How to Find an Unknown Charge from an Electric-Field Measurement

Suppose a field probe is placed \(0.20\text{ m}\) from a small charged sphere. It measures an electric field strength of \(1.8\times10^4\text{ N C}^{-1}\).

What is the charge on the sphere?

If the sphere is the only source of the field, its charge magnitude is

\[ |Q|=\frac{Er^2}{k} =\frac{(1.8\times10^4)(0.20)^2}{8.99\times10^9} \]

\[ \boxed{|Q|=8.0\times10^{-8}\text{ C}} \]

The field direction tells us the sign. If it points away from the sphere, the charge is positive. If it points towards the sphere, the charge is negative.

That’s the central method for finding an unknown source charge from an electric-field measurement. However, there’s an important distinction: if the unknown charge is a particle placed inside an existing electric field, you cannot determine its charge from the field strength alone. You also need to know the force acting on it.

Let’s separate these situations before choosing an equation.

01First identify which charge is unknown

There are two different charges to keep separate.

• Source charge \(Q\): The charge producing the electric field.
• Test charge \(q\): A charge placed in the electric field to experience an electric force.

The distinction determines which equation you need.

The first situation is the main focus of this article.

Why doesn’t the test charge determine the electric field?

Electric field strength is defined as the electric force per unit positive test charge:

\[ \boxed{E=\frac{F}{q_t}} \]

Here:

• \(E\) is electric field strength in newtons per coulomb (\(\text{N C}^{-1}\)).
• \(F\) is the electric force in newtons (N).
• \(q_t\) is the positive test charge in coulombs (C).

Imagine doubling the positive test charge at a particular location.

The electric force on it doubles, too. However, the ratio \(F/q_t\) stays the same.

That means the electric field is a property of the source charges and their arrangement, not the test charge used to measure it.

In practice, the test charge is assumed to be small enough that it does not significantly disturb the original charge distribution.

This is why you can measure an electric field without needing to know the charge of the probe.

02Uniform fields and point-charge fields are different

Before calculating an unknown charge, you need to identify the type of electric field being measured.

Two important cases behave differently.

Case A: Uniform electric field

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

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

Notice that moving the probe closer to one plate does not automatically make the field stronger.

In the central region, the field remains approximately constant.

For parallel plates, the field strength can be calculated using

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

where:

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

This is not an inverse-square relationship.

If you are trying to find the charge of a particle placed in this uniform field, you need its electric force as well.

Case B: Electric field around a point charge

An isolated point charge produces a radial electric field.

The field points:

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

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

The relationship is

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

This is Coulomb’s inverse-square electric-field relationship.

Here:

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

The inverse-square relationship gives us a useful prediction.

If you double the distance from a point charge, the electric field becomes one-quarter as strong.

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

For example:

This behaviour is the key to recognising when you can use the point-charge equation.

Not every non-uniform electric field follows an inverse-square relationship. The equation applies to an isolated point source, or an appropriate spherically symmetric source viewed from outside it.

03The method: calculate an unknown source charge

Once you’ve established that the measured field comes from a single point source, rearrange the equation:

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

Multiply both sides by \(r^2\):

\[ Er^2=k|Q| \]

Divide by \(k\):

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

This is the equation you’ll use most often for this type of question.

A reliable calculation sequence

1. Identify the source charge and the measurement position.
2. Confirm that the point-source model is appropriate.
3. Record the measured field strength \(E\).
4. Convert the distance \(r\) into metres.
5. Substitute into \( |Q|=Er^2/k \).
6. Use the field direction to determine whether the source is positive or negative.
7. State the final answer in coulombs, including its sign.

One important detail: the equation calculates the magnitude of the charge. You must determine its sign separately.

If the field measurement gives only a magnitude, without a direction, you cannot determine whether the source charge is positive or negative from that measurement alone.

04What if you have measurements at several distances?

Sometimes a question provides more than one field measurement.

This is useful because you can check whether the field actually follows an inverse-square relationship.

For a single point source,

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

Rearranging gives

\[ Er^2=k|Q| \]

The right-hand side is constant if the source charge stays unchanged.

Therefore, for measurements taken at different distances from the same isolated point source:

\[ \boxed{Er^2=\text{constant}} \]

If the values of \(Er^2\) are approximately equal, the measurements are consistent with an inverse-square field.

05The important complication: other electric fields

So far, we’ve assumed that one point charge produces the entire measured electric field.

What happens if another charge is nearby?

The field probe measures the combined electric field from all the sources.

This is called the principle of superposition.

\[ \boxed{\vec E_{\text{net}}=\vec E_1+\vec E_2+\cdots} \]

Electric fields are vectors, so you must account for their directions when combining them.

This creates an important limitation.

You cannot automatically substitute the total measured field into

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

if other sources also contribute to the measurement.

Instead, you must determine the field contribution from the unknown source first.

For example, if the known source produces a field to the left and the measured net field points to the right, the unknown source must produce a sufficiently strong rightward field to overcome the known contribution.

This is particularly important when an unknown point charge sits inside a uniform background field.

The total field might not follow an inverse-square relationship, even though the contribution from the point charge does.

We’ll test this in the final question.

06Common mistakes to avoid

A useful final check is to ask whether your answer makes physical sense.

If you double the distance from an isolated point charge while keeping the source unchanged, the field should decrease to one-quarter of its original value. If your calculation predicts otherwise, revisit your substitution.

07Questions and solutions

The following questions build from a straightforward source-charge calculation to situations involving uniform fields, test charges, and superposition.

Question 1

An electric-field probe is positioned \(0.25\text{ m}\) from an isolated point charge.

The probe measures a field strength of \(3.6\times10^4\text{ N C}^{-1}\), directed towards the source charge.

Calculate the magnitude and sign of the source charge.

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

Solution 1

The source charge is

\[ \boxed{Q=-2.5\times10^{-7}\text{ C}} \]

The field points towards the source, so the charge must be negative.

First, calculate its magnitude using

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

Substitute the given values:

\[ \begin{aligned} |Q|&=\frac{(3.6\times10^4)(0.25)^2} {8.99\times10^9}\\ &=\frac{(3.6\times10^4)(0.0625)} {8.99\times10^9}\\ &=\frac{2250}{8.99\times10^9}\\ &=2.50\times10^{-7}\text{ C} \end{aligned} \]

The field direction points towards the source, indicating a negative charge.

Therefore,

\[ \boxed{Q=-2.5\times10^{-7}\text{ C}} \]

The tempting mistake is to report the positive magnitude calculated by the equation. Remember that the magnitude equation does not include the sign. You must interpret the direction separately.

Question 2

Two large parallel plates are separated by \(12\text{ cm}\). The left plate is positively charged, and the right plate is negatively charged.

A field probe measures \(2.4\times10^3\text{ N C}^{-1}\) to the right at two positions:

• Position A is \(2.0\text{ cm}\) from the positive plate.
• Position B is \(8.0\text{ cm}\) from the positive plate.

A student argues that the field at B should be weaker because it is four times farther from the positive plate.

They attempt to apply the inverse-square equation to both measurements.

Explain why this reasoning is incorrect. Identify the appropriate field model and explain whether the measured field strength alone is sufficient to calculate the total charge on either plate.

Solution 2

The field is approximately uniform, so the inverse-square equation is not appropriate.

The key issue is that the plates are extended charged surfaces, not isolated point charges.

For a point source, the field strength depends on the square of the radial distance from that source:

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

However, the central region between large parallel plates has an approximately constant electric field.

The field is therefore expected to have the same magnitude at positions A and B, provided both positions are sufficiently far from the plate edges.

The student’s mistake is treating the distance from a plate as the radial distance from a point charge.

Those are different physical situations.

The appropriate model for the central region between the plates is

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

The field measurement alone is not enough to determine the total charge on a plate. Additional information, such as the plate area and an appropriate model of the charge distribution, is needed.

This is a useful distinction:

• A point charge produces an inverse-square field.
• Large parallel plates produce an approximately uniform field between them.

Choosing the correct model matters more than simply recognising an equation.

Question 3

An unknown particle is placed in a uniform electric field of strength \(3.0\times10^4\text{ N C}^{-1}\), directed west.

An independent force measurement shows that the electric force on the particle is \(6.0\times10^{-3}\text{ N}\), directed east.

Determine the signed charge of the particle.

Solution 3

The particle has a negative charge:

\[ \boxed{q=-2.0\times10^{-7}\text{ C}} \]

This is a different problem from finding the charge that produces a point-source field.

Here, we know the electric field and the force acting on an unknown test charge.

The relevant relationship is

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

Choose east as the positive direction.

The signed quantities are therefore

\[ F_x=+6.0\times10^{-3}\text{ N} \]

\[ E_x=-3.0\times10^4\text{ N C}^{-1} \]

Rearrange the equation:

\[ q=\frac{F_x}{E_x} \]

Substitute:

\[ \begin{aligned} q&=\frac{6.0\times10^{-3}} {-3.0\times10^4}\\ &=-2.0\times10^{-7}\text{ C} \end{aligned} \]

The negative result tells us that the particle’s force is opposite to the electric field.

This is physically consistent because negative charges experience forces opposite to the electric-field direction.

The important trap is confusing the source charge with the charge being placed in the field. We used \(F=qE\) because the unknown is the test particle, not the source of the field.

Question 4

An unknown point charge \(Q\) is located at \(x=0\).

A field probe is placed at \(x=0.30\text{ m}\). A second, known charge of \(+2.0\times10^{-8}\text{ C}\) is located at \(x=0.60\text{ m}\).

The probe measures a net electric field of \(1.0\times10^3\text{ N C}^{-1}\) directed to the right.

Determine the unknown charge \(Q\), including its sign.

Assume the charges are stationary and that the measured field is the vector sum of the contributions from these two charges.

Solution 4

The unknown charge is

\[ \boxed{Q=+3.0\times10^{-8}\text{ C}} \]

We cannot use the measured net field directly in the point-charge equation because the known positive charge also contributes to the field at the probe.

We must first isolate the contribution from the unknown charge.

Step 1: Calculate the field produced by the known charge.

The known charge is positive and sits to the right of the probe.

Its electric field at the probe points away from the known charge, so it points left.

The magnitude is

\[ E_{\text{known}}=\frac{k|Q_{\text{known}}|}{r^2} \]

\[ \begin{aligned} E_{\text{known}} &=\frac{(8.99\times10^9)(2.0\times10^{-8})} {(0.30)^2}\\ &=2.0\times10^3\text{ N C}^{-1} \end{aligned} \]

Therefore, the known charge produces a field of approximately \(2.0\times10^3\text{ N C}^{-1}\) to the left.

Step 2: Determine the field produced by the unknown charge.

Take right as positive.

The net field is \(+1.0\times10^3\text{ N C}^{-1}\), while the known contribution is \(-2.0\times10^3\text{ N C}^{-1}\).

Using superposition:

\[ E_{\text{net}}=E_Q+E_{\text{known}} \]

Rearrange:

\[ E_Q=E_{\text{net}}-E_{\text{known}} \]

\[ \begin{aligned} E_Q&=1.0\times10^3-(-2.0\times10^3)\\ &=3.0\times10^3\text{ N C}^{-1} \end{aligned} \]

The unknown charge must produce a field pointing right at the probe.

Step 3: Calculate the unknown charge.

The probe is \(0.30\text{ m}\) from the unknown source.

Use

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

\[ \begin{aligned} |Q|&=\frac{(3.0\times10^3)(0.30)^2} {8.99\times10^9}\\ &=\frac{270}{8.99\times10^9}\\ &=3.0\times10^{-8}\text{ C} \end{aligned} \]

Step 4: Determine the sign.

The probe is to the right of the unknown source.

For the field there to point right, the source must be positive.

Therefore,

\[ \boxed{Q=+3.0\times10^{-8}\text{ C}} \]

The important lesson is that a field measurement can contain contributions from several sources. You must isolate the field produced by the unknown charge before using the inverse-square equation.

Question 5

An unknown point charge \(Q\) is located at the origin of a horizontal \(x\)-axis.

A uniform background electric field also exists throughout the region. Its direction and strength are unknown.

A field probe records the following illustrative measurements at three positions to the right of the unknown charge.

Assume that the only contributions to the measured field are:

• The field produced by the unknown point charge at the origin.
• A uniform background electric field with constant magnitude and direction.

Answer the following:

a. Determine the sign and magnitude of the unknown point charge.

b. Determine the magnitude and direction of the uniform background field.

c. Find every position along the \(x\)-axis where the net electric field is zero. Explain the direction of the net field in the different regions of the axis.

d. Verify whether the third measurement is consistent with your results.

Solution 5

The unknown source charge is positive, with magnitude

\[ \boxed{Q=+8.0\times10^{-8}\text{ C}} \]

The uniform background field has magnitude

\[ \boxed{E_0=2.0\times10^3\text{ N C}^{-1}} \]

and points to the left.

The net electric field is zero at exactly one position on the \(x\)-axis:

\[ \boxed{x=+0.60\text{ m}} \]

This question is more difficult because the measured field is not produced by the point charge alone.

A tempting approach is to substitute each measured net field directly into the inverse-square equation.

Let’s see why that fails.

Using the first measurement:

\[ |Q|=\frac{(1.60\times10^4)(0.20)^2} {8.99\times10^9} =7.1\times10^{-8}\text{ C} \]

Using the second measurement:

\[ |Q|=\frac{(2.50\times10^3)(0.40)^2} {8.99\times10^9} =4.4\times10^{-8}\text{ C} \]

These results disagree.

The source charge has not changed between measurements, so it cannot have two different values.

The problem is that we have applied the inverse-square equation to the total field rather than the field produced by the point charge.

We need to separate the two contributions.

Step 1: Establish a sign convention.

Choose right as the positive direction.

Let:

• \(E_Q\) represent the signed field produced by the unknown point charge.
• \(E_0\) represent the signed uniform background field.

The net field is

\[ E_{\text{net}}=E_Q+E_0 \]

Because the point charge is at the origin, the field it produces at \(x=0.40\text{ m}\) is one-quarter of its strength at \(x=0.20\text{ m}\).

This follows from the inverse-square relationship:

\[ \frac{E_Q(0.40)}{E_Q(0.20)} = \left(\frac{0.20}{0.40}\right)^2 =\frac14 \]

Step 2: Set up two equations using the first two measurements.

Let \(S\) be the field strength produced by the unknown point charge at \(x=0.20\text{ m}\).

Since the distance doubles at \(x=0.40\text{ m}\), the point-charge contribution becomes \(S/4\).

The first measurement gives

\[ E_0+S=1.60\times10^4 \]

The second measurement gives

\[ E_0+\frac{S}{4}=2.50\times10^3 \]

Subtract the second equation from the first:

\[ \begin{aligned} S-\frac{S}{4} &=1.60\times10^4-2.50\times10^3\\ \frac34S&=1.35\times10^4\\ S&=1.80\times10^4\text{ N C}^{-1} \end{aligned} \]

The point charge produces a field of \(1.80\times10^4\text{ N C}^{-1}\) to the right at \(x=0.20\text{ m}\).

Step 3: Determine the uniform background field.

Substitute \(S\) into the first equation:

\[ E_0+1.80\times10^4=1.60\times10^4 \]

Therefore,

\[ \boxed{E_0=-2.0\times10^3\text{ N C}^{-1}} \]

The negative sign means the uniform background field points left.

Its magnitude is

\[ \boxed{2.0\times10^3\text{ N C}^{-1}\text{ left}} \]

Step 4: Calculate the unknown point charge.

At \(x=0.20\text{ m}\), the point-charge contribution is

\[ E_Q=1.80\times10^4\text{ N C}^{-1} \]

Use the inverse-square equation:

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

\[ \begin{aligned} |Q|&=\frac{(1.80\times10^4)(0.20)^2} {8.99\times10^9}\\ &=\frac{720}{8.99\times10^9}\\ &=8.01\times10^{-8}\text{ C} \end{aligned} \]

At a position to the right of the source, the point-charge field points right.

Therefore, the source charge is positive:

\[ \boxed{Q=+8.0\times10^{-8}\text{ C}} \]

Step 5: Find where the net electric field becomes zero.

The net field is zero when the point-charge contribution exactly balances the uniform background field.

The two fields must have equal magnitudes and opposite directions.

At \(x=0.20\text{ m}\), the point-charge field is \(1.80\times10^4\text{ N C}^{-1}\).

The background field has magnitude \(2.0\times10^3\text{ N C}^{-1}\).

Because the point-charge field follows an inverse-square relationship,

\[ E_Q(r)=1.80\times10^4 \left(\frac{0.20}{r}\right)^2 \]

Set this equal to the background field magnitude:

\[ 1.80\times10^4 \left(\frac{0.20}{r}\right)^2 =2.0\times10^3 \]

Divide both sides:

\[ \left(\frac{0.20}{r}\right)^2=\frac19 \]

Taking the positive square root for a position to the right:

\[ \frac{0.20}{r}=\frac13 \]

Therefore,

\[ \boxed{r=0.60\text{ m}} \]

The net field is zero at \(x=+0.60\text{ m}\).

There is no cancellation point to the left of the origin. A positive point charge produces a leftward field on its left-hand side, which is the same direction as the uniform background field.

The complete direction pattern is therefore:

• For \(x<0\), both fields point left, so the net field points left.
• For \(0<x<0.60\text{ m}\), the point-charge field is stronger, so the net field points right.
• At \(x=0.60\text{ m}\), the fields cancel.
• For \(x>0.60\text{ m}\), the uniform field is stronger, so the net field points left.

Step 6: Check the third measurement.

At \(x=0.80\text{ m}\), the point-charge field is

\[ \begin{aligned} E_Q&=1.80\times10^4 \left(\frac{0.20}{0.80}\right)^2\\ &=1.125\times10^3\text{ N C}^{-1} \end{aligned} \]

It points right.

The uniform background field is \(2.0\times10^3\text{ N C}^{-1}\) to the left.

Therefore, the net field is

\[ \begin{aligned} E_{\text{net}} &=1.125\times10^3-2.0\times10^3\\ &=-8.75\times10^2\text{ N C}^{-1} \end{aligned} \]

The negative sign means the field points left.

Thus,

\[ \boxed{E_{\text{net}}=8.75\times10^2\text{ N C}^{-1} \text{ left}} \]

This exactly matches the third measurement.

The important insight is that an inverse-square field does not guarantee that the total measured field follows an inverse-square relationship. Other contributions can change both its magnitude and direction.

08Where to go next

Once you can identify an unknown source charge from a field measurement, the next useful skill is working out what that field does to another charge.

For the relationship between electric field strength and the force on a charged particle, read How to Calculate Electric Field from Force on a Test Charge.

If you’re working with parallel plates instead, How to Calculate Electric Field Strength Between Parallel Plates develops the uniform-field method.

Keep the three relationships distinct:

\[ E=\frac{F}{q_t} \]

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

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

They describe different ways of determining electric field strength. Choosing the right one is the first step towards getting the charge calculation right.