Why Electric Potential Has No Direction but Electric Field Does

Imagine a voltmeter reads +12 V at point A and +4 V at point B. The readings tell you the electric potential at each point, but neither reading points in a particular direction. An electric field arrow, on the other hand, has a definite direction.

Why?

The key difference is that **electric potential is a scalar quantity, while electric field is a vector quantity**. Potential tells you the electric potential energy per unit charge at a location. Electric field tells you the force per unit positive charge, including the direction of that force.

This distinction is easy to state but has some important consequences. In particular, a negative charge can move in the opposite direction to the electric field, even though the electric field itself never changes direction just because you swap the test charge.

## 1. The fundamental difference: a value versus a direction

Think about a weather report giving the temperature in Sydney as 24 °C. The temperature has a numerical value, but it doesn't point north, south, east, or west.

Now imagine the weather report also gives the wind velocity as 15 km/h towards the south-west. This requires both a magnitude and a direction.

Electric potential works like temperature: it has a value at each point, but no direction. Electric field works like velocity: it has both magnitude and direction.

[IMAGE: An annotated comparison showing a temperature reading of 24 degrees at a location as a scalar value, beside a wind velocity arrow pointing south-west with a labelled magnitude. Transition visually to an electric potential value at a point and an electric field vector arrow at the same point.]

There is one important difference between temperature and electric potential: electric potential can be positive, negative, or zero. Its sign depends on the chosen reference potential. That still doesn't give it a direction.

| Quantity | What it describes | Type | SI unit |
|---|---|---|---|
| Electric potential, \(V\) | Electric potential energy per unit charge | Scalar | Volt (V) |
| Electric field, \(\vec E\) | Force per unit positive charge | Vector | N C\(^{-1}\) or V m\(^{-1}\) |

A scalar can have a negative value. A vector has a magnitude and a direction. Don't confuse the sign of a scalar with the direction of a vector.

That comparison is about mathematical type, not mechanism. An electric field isn’t literally a wind. The useful point is that one quantity has a value at a location, while the other also needs a direction.

02Electric potential is energy per unit charge

Electric potential, \(V\), is defined as electric potential energy per unit charge:

\[ V=\frac{U}{q} \]

where:

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

One volt is equivalent to one joule per coulomb:

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

This definition tells you something important. Electric potential describes a property of a location in an electric field. It doesn’t describe how a particular charged particle will move.

A proton and an electron placed at the same location experience the same electric potential. However, their electric potential energies have opposite signs because their charges have opposite signs.

The relationship is:

\[ U=qV \]

This is why electric potential and electric potential energy are not interchangeable.

What does a negative potential mean?

Electric potential can be positive, negative, or zero.

The sign depends on the reference point chosen as zero potential.

For example, suppose:

• Point A has a potential of \(+12\text{ V}\).
• Point B has a potential of \(+4\text{ V}\).

Moving from A to B gives a potential difference of:

\[ \Delta V=V_f-V_i \]

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

The negative sign tells you that the final point has a lower potential than the starting point.

It does not mean that the potential has a direction.

If you travel from B to A instead, the potential difference becomes \(+8\text{ V}\). The physical locations haven’t changed. You’ve simply reversed the order in which you compare them.

This is an important distinction:

• A negative potential means the location is below the chosen reference potential.
• A negative potential difference means the final potential is lower than the initial potential.
• Neither statement makes electric potential a vector.

03Why electric field needs a direction

Electric field describes the force experienced per unit positive charge.

Its direction is defined as the direction of the force on a positive test charge.

The relationship between electric force and electric field is:

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

Here:

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

The arrow above \(F\) and \(E\) indicates that these are vectors.

The sign of \(q\) determines whether the force acts with or against the electric field.

Consider the two possible test charges:

Positive charge:

If \(q>0\), the electric force acts in the same direction as the electric field.

Negative charge:

If \(q<0\), the electric force acts in the opposite direction to the electric field.

The electric field itself hasn’t changed. Only the direction of the force on the test charge has changed.

This is why you should never define electric field direction by looking at the direction an electron moves.

The field direction is established independently of the charge you place in it.

The connection between electric potential and electric field

Here’s where the two quantities become connected.

Electric field points in the direction in which electric potential decreases most rapidly.

For a uniform field along the \(x\)-axis:

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

where:

• \(E_x\) is the electric field component along the \(x\)-axis, in V m\(^{-1}\).
• \(\Delta V\) is the signed change in potential, in volts.
• \(\Delta x\) is the signed displacement along the \(x\)-axis, in metres.

The negative sign is important. It tells you that potential decreases in the direction of the electric field.

For example, if potential decreases as you move east, the electric field points east.

If potential increases as you move east, the electric field points west.

For a non-uniform field, a calculation over a finite distance gives an average field component over that interval. The local field direction can change from point to point.

Notice what has happened here: electric potential is still a scalar. We find the vector direction of the electric field by examining how that scalar changes across space.

04How an experiment reveals the direction of an electric field

How can you experimentally find a direction using measurements that don’t have directions?

The answer is to measure electric potential at many different positions.

A useful practical method is electric field mapping using conductive paper.

The experimental setup

Imagine a rectangular sheet of semi-conductive paper sitting flat on a laboratory bench.

Two parallel metal strips are placed on opposite sides of the sheet.

A low-voltage DC power supply connects the positive terminal to the left strip and the negative terminal to the right strip. Suppose the left strip is maintained at \(6.0\text{ V}\), while the right strip is at \(0.0\text{ V}\).

A digital voltmeter is connected as follows:

1. Its COM lead connects to the right-hand electrode, establishing the \(0.0\text{ V}\) reference.
2. Its positive probe touches different positions on the conductive paper.
3. Each reading gives the potential at the probe’s location relative to the reference electrode.

What should the measurements show?

Before taking measurements, make a prediction.

If the two electrodes are parallel, and we measure points along the centreline between them, how should the potential change?

In the ideal uniform-field region, it should decrease steadily from \(6.0\text{ V}\) to \(0.0\text{ V}\).

To investigate, move the probe across the sheet and record readings at regular intervals.

The following is an illustrative dataset, not a record of an actual experiment.

The readings decrease as the probe moves to the right.

The voltmeter itself never reports an electric field direction. It only reports potential differences.

However, the pattern of readings allows us to infer the field direction.

Why are equipotential lines perpendicular to electric field lines?

The experiment can be extended by finding several points that have exactly the same potential.

For example, move the probe around the sheet until it reads \(4.0\text{ V}\). Mark that position.

Find several more positions with the same reading and connect them. This creates an equipotential line.

Repeat the process for other potential values.

An equipotential line is a line connecting points with the same electric potential.

Why must the electric field be perpendicular to these lines?

Imagine moving a positive test charge along an equipotential line.

Its potential doesn’t change, so:

\[ \Delta V=0 \]

The electric field cannot have a component along that line. Otherwise, moving along the field would produce a change in potential.

Therefore, the electric field must be perpendicular to the equipotential line.

It also points from higher potential towards lower potential.

This gives you two useful rules:

• Electric field lines cross equipotential lines at right angles.
• Electric field lines point from higher electric potential to lower electric potential.

The closer the equipotential lines are, the stronger the electric field, provided you are comparing equal potential differences and measuring their perpendicular separation.

The reason is straightforward: the same potential change occurring over a shorter distance means a greater change in potential per metre.

What could make the experiment inaccurate?

A single measurement between the electrodes only tells you the total potential difference. It doesn’t reveal how potential varies sideways or how the field behaves around differently shaped electrodes.

Mapping many points improves the investigation because it reveals the spatial pattern.

However, the experiment still has limitations:

• Conductive paper is a model of a two-dimensional potential distribution, not a direct measurement of an isolated static-charge arrangement.
• Uneven conductivity can distort the readings.
• Poor probe contact can produce inconsistent voltmeter measurements.
• Finite electrode sizes create edge effects, so the field near the edges may not be uniform.

The experiment supports the relationship between potential and field direction, but the field arrows are inferred from the measured potential pattern rather than directly measured by the voltmeter.

05Why zero electric potential does not mean zero electric field

Here’s a particularly tempting misconception.

Suppose a voltmeter reads \(0\text{ V}\) at point P.

Does that mean the electric field at P must be zero?

No.

Electric potential tells you the value at a location. Electric field depends on how potential changes around that location.

A point can have zero potential while sitting in a region where potential changes rapidly with position.

This is also connected to the arbitrary choice of zero potential.

Imagine changing the reference so that every potential reading increases by \(50\text{ V}\).

The new potential is:

\[ V’=V+50 \]

The potential difference between two points becomes:

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

The potential difference is unchanged.

Since the electric field depends on the spatial change in potential, the field is unchanged too.

This means that:

• Changing the zero of potential changes the numerical potential values.
• It does not change potential differences.
• It does not change the electric field.
• It does not change the force experienced by a given charge.

The zero of potential is a reference choice, not a special physical location where the electric field must disappear.

07A reliable way to approach electric potential and field questions

When you encounter an HSC question involving potential and electric field, work through these steps in order.

A few final checks can prevent common errors.

Check 1: Don’t confuse potential with potential energy.

Potential is energy per unit charge. Potential energy depends on the actual charge placed in the field.

Check 2: Don’t use the sign of the charge to determine field direction.

The field direction is defined using a positive test charge. A negative particle experiences a force in the opposite direction.

Check 3: Don’t use the absolute potential to determine field strength.

A potential of \(100\text{ V}\) doesn’t automatically mean a stronger field than a potential of \(10\text{ V}\). You need to know how quickly potential changes with distance.

Check 4: Keep the signs in your calculations.

Use:

\[ \Delta V=V_f-V_i \]

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

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

Don’t replace signed quantities with magnitudes halfway through a calculation.

For uniform parallel-plate fields, the magnitude relationship is also commonly written as \(E=|\Delta V|/d\). For a fuller treatment of this calculation, see Electric Field and Potential Difference Using E = V/d.

A note on calculating particle speeds

Sometimes you will use the change in electric potential energy to calculate a particle’s final speed.

If the electric field is the only force doing work:

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

You can then use the classical kinetic energy equation:

\[ K=\frac12 mv^2 \]

However, this equation is a non-relativistic approximation. It assumes that the particle’s speed is much smaller than the speed of light:

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

There is no sharp speed cut-off, but as a particle approaches a noticeable fraction of \(c\), relativistic corrections become important. A calculated speed equal to or greater than \(c\) is physically impossible and indicates that the non-relativistic model is no longer valid.

07Questions and solutions

Question 1

Points A and B lie along a straight line in a uniform electric field.

• Point A has a potential of \(+12\text{ V}\).
• Point B has a potential of \(+4\text{ V}\).
• B is \(0.040\text{ m}\) east of A.

A student claims that the potential at A points towards B because A has a higher potential.

(a) Calculate the signed potential difference when moving from A to B.

(b) Explain why the student’s statement is incorrect.

(c) Calculate the electric field component and state its direction. Take east as positive.

Solution 1

The potential difference is \(-8.0\text{ V}\), and the electric field is \(200\text{ V m}^{-1}\) east.

Part (a): Calculate the potential difference.

Use:

\[ \Delta V=V_f-V_i \]

Substitute:

\[ \Delta V=4-12 \]

\[ \Delta V=-8.0\text{ V} \]

The negative sign means that B is at a lower potential than A.

Part (b): Explain the misconception.

The student’s statement is incorrect because electric potential is a scalar quantity.

A potential of \(+12\text{ V}\) describes the electric potential at A. It doesn’t have a direction.

The potential difference tells us how the potential changes between two points, but it is still a signed scalar.

Part (c): Calculate the electric field.

For a uniform field:

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

Substitute:

\[ E_x=-\frac{-8.0}{0.040} \]

\[ E_x=+200\text{ V m}^{-1} \]

The positive sign means the field points east.

This agrees with the physical rule that electric potential decreases in the direction of the electric field.

Question 2

An experimenter maps the equipotential lines in two different regions of a conductive sheet.

In both regions, the two neighbouring equipotential lines have potentials of \(10\text{ V}\) and \(6\text{ V}\).

The measurements are:

• At region P, the \(6\text{ V}\) line is \(2.0\text{ cm}\) east of the \(10\text{ V}\) line.
• At region Q, the \(6\text{ V}\) line is \(0.50\text{ cm}\) north of the \(10\text{ V}\) line.

Assume the field is approximately uniform between each pair of lines, which are locally parallel.

(a) Calculate the approximate electric field strength at P.

(b) Calculate the approximate electric field strength at Q.

(c) State the direction of the electric field at each location.

(d) Explain why the electric field is stronger in one region, even though the potential difference is identical.

Solution 2

The electric field is \(200\text{ V m}^{-1}\) east at P and \(800\text{ V m}^{-1}\) north at Q. The field is four times stronger at Q.

Part (a): Calculate the field at P.

The potential difference is:

\[ |\Delta V|=|6-10|=4.0\text{ V} \]

Convert the separation into metres:

\[ d=2.0\text{ cm}=0.020\text{ m} \]

For an approximately uniform field:

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

\[ E_P=\frac{4.0}{0.020} \]

\[ E_P=200\text{ V m}^{-1} \]

Part (b): Calculate the field at Q.

Convert the separation:

\[ d=0.50\text{ cm}=0.0050\text{ m} \]

Substitute:

\[ E_Q=\frac{4.0}{0.0050} \]

\[ E_Q=800\text{ V m}^{-1} \]

Part (c): Determine the directions.

At P, potential decreases from \(10\text{ V}\) to \(6\text{ V}\) towards the east.

Therefore, the field points east.

At Q, potential decreases from \(10\text{ V}\) to \(6\text{ V}\) towards the north.

Therefore, the field points north.

Part (d): Explain the difference in field strength.

Both regions have the same potential difference, but the distance over which that change occurs is different.

At Q, the same \(4.0\text{ V}\) change occurs over a shorter distance.

This produces a greater change in potential per metre, so the electric field is stronger.

The important principle is that field strength depends on the rate of change of potential with distance, not simply the potential difference itself.

Question 3

A particle with charge

\[ q=-3.0\times10^{-6}\text{ C} \]

moves from point X, at a potential of \(-10\text{ V}\), to point Y, at a potential of \(+50\text{ V}\).

The potential increases steadily along the straight path from X to Y.

Assume the electric force is the only force doing work on the particle.

(a) Calculate the potential difference.

(b) Calculate the change in electric potential energy.

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

(d) Explain why the electric force acts in the direction of motion, even though the particle moves towards higher electric potential.

Solution 3

The potential difference is \(+60\text{ V}\), the particle loses \(1.8\times10^{-4}\text{ J}\) of electric potential energy, and the electric field does positive work of \(1.8\times10^{-4}\text{ J}\).

Part (a): Calculate the potential difference.

\[ \Delta V=V_f-V_i \]

\[ \Delta V=50-(-10) \]

\[ \Delta V=+60\text{ V} \]

The particle moves towards higher electric potential.

Part (b): Calculate the change in potential energy.

Use:

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

Substitute the signed charge:

\[ \Delta U=(-3.0\times10^{-6})(60) \]

\[ \Delta U=-1.8\times10^{-4}\text{ J} \]

The particle loses electric potential energy.

Part (c): Calculate the work done by the field.

The relationship is:

\[ W_{\text{field}}=-\Delta U \]

Therefore:

\[ W_{\text{field}}=+1.8\times10^{-4}\text{ J} \]

The electric field does positive work.

Part (d): Explain the force direction.

Electric potential increases towards Y.

Therefore, the electric field points towards X, where potential is lower.

However, the particle has a negative charge. Its electric force acts opposite to the electric field.

Consequently, the force points towards Y, in the same direction as the particle’s displacement.

This is why a negative charge can naturally accelerate towards higher electric potential while losing electric potential energy.

Question 4

A positive charge of \(+4.0\times10^{-6}\text{ C}\) is moved from point A to point B along a path that passes through point C.

The electric potentials are:

An external guide constrains the charge to follow the route A → C → B.

(a) Calculate the change in electric potential energy from A to C.

(b) Calculate the change in electric potential energy from C to B.

(c) Calculate the total work done by the electric field from A to B.

(d) A student argues that because A and B have the same potential, the electric field must be zero everywhere along the route. Explain why this conclusion is incorrect.

Solution 4

The charge loses \(2.4\times10^{-5}\text{ J}\) of potential energy from A to C, then gains the same amount from C to B. The total work done by the electric field is zero, but this does not mean the field is zero.

Part (a): Calculate the change from A to C.

First calculate the potential difference:

\[ \Delta V_{AC}=2.0-8.0=-6.0\text{ V} \]

Use:

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

\[ \Delta U_{AC}=(4.0\times10^{-6})(-6.0) \]

\[ \Delta U_{AC}=-2.4\times10^{-5}\text{ J} \]

The charge loses electric potential energy along this section.

Part (b): Calculate the change from C to B.

The potential difference is:

\[ \Delta V_{CB}=8.0-2.0=+6.0\text{ V} \]

Therefore:

\[ \Delta U_{CB}=(4.0\times10^{-6})(+6.0) \]

\[ \Delta U_{CB}=+2.4\times10^{-5}\text{ J} \]

The charge gains the same amount of electric potential energy along this section.

Part (c): Calculate the total work done.

The total change in potential energy is:

\[ \Delta U_{\text{total}} =-2.4\times10^{-5} +2.4\times10^{-5} \]

\[ \Delta U_{\text{total}}=0\text{ J} \]

Since:

\[ W_{\text{field}}=-\Delta U \]

the total work is:

\[ W_{\text{field}}=0\text{ J} \]

Part (d): Explain why the field need not be zero.

The student’s conclusion is incorrect because equal potential at two endpoints does not mean that potential is constant everywhere between them.

The electric field can do positive work along one section and negative work along another.

In this example, the field does positive work from A to C, then negative work from C to B.

These contributions cancel.

The key principle is that the work done by an electrostatic field depends on the initial and final potentials, not on the particular route taken between them.

The charge can therefore experience a non-zero electric field even though its total change in potential energy is zero.

Question 5

A small region of space is mapped using a voltmeter.

The following potential measurements are taken relative to an initially chosen reference:

Take east as positive \(x\) and north as positive \(y\). Assume the potential varies linearly across this region.

A second measurement system uses a different reference potential. It reports readings that are exactly \(50.0\text{ V}\) higher at every position.

A particle with charge

\[ q=-4.0\times10^{-6}\text{ C} \]

is initially at rest at P.

(a) Calculate the electric field components at P, using the potential measurements on either side of P. State the field’s magnitude and direction.

(b) Calculate the magnitude and direction of the electric force on the particle.

(c) If the particle moves \(1.0\text{ cm}\) north from P, calculate the change in electric potential energy and the work done by the electric field.

(d) Determine whether the second measurement system produces a different electric field or force. Explain your answer.

(e) The second measurement system reports that P has a potential of \(+50.0\text{ V}\). A student claims that the field at P must therefore be different from the field calculated using the original \(0.0\text{ V}\) reading. Identify the error in this reasoning.

Solution 5

The electric field at P is \(200\text{ V m}^{-1}\) south, and the particle experiences a force of \(8.0\times10^{-4}\text{ N}\) north. Its electric potential energy decreases by \(8.0\times10^{-6}\text{ J}\) when it moves north. Changing the reference potential does not change any of these results.

Part (a): Calculate the electric field components.

We need to examine how potential changes in each coordinate direction.

First, consider the east-west direction.

The potential at the east point and west point is identical:

\[ \Delta V_x=0.0-0.0=0.0\text{ V} \]

The separation between these points is:

\[ \Delta x=0.020\text{ m} \]

Therefore:

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

\[ E_x=-\frac{0}{0.020}=0\text{ V m}^{-1} \]

There is no east-west field component.

Now consider the north-south direction.

The potential difference is:

\[ \Delta V_y=2.0-(-2.0) \]

\[ \Delta V_y=+4.0\text{ V} \]

The separation is again:

\[ \Delta y=0.020\text{ m} \]

Therefore:

\[ E_y=-\frac{4.0}{0.020} \]

\[ E_y=-200\text{ V m}^{-1} \]

The negative sign means the field points south.

Combining the components:

\[ E_x=0 \]

\[ E_y=-200\text{ V m}^{-1} \]

The electric field has a magnitude of \(200\text{ V m}^{-1}\), directed south.

Part (b): Calculate the electric force.

Use:

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

The charge is negative, so the force acts opposite to the electric field.

Its magnitude is:

\[ F=|q|E \]

\[ F=(4.0\times10^{-6})(200) \]

\[ F=8.0\times10^{-4}\text{ N} \]

The force points north.

Since the particle starts from rest, its initial acceleration is also north.

Part (c): Calculate the energy change for a northward displacement.

The particle moves from P to the point 1.0 cm north.

The signed potential difference is:

\[ \Delta V=2.0-0.0 \]

\[ \Delta V=+2.0\text{ V} \]

Use the signed charge:

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

\[ \Delta U=(-4.0\times10^{-6})(+2.0) \]

\[ \Delta U=-8.0\times10^{-6}\text{ J} \]

The particle loses electric potential energy.

The work done by the field is:

\[ W_{\text{field}}=-\Delta U \]

\[ W_{\text{field}}=+8.0\times10^{-6}\text{ J} \]

The positive work agrees with the fact that the particle moves in the direction of its electric force.

Notice that the particle moves towards higher electric potential, but its potential energy decreases. Its negative charge reverses the relationship between potential and potential energy.

Part (d): Consider the second measurement system.

The second system adds \(50.0\text{ V}\) to every reading.

The new values are:

The potential differences are unchanged.

For example, the north-south difference remains:

\[ \Delta V=52.0-48.0=4.0\text{ V} \]

Therefore, the electric field remains:

\[ E_y=-\frac{4.0}{0.020} =-200\text{ V m}^{-1} \]

The force is also unchanged because the charge and electric field are unchanged.

The potential energy difference remains:

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

\[ \Delta U=-8.0\times10^{-6}\text{ J} \]

Changing the reference potential does not change the physical energy difference.

Part (e): Identify the student’s error.

The student is confusing the absolute value of electric potential with the spatial change in potential.

The electric field depends on how potential changes between nearby positions, not on the numerical value assigned to one point.

Adding a constant to every potential reading shifts the reference but leaves every potential difference unchanged.

Therefore, the field at P is still \(200\text{ V m}^{-1}\) south.

This is the central idea of the entire topic: electric potential is a scalar value at a point, while electric field is a vector describing the directional change of potential through space.

Understanding that distinction makes equipotential maps, field diagrams, and charged-particle calculations much easier to interpret.