\times10^6\text{ m/s} \end{aligned} \]

The electron reaches the destination because its final kinetic energy is still positive.

The tempting mistake is to assume that an electron must accelerate towards lower potential because that is what a positive charge does. The negative charge reverses the force direction.

Question 3

A small positively charged bead is placed between two horizontal parallel plates.

The lower plate is at \(+400\text{ V}\), and the upper plate is at \(0\text{ V}\). The separation is \(0.10\text{ m}\).

The bead has:

• Charge: \(+2.0\times10^{-6}\text{ C}\)
• Mass: \(1.0\times10^{-3}\text{ kg}\)

The bead is free to move vertically.

Take gravitational acceleration as \(g=9.8\text{ m/s}^2\).

Calculate:

a. The electric field strength and direction.

b. The electric force on the bead.

c. The bead’s net acceleration.

d. The minimum electric field strength required for the bead to accelerate upwards.

Solution 3

The electric force acts upwards, towards lower potential, but the bead’s actual acceleration is downwards because gravity is stronger. The minimum field strength needed to overcome gravity is \(4.9\times10^3\text{ N/C}\), with a greater field required for upward acceleration.

Part a: Electric field

The potential decreases from the bottom plate to the top plate.

Therefore, the electric field points upwards.

Its magnitude is:

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

\[ \begin{aligned} E&=\frac{400}{0.10}\\ &=4.0\times10^3\text{ N/C} \end{aligned} \]

Part b: Electric force

The bead is positively charged, so the electric force points upwards.

\[ F_E=qE \]

\[ \begin{aligned} F_E&=(2.0\times10^{-6})(4.0\times10^3)\\ &=8.0\times10^{-3}\text{ N} \end{aligned} \]

Part c: Net acceleration

Gravity acts downwards.

The gravitational force is:

\[ F_g=mg \]

\[ \begin{aligned} F_g&=(1.0\times10^{-3})(9.8)\\ &=9.8\times10^{-3}\text{ N} \end{aligned} \]

Taking upwards as positive:

\[ F_{\text{net}}=F_E-F_g \]

\[ \begin{aligned} F_{\text{net}} &=8.0\times10^{-3}-9.8\times10^{-3}\\ &=-1.8\times10^{-3}\text{ N} \end{aligned} \]

The negative result means the net force is downwards.

Using Newton’s second law:

\[ \begin{aligned} a&=\frac{F_{\text{net}}}{m}\\ &=\frac{-1.8\times10^{-3}}{1.0\times10^{-3}}\\ &=-1.8\text{ m/s}^2 \end{aligned} \]

The bead accelerates downwards at \(1.8\text{ m/s}^2\).

Notice that this does not contradict the relationship between electric potential and acceleration.

The electric force still acts towards lower potential. It is simply not the only force acting on the bead.

Part d: Minimum field strength for upward acceleration

For the bead to have zero net force, the electric force must equal its weight:

\[ qE=mg \]

Rearranging:

\[ E=\frac{mg}{q} \]

Substituting:

\[ \begin{aligned} E&=\frac{(1.0\times10^{-3})(9.8)}{2.0\times10^{-6}}\\ &=4.9\times10^3\text{ N/C} \end{aligned} \]

At exactly this field strength, the bead has zero net acceleration.

For upward acceleration, the electric field must be greater than \(4.9\times10^3\text{ N/C}\).

Question 4

A positively charged particle moves along a straight horizontal path through a non-uniform electric field.

Its electric potential varies with position as follows:

Assume the potential changes linearly between each pair of points.

The particle has:

• Charge: \(+3.0\,\mu\text{C}\)
• Mass: \(6.0\times10^{-6}\text{ kg}\)

It is released from rest at A. Ignore other forces.

Calculate:

a. The electric field direction in regions AB and BC.

b. The direction of the particle’s acceleration in each region.

c. Its kinetic energy at B and C.

d. Its speed at C.

Explain why the particle can continue travelling to the right even though its acceleration points to the left in region BC.

Solution 4

The particle accelerates to the right in AB and to the left in BC. It still reaches C with a kinetic energy of \(1.8\times10^{-5}\text{ J}\) and a speed of approximately \(2.45\text{ m/s}\).

Part a: Electric field directions

For region AB:

\[ \begin{aligned} E_{AB} &=-\frac{6-24}{0.10}\\ &=+180\text{ N/C} \end{aligned} \]

The field points to the right.

For region BC:

\[ \begin{aligned} E_{BC} &=-\frac{18-6}{0.10}\\ &=-120\text{ N/C} \end{aligned} \]

The field points to the left.

Part b: Acceleration directions

The particle has positive charge, so its electric force acts in the same direction as the electric field.

Since the electric force is the only significant force:

• In AB, acceleration is to the right.
• In BC, acceleration is to the left.

Part c: Kinetic energy at B and C

The particle starts from rest, so:

\[ K_A=0 \]

Between A and B:

\[ \Delta V_{AB}=6-24=-18\text{ V} \]

Using:

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

\[ \begin{aligned} \Delta K_{AB} &=-(3.0\times10^{-6})(-18)\\ &=+5.4\times10^{-5}\text{ J} \end{aligned} \]

Therefore:

\[ K_B=5.4\times10^{-5}\text{ J} \]

Between B and C:

\[ \Delta V_{BC}=18-6=+12\text{ V} \]

The change in kinetic energy is:

\[ \begin{aligned} \Delta K_{BC} &=-(3.0\times10^{-6})(12)\\ &=-3.6\times10^{-5}\text{ J} \end{aligned} \]

Therefore:

\[ \begin{aligned} K_C&=K_B+\Delta K_{BC}\\ &=5.4\times10^{-5}-3.6\times10^{-5}\\ &=1.8\times10^{-5}\text{ J} \end{aligned} \]

Part d: Final speed

Using:

\[ K_C=\frac12mv_C^2 \]

Rearranging:

\[ v_C=\sqrt{\frac{2K_C}{m}} \]

\[ \begin{aligned} v_C&=\sqrt{\frac{2(1.8\times10^{-5})}{6.0\times10^{-6}}}\\ &=\sqrt{6}\\ &\approx2.45\text{ m/s} \end{aligned} \]

The particle reaches C with a speed of approximately \(2.45\text{ m/s}\).

Why doesn’t it stop in region BC?

It enters BC with \(5.4\times10^{-5}\text{ J}\) of kinetic energy. Although the electric force opposes its motion in this region, it only loses \(3.6\times10^{-5}\text{ J}\).

It still has positive kinetic energy at C.

This question demonstrates an important distinction: the electric field determines the local acceleration direction, while the potential difference between two points determines the overall change in kinetic energy.

Question 5

An ideal one-dimensional electrostatic region has an electric potential that changes with position as follows:

• At \(x=0\text{ m}\), \(V=40\text{ V}\).
• The potential decreases smoothly to a local minimum of \(10\text{ V}\) at \(x=1.0\text{ m}\).
• It then increases smoothly to a local maximum of \(55\text{ V}\) at \(x=2.0\text{ m}\).
• Finally, it decreases smoothly to \(5\text{ V}\) at \(x=3.0\text{ m}\).

There are no other local maxima or minima.

A positively charged particle enters at \(x=0\text{ m}\), travelling to the right.

Its properties are:

• Charge: \(+2.0\times10^{-6}\text{ C}\)
• Mass: \(4.0\times10^{-6}\text{ kg}\)
• Initial kinetic energy: \(4.0\times10^{-5}\text{ J}\)

Assume the electric force is the only force acting.

Answer the following:

a. Determine the electric acceleration direction in each of the three regions.

b. Calculate the particle’s kinetic energy at \(x=1.0\text{ m}\), \(x=2.0\text{ m}\), and \(x=3.0\text{ m}\).

c. Calculate its speed at \(x=2.0\text{ m}\) and \(x=3.0\text{ m}\). Does it reach \(x=3.0\text{ m}\)?

d. Determine the minimum initial kinetic energy required for the particle to pass the potential maximum at \(x=2.0\text{ m}\) with non-zero speed.

e. Suppose the initial kinetic energy were instead \(2.0\times10^{-5}\text{ J}\). Determine the electric potential at which the particle turns around. Explain why its exact turning-point position cannot be calculated from the information provided.

Solution 5

The particle reaches \(x=3.0\text{ m}\) with positive kinetic energy. Its acceleration points right, then left, then right. The minimum initial kinetic energy needed to cross the \(55\text{ V}\) maximum with non-zero speed is greater than \(3.0\times10^{-5}\text{ J}\).

The key is to distinguish between the local potential gradient and the total change in potential energy.

Part a: Acceleration direction in each region

For a positive charge, acceleration follows the electric field if no other forces act.

The electric field points towards decreasing electric potential.

Therefore:

• From \(x=0\) to \(x=1.0\text{ m}\), potential decreases, so acceleration points right.
• From \(x=1.0\) to \(x=2.0\text{ m}\), potential increases, so acceleration points left.
• From \(x=2.0\) to \(x=3.0\text{ m}\), potential decreases, so acceleration points right.

At the smooth local minimum and maximum themselves, the potential gradient is zero, so the electric field is zero at those exact positions.

The particle does not have the same acceleration direction throughout its journey.

Part b: Kinetic energy at each position

Use conservation of energy:

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

Rearranging for the final kinetic energy:

\[ K_f=K_i-q(V_f-V_i) \]

Equivalently:

\[ K_f=K_i+q(V_i-V_f) \]

The initial kinetic energy is:

\[ K_0=4.0\times10^{-5}\text{ J} \]

The charge is:

\[ q=+2.0\times10^{-6}\text{ C} \]

At \(x=1.0\text{ m}\), the potential is \(10\text{ V}\).

\[ \begin{aligned} K_1&=K_0+q(V_0-V_1)\\ &=4.0\times10^{-5}\\ &\quad +(2.0\times10^{-6})(40-10)\\ &=4.0\times10^{-5}+6.0\times10^{-5}\\ &=1.0\times10^{-4}\text{ J} \end{aligned} \]

At \(x=2.0\text{ m}\), the potential is \(55\text{ V}\).

\[ \begin{aligned} K_2&=K_0+q(V_0-V_2)\\ &=4.0\times10^{-5}\\ &\quad +(2.0\times10^{-6})(40-55)\\ &=4.0\times10^{-5}-3.0\times10^{-5}\\ &=1.0\times10^{-5}\text{ J} \end{aligned} \]

At \(x=3.0\text{ m}\), the potential is \(5\text{ V}\).

\[ \begin{aligned} K_3&=K_0+q(V_0-V_3)\\ &=4.0\times10^{-5}\\ &\quad +(2.0\times10^{-6})(40-5)\\ &=4.0\times10^{-5}+7.0\times10^{-5}\\ &=1.1\times10^{-4}\text{ J} \end{aligned} \]

The kinetic energies are therefore:

Part c: Final speeds and whether the particle reaches the destination

At \(x=2.0\text{ m}\):

\[ v_2=\sqrt{\frac{2K_2}{m}} \]

\[ \begin{aligned} v_2&=\sqrt{\frac{2(1.0\times10^{-5})}{4.0\times10^{-6}}}\\ &=\sqrt{5}\\ &\approx2.24\text{ m/s} \end{aligned} \]

At \(x=3.0\text{ m}\):

\[ v_3=\sqrt{\frac{2K_3}{m}} \]

\[ \begin{aligned} v_3&=\sqrt{\frac{2(1.1\times10^{-4})}{4.0\times10^{-6}}}\\ &=\sqrt{55}\\ &\approx7.42\text{ m/s} \end{aligned} \]

The particle reaches \(x=3.0\text{ m}\).

It still has \(1.0\times10^{-5}\text{ J}\) of kinetic energy at the potential maximum.

This is the crucial detail.

Between \(x=1.0\text{ m}\) and \(x=2.0\text{ m}\), the electric force points left while the particle travels right. It therefore slows down.

However, it has enough kinetic energy to pass the potential maximum without stopping.

The tempting argument is that the final potential is lower than the initial potential, so the particle must reach the destination. That conclusion is not automatically correct. We must check whether the particle has enough kinetic energy to get past every intermediate potential-energy barrier.

Part d: Minimum initial kinetic energy to cross the potential maximum

The particle starts at \(40\text{ V}\), and the highest potential along its path is \(55\text{ V}\).

For a positive charge, the increase in electric potential energy required to reach the maximum is:

\[ \Delta U=q(V_{\max}-V_i) \]

Substituting:

\[ \begin{aligned} \Delta U &=(2.0\times10^{-6})(55-40)\\ &=3.0\times10^{-5}\text{ J} \end{aligned} \]

Therefore, the particle needs more than \(3.0\times10^{-5}\text{ J}\) of initial kinetic energy to pass the maximum with non-zero speed.

At exactly \(3.0\times10^{-5}\text{ J}\), it reaches the maximum with zero kinetic energy.

The threshold is determined by the highest potential encountered, not simply the potential at the final position.

Part e: Turning point with a smaller initial kinetic energy

Now suppose the particle starts with:

\[ K_0=2.0\times10^{-5}\text{ J} \]

At the turning point, its instantaneous speed is zero, so:

\[ K_f=0 \]

Using conservation of energy:

\[ K_f=K_0+q(V_0-V_f) \]

Substituting:

\[ 0=2.0\times10^{-5} +(2.0\times10^{-6})(40-V_f) \]

Rearranging:

\[ \begin{aligned} 2.0\times10^{-6}(V_f-40) &=2.0\times10^{-5}\\ V_f-40&=10\\ V_f&=50\text{ V} \end{aligned} \]

The particle turns around when the potential reaches \(50\text{ V}\).

This occurs somewhere between \(x=1.0\text{ m}\) and \(x=2.0\text{ m}\), where the potential is increasing.

We cannot calculate the exact position because we have not been given the mathematical relationship between potential and position in that region. We know the turning-point potential, but not the precise position at which that potential occurs.

This final distinction is important: energy conservation tells us the potential at which the particle stops, while the shape of the potential profile determines where that happens.

08Where to go next

Once you can connect electric potential to the direction of acceleration, you can start analysing complete particle trajectories rather than just identifying force directions.

The next useful step is combining potential differences with motion equations to calculate how charged particles behave between parallel plates, including their final speeds, displacements, and deflection angles.