How to Calculate Electric Field from Force on a Test Charge

A \(+2.0\,\mu\text{C}\) test charge is placed at point P. It experiences an electric force of \(0.040\text{ N}\) towards the east.

What is the electric field strength at P?

You can calculate it by dividing the electric force by the test charge:

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

Substituting:

\[ E=\frac{0.040}{2.0\times10^{-6}} =2.0\times10^4\text{ N/C} \]

The electric field is \(2.0\times10^4\text{ N/C}\) east.

Now imagine replacing the positive test charge with a negative charge of the same magnitude. The force would point west instead. However, the electric field at P would remain unchanged.

This is the key idea: the electric field belongs to the arrangement of source charges, not to the test charge used to measure it.

01The equation for calculating electric field from force

Electric field strength describes the electric force experienced per coulomb of test charge at a particular location.

The vector equation is:

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

Where:

• \(\vec E\) is the electric field strength, measured in newtons per coulomb (N/C).
• \(\vec F_E\) is the electric force acting on the test charge, measured in newtons (N).
• \(q\) is the signed test charge, measured in coulombs (C).

For the magnitude of the electric field, use:

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

The absolute value is important because field strength as a magnitude cannot be negative. You determine the direction separately.

Understanding the direction

The electric field direction is defined as the direction of the force on a positive test charge.

This gives you two possibilities:

For example, suppose a negative test charge experiences an electric force towards the left.

The electric field must point towards the right.

The negative charge reverses the direction of the force, not the direction of the electric field itself.

A useful way to avoid sign mistakes is to calculate the magnitude first, then determine the direction using the sign of the charge.

02Why electric field strength does not depend on the test charge

Consider two test charges placed at exactly the same location in an electric field.

• Charge A is \(+2.0\,\mu\text{C}\).
• Charge B is \(+4.0\,\mu\text{C}\).

Both charges are positive, and both experience the same electric field.

Which one experiences the greater force?

Charge B experiences twice the force because it has twice the charge.

This follows from rearranging the electric field equation:

\[ F_E=|q|E \]

If the electric field remains constant, doubling the test charge doubles the force.

However, dividing the force by the charge gives the same field strength for both particles.

The important distinction is between the field and the force.

The field describes the electrical conditions at a location. The force describes how strongly a particular charge responds to those conditions.

A useful comparison is gravitational field strength.

Two objects with different masses experience different gravitational forces at the same location, but the gravitational field strength remains the same. Electric fields work similarly, although electric charges can be positive or negative, so the force can also reverse direction.

What actually creates the electric field?

The electric field is produced by source charges.

A test charge is simply a way of measuring the field at a particular location. Ideally, it must be small enough that it does not significantly disturb the original arrangement of source charges.

It helps to distinguish two different quantities:

• \(Q\): the source charge responsible for producing the electric field.
• \(q\): the test charge used to measure the field.

Confusing these two charges is a common source of mistakes when questions introduce Coulomb’s law.

03Uniform electric fields versus inverse-square electric fields

Here is an important distinction.

If you move a test charge further away from the source of an electric field, does the field strength decrease?

Sometimes it does. Sometimes it doesn’t.

The answer depends on how the electric field is produced.

Uniform electric fields

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

A common approximation is the field between two large, parallel charged plates.

Within the central region between the plates, the electric field is approximately uniform.

This means:

• Moving a test charge to another position within the region does not change the field strength.
• A positive test charge experiences a force in the direction of the field.
• A negative test charge experiences a force in the opposite direction.

For ideal parallel plates, the field strength can also be calculated using:

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

Where:

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

The electric field remains approximately constant provided you stay away from the edges, where the field becomes distorted.

Inverse-square electric fields

Now consider an isolated point source charge.

Unlike the uniform field between parallel plates, the electric field around a point charge becomes weaker as you move further away.

The magnitude is calculated using:

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

Where:

• \(E\) is electric field strength in N/C.
• \(k\) 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 source charge to the location where the field is being calculated, in metres (m).

The field points radially away from a positive source charge and towards a negative source charge.

The inverse-square relationship means:

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

So, if you double the distance from an isolated point charge:

\[ E_{\text{new}}=\frac{E_{\text{old}}}{2^2} \]

\[ \boxed{E_{\text{new}}=\frac{E_{\text{old}}}{4}} \]

The field strength becomes one-quarter of its original value.

This is very different from a uniform electric field, where moving between positions within the uniform region does not change the field strength.

Comparing the two models

There is an important detail here.

The equation \(E=F_E/|q|\) works for both types of electric field. It calculates the field strength at the particular location where the force is measured.

The inverse-square equation is a different method. It calculates the field produced by a point source charge when its charge and distance are known.

You should not assume that every electric field follows an inverse-square relationship.

04Choosing the correct equation in an HSC question

When you see an electric field question, first identify what information you have been given.

A particularly important point is that a force measurement gives you the electric field at one location.

It does not automatically tell you whether the field is uniform throughout the surrounding region.

For multiple source charges, the resultant field is the vector sum of their individual contributions:

\[ \vec E_{\text{net}}=\sum_i\vec E_i \]

This means you must consider direction when combining fields. Simply adding their magnitudes can give the wrong answer.

Three mistakes worth avoiding

Mistake 1: Ignoring the sign of the test charge.

If a negative test charge experiences a force to the left, the field points right. Do not automatically use the force direction as the field direction.

Mistake 2: Using the test charge in the inverse-square equation.

The equation \(E=k|Q|/r^2\) uses the source charge \(Q\), not the test charge \(q\).

Mistake 3: Assuming the field is always uniform or always inverse-square.

The field’s behaviour depends on the arrangement of source charges. The inverse-square equation specifically describes an isolated point source.

One final qualification: if a question gives you the net force on a particle rather than the electric force alone, you must account for other forces first. For example, gravity may contribute to the net force. You cannot automatically divide the total force by the charge and call the result the electric field.

05Questions and solutions

The following questions move from straightforward field calculations to more demanding problems involving source charges, competing models, and electric field superposition.

Question 1

A positive test charge of \(+5.0\,\mu\text{C}\) is placed at point A.

It experiences an electric force of \(0.15\text{ N}\) towards the east.

Calculate:

a. The electric field strength at A.

b. The direction of the electric field.

Solution 1

The electric field at A is:

\[ \boxed{E=3.0\times10^4\text{ N/C east}} \]

Step 1: Convert the charge into coulombs.

\[ q=5.0\times10^{-6}\text{ C} \]

Step 2: Apply the electric field equation.

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

Substituting:

\[ \begin{aligned} E&=\frac{0.15}{5.0\times10^{-6}}\\ &=3.0\times10^4\text{ N/C} \end{aligned} \]

Step 3: Determine the direction.

The test charge is positive, so the electric field points in the same direction as the electric force.

The force acts east, so the field also points east.

A common mistake is to report the magnitude without the direction. Electric field strength is a vector quantity, so both are needed for a complete answer.

Question 2

A test charge of \(-2.5\text{ nC}\) is placed at point A.

It experiences an electric force of \(6.0\,\mu\text{N}\) downwards.

a. Calculate the electric field strength at A, including its direction.

b. A second test charge of \(+7.5\text{ nC}\) is placed at the same location. Calculate the magnitude and direction of the electric force on this charge.

Assume the test charges are sufficiently small that they do not disturb the original electric field.

Solution 2

The electric field is \(2.4\times10^3\text{ N/C}\) upwards. The second test charge experiences a force of \(1.8\times10^{-5}\text{ N}\) upwards.

Step 1: Calculate the field strength.

Convert the charge:

\[ q=-2.5\times10^{-9}\text{ C} \]

Use the magnitude equation:

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

Substitute:

\[ \begin{aligned} E&=\frac{6.0\times10^{-6}} {2.5\times10^{-9}}\\ &=2.4\times10^3\text{ N/C} \end{aligned} \]

Step 2: Determine the field direction.

The original test charge is negative.

Therefore, the electric field acts opposite to the electric force.

The force points downwards, so:

\[ \boxed{\vec E=2.4\times10^3\text{ N/C upwards}} \]

Step 3: Calculate the force on the second test charge.

The second charge is:

\[ q=+7.5\times10^{-9}\text{ C} \]

Use:

\[ F_E=|q|E \]

Substitute:

\[ \begin{aligned} F_E&=(7.5\times10^{-9})(2.4\times10^3)\\ &=1.8\times10^{-5}\text{ N} \end{aligned} \]

Step 4: Determine the force direction.

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

Therefore:

\[ \boxed{F_E=1.8\times10^{-5}\text{ N upwards}} \]

Notice that the field did not change when we replaced the negative test charge with a positive one.

The force changed direction because the sign of the test charge changed.

Question 3

An isolated positive point source charge \(Q\) is located at S.

Two points, A and B, lie along the same straight line extending outwards from S.

• Point A is \(0.15\text{ m}\) from S.
• Point B is \(0.30\text{ m}\) from S.

A positive test charge of \(+1.5\text{ nC}\) placed at A experiences an electric force of \(0.12\text{ mN}\) away from S.

Calculate:

a. The electric field strength at A.

b. The magnitude of the source charge \(Q\).

c. The electric field strength at B.

d. The magnitude and direction of the electric force on a test charge of \(-3.0\text{ nC}\) placed at B.

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

Solution 3

The answers are:

• At A: \(8.0\times10^4\text{ N/C}\) outwards.
• Source charge: \(+0.20\,\mu\text{C}\).
• At B: \(2.0\times10^4\text{ N/C}\) outwards.
• Force at B: \(6.0\times10^{-5}\text{ N}\) towards S.

Step 1: Calculate the electric field strength at A.

Convert the given quantities:

\[ F_E=0.12\text{ mN}=1.2\times10^{-4}\text{ N} \]

\[ q=1.5\times10^{-9}\text{ C} \]

Using:

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

Substitute:

\[ \begin{aligned} E_A&=\frac{1.2\times10^{-4}}{1.5\times10^{-9}}\\ &=8.0\times10^4\text{ N/C} \end{aligned} \]

The test charge is positive, so the field points in the same direction as the force.

Therefore, the field points away from S.

Step 2: Calculate the source charge.

We know the field strength at A and its distance from the source.

Use the inverse-square equation:

\[ E_A=\frac{k|Q|}{r_A^2} \]

Rearrange:

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

Substitute:

\[ \begin{aligned} |Q|&=\frac{(8.0\times10^4)(0.15)^2} {8.99\times10^9}\\ &=2.0\times10^{-7}\text{ C} \end{aligned} \]

The field points away from S, so the source charge must be positive.

\[ \boxed{Q=+0.20\,\mu\text{C}} \]

Step 3: Calculate the field strength at B.

Point B is twice as far from the source as point A.

For an inverse-square field:

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

Therefore:

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

Substitute:

\[ \begin{aligned} E_B&=(8.0\times10^4) \left(\frac{0.15}{0.30}\right)^2\\ &=8.0\times10^4(0.25)\\ &=2.0\times10^4\text{ N/C} \end{aligned} \]

The field still points away from the positive source charge.

Step 4: Calculate the force on the negative test charge at B.

Use:

\[ F_E=|q|E_B \]

Substitute:

\[ \begin{aligned} F_E&=(3.0\times10^{-9})(2.0\times10^4)\\ &=6.0\times10^{-5}\text{ N} \end{aligned} \]

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

Therefore:

\[ \boxed{F_E=6.0\times10^{-5}\text{ N towards S}} \]

The important reasoning is that doubling the distance reduces the field to one-quarter, not one-half. The negative test charge then reverses the force direction without changing the field itself.

Question 4

A student investigates the electric field at two points, P and R, along the same straight line.

Both points are measured from a possible isolated point source S, which lies to their west.

The following measurements are recorded:

A student makes the following claim:

“The force doubles at R, so the electric field must be twice as strong there. This also shows that the field is uniform because the test charge doubled.”

Evaluate this claim.

Your answer should:

a. Calculate the electric field at P and R.

b. Explain why the electric force doubles.

c. Determine whether the measurements are consistent with a single isolated point source.

d. Decide whether the measurements prove that the field is uniform.

Solution 4

The student’s claim is incorrect.

The electric field has the same measured value at P and R:

\[ \boxed{E_P=E_R=2.0\times10^4\text{ N/C east}} \]

However, the measurements do not prove that the field is uniform throughout the region, and they are inconsistent with a single isolated point source.

Step 1: Calculate the electric field at P.

\[ E_P=\frac{F_P}{|q_P|} \]

Substitute:

\[ \begin{aligned} E_P&=\frac{0.020}{1.0\times10^{-6}}\\ &=2.0\times10^4\text{ N/C} \end{aligned} \]

The test charge is positive, so the field points east.

Step 2: Calculate the electric field at R.

\[ E_R=\frac{F_R}{|q_R|} \]

Substitute:

\[ \begin{aligned} E_R&=\frac{0.040}{2.0\times10^{-6}}\\ &=2.0\times10^4\text{ N/C} \end{aligned} \]

Again, the positive test charge means the field points east.

Therefore, the measured field vectors are equal at P and R.

Step 3: Explain why the force doubles.

The test charge at R is twice as large as the test charge at P.

Since:

\[ F_E=|q|E \]

doubling the test charge doubles the force if the field remains unchanged.

The force has doubled because the test charge has doubled, not because the measured electric field has become stronger.

Step 4: Test the isolated point-source model.

If S were the only source charge, doubling the distance from \(0.20\text{ m}\) to \(0.40\text{ m}\) would reduce the electric field to one-quarter.

The predicted field at R would be:

\[ E_R=\frac{2.0\times10^4}{4} \]

\[ E_R=5.0\times10^3\text{ N/C} \]

But the measured value is:

\[ E_R=2.0\times10^4\text{ N/C} \]

These values do not agree.

Therefore, the measurements cannot be explained by a single isolated point source charge.

Step 5: Decide whether the field is uniform.

The measurements show equal field vectors at two locations.

This is consistent with a uniform field, but it does not prove uniformity throughout the entire region.

A non-uniform field can have the same field strength and direction at two particular points while changing elsewhere.

This is an important distinction: measuring the field at two locations gives information about those locations. It does not automatically establish the behaviour of the entire field.

Question 5

Two identical positive source charges, each with a charge of \(+3.0\text{ nC}\), are fixed on a horizontal line.

They are separated by \(0.40\text{ m}\), with midpoint M exactly halfway between them.

A small positive test charge of \(+1.0\text{ nC}\) is placed at M.

Assume the source charges remain fixed, the test charge does not disturb them, and no forces other than the electric forces need to be considered.

Answer the following:

a. Calculate the resultant electric field at M and the electric force on the test charge.

b. The positive test charge is moved \(1.0\text{ cm}\) to the right of M, along the horizontal line. Calculate the resultant electric field and electric force at this new position.

c. Instead, return the charge to M and move it \(1.0\text{ cm}\) vertically upwards. Calculate the resultant electric field and electric force at this new position.

d. Is M a stable equilibrium position for the positive test charge? Explain what happens if the test charge is negative instead.

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

Solution 5

The resultant electric field at M is zero, but M is not a stable equilibrium position for either sign of test charge.

The key is to examine what happens when the charge is displaced in different directions.

Part a: Electric field and force at M

At the midpoint, each source charge is exactly \(0.20\text{ m}\) away.

Both source charges are positive, so their electric fields point away from them.

The field produced by the left source points right at M.

The field produced by the right source points left at M.

Their magnitudes are equal because the source charges and distances are identical.

For either source:

\[ E_1=E_2=\frac{kQ}{r^2} \]

Substituting:

\[ \begin{aligned} E_1&=\frac{(8.99\times10^9)(3.0\times10^{-9})} {(0.20)^2}\\ &=674\text{ N/C} \end{aligned} \]

The two fields cancel:

\[ E_{\text{net}}=674-674 \]

\[ \boxed{E_{\text{net}}=0\text{ N/C}} \]

The force on the positive test charge is:

\[ F_E=|q|E_{\text{net}} \]

\[ \boxed{F_E=0\text{ N}} \]

There is an important distinction here.

The electric field from each individual source charge is not zero. The resultant field is zero because the two non-zero field vectors point in opposite directions and cancel.

Part b: Displacing the charge horizontally

Move the positive test charge \(0.010\text{ m}\) to the right of M.

It is now:

• \(0.21\text{ m}\) from the left source charge.
• \(0.19\text{ m}\) from the right source charge.

The charge is closer to the right source.

Because the field follows an inverse-square relationship, the right source produces a stronger field at this location.

The left source produces a field pointing right, while the right source produces a field pointing left.

Taking right as positive:

\[ E_x=kQ\left( \frac{1}{(0.21)^2} -\frac{1}{(0.19)^2} \right) \]

Substituting:

\[ \begin{aligned} E_x&=(8.99\times10^9)(3.0\times10^{-9})\\ &\quad\times \left(\frac{1}{0.21^2}-\frac{1}{0.19^2}\right)\\ &=-1.36\times10^2\text{ N/C} \end{aligned} \]

The negative result means the field points left.

Therefore:

\[ \boxed{E_{\text{net}}=1.36\times10^2\text{ N/C left}} \]

Now calculate the force on the positive test charge.

\[ F_E=|q|E \]

\[ \begin{aligned} F_E&=(1.0\times10^{-9})(1.36\times10^2)\\ &=1.36\times10^{-7}\text{ N} \end{aligned} \]

Because the test charge is positive, the force acts in the same direction as the field.

\[ \boxed{F_E=1.36\times10^{-7}\text{ N left}} \]

Notice what this means.

The test charge was displaced to the right, but the resultant force points left, towards M.

The electric force is trying to return the positive test charge to its original position.

Part c: Displacing the charge vertically

Now return the positive test charge to M and move it \(0.010\text{ m}\) vertically upwards.

The two source charges are still identical and symmetrically positioned.

The horizontal components of their electric fields cancel each other.

However, their vertical components both point upwards, so they add.

The distance from either source charge to the displaced test charge is:

\[ r=\sqrt{(0.20)^2+(0.010)^2} \]

\[ r=\sqrt{0.0401} \]

\[ r\approx0.20025\text{ m} \]

For each source charge, the vertical component of the electric field is:

\[ E_y=\frac{kQ}{r^2}\frac{0.010}{r} \]

There are two identical sources, so:

\[ E_{\text{net},y}=\frac{2kQ(0.010)}{r^3} \]

Substituting:

\[ \begin{aligned} E_{\text{net},y} &=\frac{2(8.99\times10^9)(3.0\times10^{-9})(0.010)} {(0.20025)^3}\\ &\approx67.1\text{ N/C} \end{aligned} \]

Both vertical components point upwards.

Therefore:

\[ \boxed{E_{\text{net}}=67.1\text{ N/C upwards}} \]

The electric force on the positive test charge is:

\[ \begin{aligned} F_E&=|q|E\\ &=(1.0\times10^{-9})(67.1)\\ &=6.71\times10^{-8}\text{ N} \end{aligned} \]

The force is also upwards.

\[ \boxed{F_E=6.71\times10^{-8}\text{ N upwards}} \]

Unlike the horizontal displacement, this force acts away from M.

Part d: Is the equilibrium stable?

No. The midpoint is not a stable equilibrium position for either sign of test charge.

For a positive test charge:

• A small horizontal displacement produces a restoring force towards M.
• A small vertical displacement produces a force further away from M.

The two directions give different results.

For a negative test charge, both force directions reverse:

• A small horizontal displacement produces a force further away from M.
• A small vertical displacement produces a restoring force towards M.

Neither sign of test charge experiences a restoring force in every direction.

This exposes an important misconception: zero resultant force at a particular point does not automatically mean that the position is stable.

If the test charge is placed exactly at M and remains undisturbed, its electric force is zero. But if it is displaced slightly, the resulting force depends on the direction of displacement.

That is why examining the field from each source separately, and then adding the vectors, is essential.

Once you can calculate the electric field at a particular location, you can use it to predict the force on any other test charge placed there.

For the next step, see Force on a Charge in an Electric Field: HSC Physics Guide.

If the question then asks how the charged particle moves, continue to Acceleration of a Charge in an Electric Field: HSC Physics.