How to Compare Electric Field Strengths from Point Charges

Two points sit on the same radial line from a positive point charge. Point B is twice as far from the charge as point A. A student predicts that the electric field at B must be half as strong. Another predicts it will be one-quarter as strong.

Which prediction is correct?

The second. Around an isolated point charge, electric field strength follows an inverse-square relationship. Doubling the distance reduces the field strength to one-quarter of its original value.

This gives you the starting point for comparing field strengths from point-charge diagrams:

\[ E=\frac{k|Q|}{r^2} \]

If both points are around the same source charge, compare their distances. If the source charges are different, you must account for both the charge magnitudes and the distances.

The important part is knowing which information in a diagram actually represents field strength.

01Understand what electric field strength measures

Before comparing diagrams, separate two quantities that are easy to confuse: electric field strength and electric force.

An electric field describes the electrical conditions at a particular location. It tells you how much electric force would act per coulomb of positive test charge placed there.

The mathematical definition is:

\[ \vec E=\frac{\vec F_E}{q} \]

Where:

• \(\vec E\) is electric field strength, measured in newtons per coulomb (\(\text{N C}^{-1}\)).
• \(\vec F_E\) is the electric force on the test charge, measured in newtons (N).
• \(q\) is the signed test charge, measured in coulombs (C).

For the magnitude of the field, use:

\[ E=\frac{F_E}{|q|} \]

Here is the important distinction: the electric field is produced by the source charges, not by the test charge.

Imagine placing two different positive test charges at exactly the same point.

• Test charge A is \(+1.0\,\mu\text{C}\).
• Test charge B is \(+2.0\,\mu\text{C}\).

Test charge B experiences twice the electric force because it has twice the charge. However, the electric field at that location has not changed.

The test charge measures the field. It does not determine the field.

This also explains why the sign of a test charge matters when determining force direction.

When comparing field strengths from diagrams, therefore, do not assume that a larger force means a stronger field. You must account for the charge experiencing that force.

02Check whether the field is uniform or inverse-square

Not every electric field behaves in the same way when you move further away from a source.

For this topic, you need to distinguish two common models.

Point-charge fields

An isolated point charge produces a radial electric field.

The field points:

• Away from a positive source charge.
• Towards a negative source charge.

Its magnitude decreases with the square of the distance from the source.

\[ E=\frac{k|Q|}{r^2} \]

Here:

• \(E\) is electric field strength in \(\text{N C}^{-1}\).
• \(k\) is Coulomb’s constant, approximately \(8.99\times10^9\ \text{N m}^2\text{C}^{-2}\).
• \(Q\) is the source charge in coulombs (C).
• \(r\) is the distance from the centre of the source charge to the point being considered, in metres (m).

Notice that the equation uses the magnitude of the source charge. Its sign determines the field direction, not whether the magnitude is positive or negative.

The inverse-square relationship means:

A useful way to picture this is to imagine light spreading out from a small lamp.

At twice the distance, the same light spreads over four times the spherical surface area. The brightness per unit area becomes one-quarter as large.

An electric field has a similar geometric spreading pattern. The analogy explains the mathematics, although electric field lines are representations of a field, not actual rays of light or physical threads.

Uniform electric fields

A uniform electric field has constant magnitude and direction throughout a particular region.

A common example is the central region between two large, oppositely charged parallel plates.

In this region, the field is approximately constant:

\[ E\approx\frac{V}{d} \]

Where:

• \(V\) is the potential difference between the plates, in volts (V).
• \(d\) is the perpendicular separation between the plates, in metres (m).

Moving a test charge from one position to another within the central region does not change the field strength.

This is different from a point-charge field, where increasing the distance from the source reduces the field.

The distinction matters because using the wrong equation can give you a completely incorrect answer.

For a more detailed comparison, see Point Charges vs Parallel Plates: Choosing the Right Field Model.

03Read field-line diagrams correctly

Electric field diagrams communicate two important things:

1. The direction of the electric field.
2. The relative strength of the electric field.

However, these are represented in different ways.

Field-line direction

The arrows on electric field lines show the direction of the electric field.

Remember that this is defined as the direction of the force on a positive test charge.

For a positive source charge, field lines point outwards. For a negative source charge, they point inwards.

Field-line density

The closer the field lines are together, the stronger the electric field.

This is called field-line density.

Consider an isolated positive point charge. Its field lines spread outwards as you move away from the source.

Near the charge, the lines are crowded together. Further away, they become more widely separated.

The electric field is therefore stronger closer to the source.

However, there is an important limitation.

A field-line diagram is a drawing convention. The number of lines an artist chooses to draw is not automatically a numerical measurement of field strength.

You can compare the total number of lines representing two different source charges only when the diagrams use the same field-line convention.

For example, if two diagrams explicitly use the same convention, with each line representing the same amount of source charge, a source with 12 lines represents twice the charge magnitude of a source with 6 lines.

Without that shared convention, counting lines across separate diagrams is unreliable.

What can you actually use?

One more detail: most field-line diagrams are two-dimensional representations of a three-dimensional field. You should use their line density to judge relative strength, not measure the gaps on the page and treat them as exact numerical values.

For more practice interpreting direction and field-line density, see Electric Field Lines Around a Point Charge: HSC Physics Guide.

04Where does the inverse-square relationship come from?

The inverse-square rule is not simply a convenient mathematical assumption. It is supported by measurements of electric force.

One important historical investigation was Charles-Augustin de Coulomb’s torsion-balance experiment in 1785.

The measurement problem

Scientists already knew that charged objects could attract or repel each other.

The unanswered question was how quickly the electric force weakened as the separation between charged objects increased.

Was the force inversely proportional to distance, distance squared, or some other power?

Measuring such small forces with an ordinary balance was difficult. Coulomb needed a much more sensitive instrument.

How the apparatus worked

Coulomb developed a torsion balance.

A fine vertical fibre suspended a lightweight horizontal arm. A small charged pith ball was attached to one end of the arm, while a counterweight was attached to the other.

A second charged pith ball was mounted on a separate insulating support beside the suspended ball.

The arrangement worked as follows:

1. The two balls were given charges of the same sign.
2. The second ball was brought near the suspended ball.
3. The balls repelled each other, causing the horizontal arm to rotate.
4. This rotation twisted the fine suspension fibre.
5. The twisted fibre exerted a restoring torque, opposing the electric force.
6. At equilibrium, the calibrated twist angle provided a measure of the electric force.

By changing the separation between the centres of the two balls, Coulomb could compare the electric force at different distances.

What did Coulomb predict?

If the electric force followed an inverse-square relationship, halving the separation should make the force four times greater.

Similarly, doubling the separation should reduce the force to one-quarter.

The torsion balance allowed Coulomb to test this prediction by measuring the change in the restoring torque needed to balance the electric repulsion.

What did the measurements show?

The measurements supported the inverse-square relationship.

When the separation was halved, the torsion required to balance the repulsive force increased by approximately a factor of four.

This was evidence that:

\[ F_E\propto\frac{1}{r^2} \]

The result became the basis of Coulomb’s law.

Why was the original method not enough?

The first arrangement worked well for repelling charges.

However, opposite charges attract. If the balls came too close, they could move together and touch, disrupting the experiment and allowing their charge to escape.

Coulomb therefore used an independent oscillation method to investigate attraction.

In this arrangement, a small charged plate was attached to a suspended needle near an oppositely charged hollow metal sphere. The needle oscillated in a narrow arc, and Coulomb measured its period at different separations.

The oscillation period depended on the electric force, allowing him to investigate the relationship using a different measurement method.

This provided an independent check of the inverse-square relationship for attraction. Charge leakage also needed to be considered when interpreting the measurements.

These historical details are described in the American Physical Society’s account of Coulomb’s work and OpenStax’s explanation of Coulomb’s law.

Connecting the experiment to electric field strength

Coulomb measured force between charged objects, not electric field strength directly.

However, electric field strength is force per unit test charge:

\[ E=\frac{F_E}{|q|} \]

For a fixed test charge, the force and field strength have the same proportional relationship with distance.

Therefore:

\[ E\propto\frac{1}{r^2} \]

This is why the inverse-square relationship is used when comparing field strengths around isolated point charges.

As a simple scale estimate, suppose a test charge experiences a force of \(0.80\text{ N}\) at a distance of \(0.20\text{ m}\).

If the distance increases to \(0.40\text{ m}\), the predicted force becomes:

\[ \begin{aligned} F_2&=F_1\left(\frac{r_1}{r_2}\right)^2\\ &=0.80\left(\frac{0.20}{0.40}\right)^2\\ &=0.20\text{ N} \end{aligned} \]

The force becomes one-quarter as large. For the same test charge, the electric field strength also becomes one-quarter as large.

These numbers are illustrative, rather than historical measurements from Coulomb’s apparatus.

05A reliable method for comparing point-charge diagrams

You can now turn the physical ideas into a repeatable exam method.

For two points, A and B, around isolated point charges, use:

\[ \frac{E_A}{E_B} = \frac{|Q_A|}{|Q_B|} \left(\frac{r_B}{r_A}\right)^2 \]

Where:

• \(E_A\) and \(E_B\) are the field magnitudes at the two points.
• \(Q_A\) and \(Q_B\) are their respective source charges.
• \(r_A\) and \(r_B\) are the distances from each source charge to its corresponding point.

The equation contains two separate comparisons.

The first compares the source charge magnitudes. The second accounts for the distances.

If both points have the same source charge

The source charge cancels:

\[ \frac{E_A}{E_B} = \left(\frac{r_B}{r_A}\right)^2 \]

This is the quickest situation.

If the source charges are different

Keep both factors:

\[ \frac{E_A}{E_B} = \frac{|Q_A|}{|Q_B|} \left(\frac{r_B}{r_A}\right)^2 \]

A larger source charge does not automatically produce a stronger field at every point. Distance matters too.

If the diagram gives field-line counts

If both diagrams explicitly use the same line-count convention, you can substitute the number of lines for the source charge magnitude.

Let \(N_A\) and \(N_B\) represent the total number of field lines drawn from the respective sources.

Then:

\[ \frac{E_A}{E_B} = \frac{N_A}{N_B} \left(\frac{r_B}{r_A}\right)^2 \]

This works only when each diagram uses the same number of field lines per unit of source charge.

06Questions and solutions

The following questions build from straightforward distance comparisons to more demanding diagram interpretations. Pay attention to what each diagram actually tells you, rather than relying on whichever object looks larger or closer.

Question 1

An isolated negative point charge of \(-3.0\,\mu\text{C}\) has two points marked on the same radial line.

• Point A is \(0.10\text{ m}\) from the source.
• Point B is \(0.30\text{ m}\) from the source.

Which point experiences the stronger electric field, and by what factor?

Solution 1

The field at A is nine times stronger than the field at B.

Because both points are around the same source charge, the source charge cancels from the comparison.

Use:

\[ \frac{E_A}{E_B} = \left(\frac{r_B}{r_A}\right)^2 \]

Substitute:

\[ \frac{E_A}{E_B} = \left(\frac{0.30}{0.10}\right)^2 \]

\[ =3^2 \]

\[ \frac{E_A}{E_B}=9 \]

Point A is therefore nine times stronger.

The tempting mistake is to say that B is three times further away, so its field must be one-third as strong. However, the distance is squared, so the field decreases by a factor of nine.

Because the source charge is negative, the field points towards the source at both locations.

Question 2

Two separate point-charge diagrams are drawn using the same spatial scale and field-line convention.

• Diagram A shows a positive source charge with 6 radial field lines. Point P is \(0.20\text{ m}\) from the source.
• Diagram B shows a negative source charge with 24 radial field lines. Point R is \(0.40\text{ m}\) from the source.

Assume each field line represents the same amount of source charge in both diagrams.

(a) Compare the magnitudes of the electric fields at P and R.

(b) Are the electric field vectors in the same direction? Explain.

Solution 2

The field magnitudes at P and R are equal, but their directions are opposite.

Part (a): Compare the magnitudes.

Because the two diagrams use the same field-line convention, the total number of lines is proportional to the source charge magnitude.

Therefore:

\[ \frac{|Q_A|}{|Q_B|} = \frac{6}{24} = \frac14 \]

Use the comparison equation:

\[ \frac{E_P}{E_R} = \frac{6}{24} \left(\frac{0.40}{0.20}\right)^2 \]

\[ =\frac14\times4 \]

\[ \frac{E_P}{E_R}=1 \]

The field magnitudes are equal.

Part (b): Compare the directions.

The source in diagram A is positive, so its field points away from the source.

The source in diagram B is negative, so its field points towards the source.

Both points are to the right of their respective sources. Therefore, the field at P points right, while the field at R points left.

The vectors have equal magnitudes but opposite directions.

The tempting approach is to count the lines and conclude that diagram B has four times the field strength.

That confuses the total source charge with the local field strength. The second source has four times the charge magnitude, but its observation point is twice as far away. The inverse-square reduction exactly cancels the larger source charge.

Question 3

An isolated positive point charge of \(+50\,\mu\text{C}\) produces an electric field at two points, P and R.

Both points are to the right of the source.

• P is \(0.20\text{ m}\) from the source and contains a positive test charge of \(+0.50\,\mu\text{C}\).
• R is \(0.40\text{ m}\) from the source and contains a negative test charge of \(-2.0\,\mu\text{C}\).

Assume the test charges do not disturb the source charge.

(a) Calculate the electric field strength at P and R.

(b) Compare the magnitudes of the electric forces acting on the two test charges.

(c) State the direction of each force and explain why the force magnitudes are equal even though the field strengths differ.

Use \(k=8.99\times10^9\ \text{N m}^2\text{C}^{-2}\).

Solution 3

The electric field at P is four times stronger than at R, but both test charges experience equal force magnitudes of approximately \(5.6\text{ N}\).

Their force directions are opposite.

Part (a): Calculate the electric fields.

First, convert the source charge into coulombs:

\[ Q=50\times10^{-6}\text{ C} \]

At P:

\[ E_P=\frac{k|Q|}{r_P^2} \]

\[ E_P= \frac{(8.99\times10^9)(50\times10^{-6})} {(0.20)^2} \]

\[ E_P=1.12\times10^7\ \text{N C}^{-1} \]

At R:

\[ E_R=\frac{k|Q|}{r_R^2} \]

\[ E_R= \frac{(8.99\times10^9)(50\times10^{-6})} {(0.40)^2} \]

\[ E_R=2.81\times10^6\ \text{N C}^{-1} \]

The field at P is four times stronger because R is twice as far from the source.

Part (b): Calculate the forces.

Use:

\[ F_E=|q|E \]

At P, the test charge is:

\[ q_P=0.50\times10^{-6}\text{ C} \]

Therefore:

\[ F_P=(0.50\times10^{-6})(1.12\times10^7) \]

\[ F_P\approx5.6\text{ N} \]

At R:

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

Therefore:

\[ F_R=(2.0\times10^{-6})(2.81\times10^6) \]

\[ F_R\approx5.6\text{ N} \]

The force magnitudes are equal.

Part (c): Determine the directions and explain the result.

The source charge is positive, so the electric field points away from it.

Both P and R are to the right of the source. Therefore, the electric field points right at both locations.

At P, the test charge is positive, so the force also points right.

At R, the test charge is negative, so the force points opposite to the field, towards the left.

Why are the force magnitudes equal?

The field at R is one-quarter as strong as at P. However, the test charge at R is four times larger in magnitude.

These effects cancel:

\[ F_R=4q_P\times\frac14E_P \]

\[ F_R=q_PE_P=F_P \]

The key lesson is that force depends on both the field and the test charge. Field strength depends on the source arrangement and the location, not on the size of the test charge.

Question 4

Two separate point-charge diagrams are labelled NOT TO SCALE.

• Diagram A contains a \(+4.0\,\mu\text{C}\) source charge. Point P is \(0.40\text{ m}\) to its right.
• Diagram B contains a \(-1.0\,\mu\text{C}\) source charge. Point R is \(0.10\text{ m}\) to its right.

The artist has drawn the field arrow at P twice as long as the arrow at R.

A student concludes that the field at P must be twice as strong as the field at R.

Evaluate this conclusion quantitatively. Explain which information in the diagrams should be used and whether the arrow lengths provide reliable evidence.

Solution 4

The student’s conclusion is incorrect. The field at R is four times stronger than the field at P.

Step 1: Identify the source charges and distances.

For diagram A:

\[ |Q_A|=4.0\,\mu\text{C} \]

\[ r_P=0.40\text{ m} \]

For diagram B:

\[ |Q_B|=1.0\,\mu\text{C} \]

\[ r_R=0.10\text{ m} \]

Step 2: Apply the full comparison equation.

\[ \frac{E_P}{E_R} = \frac{|Q_A|}{|Q_B|} \left(\frac{r_R}{r_P}\right)^2 \]

Substitute:

\[ \frac{E_P}{E_R} = \frac{4.0}{1.0} \left(\frac{0.10}{0.40}\right)^2 \]

\[ =4.0\times(0.25)^2 \]

\[ =4.0\times0.0625 \]

\[ \frac{E_P}{E_R}=0.25 \]

Therefore:

\[ E_R=4E_P \]

The field at R is four times stronger.

Step 3: Evaluate the arrow lengths.

The arrow lengths are not accompanied by a scale.

In an ordinary electric field-line diagram, arrowheads indicate direction, while the spacing of the field lines indicates relative strength.

The length of an arrow does not automatically represent field magnitude.

The artist has therefore provided a misleading visual cue. The labelled source charges and distances are the reliable information for this calculation.

There is another useful observation: the field at P points away from its positive source, while the field at R points towards its negative source. Their directions are opposite.

In an exam, do not allow the visual size of a diagram or an arrow to override the physical information supplied in the question.

Question 5

This final problem combines source charge, distance, field-line counts, test-charge magnitude, and direction.

Two experimental arrangements are compared.

Arrangement I

• An isolated positive point charge \(+Q\) produces an electric field at point P.
• P is a distance \(r\) to the right of the source.
• A positive test charge \(+q\) at P experiences a force of magnitude \(F\), directed to the right.

Arrangement II

• An isolated point charge \(Q_2\) produces an electric field at point R.
• R is a distance \(2r\) to the right of the source.
• A negative test charge \(-4q\) at R experiences a force of magnitude \(3F\), also directed to the right.
• The diagram for arrangement II shows three times as many radial field lines as arrangement I.

Both diagrams use the same field-line convention, and the test charges do not disturb the source charges. Assume there are no other source charges.

(a) Determine the magnitude of the electric field at R relative to the field at P, using the force measurements.

(b) Use the field-line counts and distances to determine the magnitude of \(Q_2\) relative to \(Q\).

(c) Determine the sign of \(Q_2\).

(d) Explain whether all the observations are consistent.

(e) Explain why arrangement II produces a larger force even though its electric field is weaker.

Solution 5

The observations are consistent, and the second source charge is \(-3Q\).

The electric field at R is three-quarters as strong as the field at P, even though the force on the test charge at R is three times larger.

This problem is difficult because three different quantities change simultaneously. We need to keep them separate.

Part (a): Determine the field-strength ratio from the force measurements.

Start with the definition of electric field strength:

\[ E=\frac{F_E}{|q|} \]

At P:

\[ E_P=\frac{F}{q} \]

At R, the test charge has magnitude \(4q\), and the force is \(3F\).

Therefore:

\[ E_R=\frac{3F}{4q} \]

Divide the two expressions:

\[ \frac{E_R}{E_P} = \frac{3F/(4q)}{F/q} \]

\[ \boxed{\frac{E_R}{E_P}=\frac34} \]

The field at R is three-quarters as strong as the field at P.

This is already an important result. The force at R is larger, but the field is weaker.

Part (b): Determine the magnitude of the second source charge.

The second diagram contains three times as many radial field lines.

Because both diagrams use the same field-line convention:

\[ \frac{|Q_2|}{|Q|}=3 \]

Therefore:

\[ |Q_2|=3Q \]

Now check whether this agrees with the different distances.

Using the point-charge equation:

\[ \frac{E_R}{E_P} = \frac{|Q_2|}{|Q|} \left(\frac{r}{2r}\right)^2 \]

Substitute the charge ratio:

\[ \frac{E_R}{E_P} = 3\left(\frac12\right)^2 \]

\[ \boxed{\frac{E_R}{E_P}=\frac34} \]

This agrees exactly with the ratio obtained from the force measurements.

Part (c): Determine the sign of the second source charge.

In arrangement II, the test charge is negative.

Its force points to the right.

A negative test charge experiences a force opposite to the electric field, so the electric field at R must point to the left.

Point R is to the right of its source charge.

For the field to point left, towards the source, the source must be negative.

Therefore:

\[ \boxed{Q_2=-3Q} \]

Part (d): Check whether the observations are consistent.

Yes.

The second source has three times the charge magnitude, which agrees with the threefold increase in field-line count.

However, the observation point is twice as far from the source.

The inverse-square relationship reduces the field by a factor of four, leaving:

\[ E_R=\frac34E_P \]

The force measurements independently produce the same field-strength ratio.

The source charge, distance, force measurements, and field-line counts are therefore mutually consistent.

Part (e): Explain why the force is larger.

The force depends on both the test charge and the electric field:

\[ F_E=|q|E \]

At R, the test charge is four times larger in magnitude than the test charge at P.

However, the field is only three-quarters as strong.

Therefore:

\[ \frac{F_R}{F_P} = 4\times\frac34 \]

\[ \boxed{\frac{F_R}{F_P}=3} \]

The force at R is three times larger, exactly as stated.

This is the trap in the question.

A student might see three times as many field lines and a three-times-larger force, then confidently conclude that the field is three times stronger.

Neither observation supports that conclusion.

The total field-line count describes the source charge magnitude, while the force also depends on the test charge. The local field strength depends on both the source charge and the distance.

07Where to go next

The most reliable habit when comparing point-charge diagrams is to separate three questions:

• What is producing the field?
• How far is the observation point from that source?
• What is the field direction, independently of the test charge?

For a single isolated point charge, the comparison is governed by:

\[ E=\frac{k|Q|}{r^2} \]

If several source charges appear in the same diagram, this single-source method is no longer sufficient on its own. You must consider the individual field contributions and combine them according to their directions to determine the resultant electric field.

That is the next step: moving from comparing individual point-charge fields to analysing the combined electric field produced by multiple charges.