How to Find the Net Electric Field at a Square’s Corner

Place three charged particles at three corners of a square. You are asked to calculate the net electric field at the fourth corner.

The tempting approach is to calculate the electric field produced by each charge and add the three answers. But what if one field points left, another points upwards, and the third points diagonally? Simply adding their magnitudes will give you the wrong result.

The correct method is to calculate each electric field separately, resolve the diagonal contribution into horizontal and vertical components, and then add the components.

This is an application of the principle of superposition of electric fields. Once you understand the geometry, these questions become much more manageable.

01The answer: how to find the net electric field at a corner of a square

Consider a square with side length \(a\). Three source charges are positioned at three corners, and we want to calculate the net electric field at the remaining corner, labelled \(P\).

Before calculating anything, notice something important about the geometry.

There are only two possible distances between a source charge and the target corner:

• An adjacent corner is a distance \(a\) from \(P\).
• The diagonally opposite corner is a distance \(\sqrt{2}a\) from \(P\).

That second distance is particularly important. A diagonal is longer than a side, so the electric field produced by a charge at the opposite corner will be weaker.

For point charges, the field strength follows an inverse-square relationship:

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

Therefore, the fields produced by the three charges are:

\[
E_1=\frac{k|Q_1|}{a^2}
\]

\[
E_2=\frac{k|Q_2|}{a^2}
\]

\[
E_3=\frac{k|Q_3|}{2a^2}
\]

The third expression contains \(2a^2\) because:

\[
(\sqrt{2}a)^2=2a^2
\]

This is one of the most common sources of lost marks in square-field questions.

But calculating the three magnitudes is only half the job. We still need their directions.

02What does electric field actually represent?

An electric field describes the electrical influence of source charges at a particular location.

It tells us the force that would act on a positive test charge placed at that location, per coulomb of test charge.

Mathematically:

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

Here:

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

The important distinction is that the electric field belongs to the arrangement of source charges. It does not depend on which test charge we use to measure it.

For example, replacing a positive test charge with a negative test charge reverses the direction of the force, but does not change the electric field at that location.

This distinction becomes especially useful when a question gives you a test charge and asks for the net electric field.

For a refresher on that relationship, see How to Calculate Electric Field from Force on a Test Charge.

The direction of each contribution

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

This gives us two rules:

• A positive source charge produces an electric field pointing away from itself.
• A negative source charge produces an electric field pointing towards itself.

For our square, with \(P\) at the bottom-left corner, the possible directions are:

Do not memorise the table as a collection of unrelated directions. Imagine placing a small positive test charge at \(P\).

A positive source charge repels it. A negative source charge attracts it.

That gives you every direction in the table.

03Why Coulomb’s experiment matters: the inverse-square relationship

Why does a charge at the diagonally opposite corner produce a weaker field?

The answer comes from the relationship between electric force and distance, established through experiments by French physicist Charles-Augustin de Coulomb.

In 1785, Coulomb used a sensitive instrument called a torsion balance to investigate the forces between charged objects.

The problem Coulomb faced was that electrostatic forces between small charged objects were extremely difficult to measure directly.

He needed an apparatus sensitive enough to detect very small forces.

His arrangement worked as follows:

1. A fine fibre was suspended vertically from the top of a glass enclosure.
2. A lightweight horizontal needle was attached to the fibre.
3. A small charged pith ball was attached to one end of the needle, with a counterweight at the other end.
4. A second charged pith ball was brought near the first.
5. The electrostatic force caused the needle to rotate, twisting the suspension fibre.

The fibre resisted this rotation. By calibrating the relationship between the twisting angle and the restoring torque, Coulomb could use the rotation to determine the electrical force.

He then changed the separation between the charged balls and measured how the force changed.

What did Coulomb predict?

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

In general:

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

The experimental measurements supported this relationship for the repulsive forces between similarly charged objects.

There was a problem with the original arrangement, however. Oppositely charged balls could attract strongly enough to jump together and stick, ending the measurement.

Coulomb therefore used an independent oscillation-based method to investigate attraction. A charged plate suspended from a needle was placed near an oppositely charged hollow sphere. By measuring the needle’s oscillation period at different separations, he could investigate the attractive force without relying on two pith balls that might collide.

The revised approach provided independent support for the inverse-square relationship in attraction as well. Coulomb’s historical experiment and its revised method.

A quick numerical illustration

Suppose two charged objects experience a force of \(0.050\text{ N}\) when separated by \(0.40\text{ m}\).

What happens if their separation is reduced to \(0.20\text{ m}\), with both charges unchanged?

Using the inverse-square relationship:

\[
F_2=F_1\left(\frac{r_1}{r_2}\right)^2
\]

\[
F_2=0.050\left(\frac{0.40}{0.20}\right)^2
\]

\[
F_2=0.050(4)
\]

\[
\boxed{F_2=0.20\text{ N}}
\]

Halving the distance increases the force by a factor of four.

This is an illustrative calculation, not a measurement from Coulomb’s historical experiment.

For an isolated point charge, the electric field also follows the inverse-square relationship:

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

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 C.
• \(r\) is the distance from the source charge to the field point in m.

This is the relationship we use for the individual point-charge contributions at the corners of a square.

04Uniform electric fields versus inverse-square electric fields

There is another distinction worth getting right before attempting more complicated questions.

Not every electric field decreases with distance.

A uniform electric field has constant magnitude and direction throughout a region. A common example is the field between two large, parallel charged plates.

For parallel plates:

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

Here, \(V\) is the potential difference between the plates in volts, and \(d\) is their separation in metres.

The approximation applies in the central region, away from the edges.

For a point charge:

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

Moving further from the source decreases the field strength.

The important thing for square questions is that the square itself does not create a uniform electric field. The fields produced by individual point charges at its corners are inverse-square fields.

If a question includes both point charges and a separately applied uniform field, calculate each contribution using the appropriate relationship and then add them as vectors.

05The method: adding electric fields at the corner of a square

Now we can put the ideas together.

Use the following method whenever you are asked to calculate the net electric field at a square’s corner.

Step 1: Identify the target corner

Choose the corner where the net electric field is required.

Call this point \(P\).

The other charged corners contain the source charges responsible for the field at \(P\).

If a test charge is placed at \(P\), do not include its own field as a contribution.

Step 2: Calculate the distance from each source charge

For a square of side length \(a\):

• Adjacent source charges are a distance \(a\) from \(P\).
• The diagonally opposite source charge is a distance \(\sqrt{2}a\) from \(P\).

Use these distances in Coulomb’s field equation.

Step 3: Calculate each electric field magnitude

For each source charge:

\[
E_i=\frac{k|Q_i|}{r_i^2}
\]

Remember that the source charge is \(Q_i\), not the test charge.

Step 4: Determine the direction of each field

Draw an arrow away from each positive source charge and towards each negative source charge.

Choose positive horizontal and vertical directions. For example:

• Right is positive \(x\).
• Up is positive \(y\).

This gives you a consistent sign convention.

Step 5: Resolve diagonal fields into components

The diagonal of a square makes an angle of \(45^\circ\) with either side.

Therefore, the horizontal and vertical components have equal magnitudes.

\[
E_x=E\cos45^\circ
\]

\[
E_y=E\sin45^\circ
\]

Since:

\[
\cos45^\circ=\sin45^\circ=\frac{1}{\sqrt2}
\]

both components have magnitude:

\[
\boxed{E_x=E_y=\frac{E}{\sqrt2}}
\]

Apply the correct positive or negative signs based on the direction of the arrows.

Step 6: Add the horizontal and vertical components separately

This is the most important calculation.

\[
E_{\text{net},x}=\sum E_x
\]

\[
E_{\text{net},y}=\sum E_y
\]

Do not add the magnitudes of fields pointing in different directions.

Step 7: Calculate the resultant magnitude and direction

Once you have the two net components, use Pythagoras’ theorem:

\[
\boxed{
E_{\text{net}}=
\sqrt{E_{\text{net},x}^2+
E_{\text{net},y}^2}
}
\]

The angle relative to the horizontal is:

\[
\theta=
\tan^{-1}\left(
\frac{|E_{\text{net},y}|}
{|E_{\text{net},x}|}
\right)
\]

Use the signs of the components to identify the correct quadrant.

For example, if \(E_x\) is negative and \(E_y\) is positive, the resultant points north of west.

Always include the direction in your final answer.

08Common mistakes when calculating net electric fields

Before attempting the practice questions, check that you can avoid these mistakes.

One final check: your resultant should never be greater than the sum of the individual field magnitudes. If it is, revisit your component calculations.

09Questions and solutions

The following questions build from a straightforward vector addition problem to more demanding combinations of electric fields.

Use:

\[
k=9.0\times10^9\text{ N m}^2\text{/C}^2
\]

Unless otherwise stated, assume static point charges in air, negligible disturbance by any test charge, and no additional electric fields.

Question 1

A square has side length \(0.20\text{ m}\).

A charge of \(+3.0\,\mu\text{C}\) is placed at the bottom-right corner, and an identical charge is placed at the top-left corner. The bottom-left corner is labelled \(P\).

Calculate the magnitude and direction of the net electric field at \(P\).

Solution 1

The net electric field is \(9.55\times10^5\text{ N/C}\), directed \(45^\circ\) south of west.

Both charges produce fields pointing away from themselves.

Therefore:

• The bottom-right charge produces a leftward field.
• The top-left charge produces a downward field.

Step 1: Calculate the magnitude of each field.

Both source charges are a distance of \(0.20\text{ m}\) from \(P\).

\[
E_1=E_2=
\frac{(9.0\times10^9)(3.0\times10^{-6})}
{(0.20)^2}
\]

\[
E_1=E_2=6.75\times10^5\text{ N/C}
\]

Step 2: Add the components.

The horizontal component is:

\[
E_x=-6.75\times10^5\text{ N/C}
\]

The vertical component is:

\[
E_y=-6.75\times10^5\text{ N/C}
\]

Step 3: Calculate the resultant.

\[
E_{\text{net}}=
\sqrt{(6.75\times10^5)^2+
(6.75\times10^5)^2}
\]

\[
\boxed{
E_{\text{net}}=9.55\times10^5\text{ N/C}
}
\]

Step 4: Find the direction.

\[
\theta=\tan^{-1}\left(\frac{6.75}{6.75}\right)
\]

\[
\theta=45^\circ
\]

The equal horizontal and vertical components explain why the resultant lies exactly halfway between west and south.

The key is that the two field magnitudes are equal, but their directions are perpendicular. Adding the magnitudes directly would produce an incorrect answer.

Question 2

A square has side length \(0.25\text{ m}\). Three point charges are positioned at its corners:

• Bottom-right: \(+2.0\,\mu\text{C}\)
• Top-left: \(-3.0\,\mu\text{C}\)
• Top-right: \(+4.0\,\mu\text{C}\)

Calculate the net electric field at the bottom-left corner.

Give your answer in scientific notation, including its direction relative to west.

Solution 2

The net electric field is approximately \(5.42\times10^5\text{ N/C}\), directed \(24.9^\circ\) north of west.

The negative charge at the top-left corner is the important change from Question 1. Its field at \(P\) points upwards, not downwards.

Step 1: Calculate the field magnitudes.

For the bottom-right charge:

\[
E_1=
\frac{(9.0\times10^9)(2.0\times10^{-6})}
{(0.25)^2}
\]

\[
E_1=2.88\times10^5\text{ N/C}
\]

For the top-left charge:

\[
E_2=
\frac{(9.0\times10^9)(3.0\times10^{-6})}
{(0.25)^2}
\]

\[
E_2=4.32\times10^5\text{ N/C}
\]

For the diagonally opposite charge:

\[
E_3=
\frac{(9.0\times10^9)(4.0\times10^{-6})}
{2(0.25)^2}
\]

\[
E_3=2.88\times10^5\text{ N/C}
\]

Step 2: Determine the directions.

The positive bottom-right charge produces a leftward field.

The negative top-left charge produces an upward field.

The positive diagonal charge produces a down-left field.

Step 3: Resolve the diagonal contribution.

\[
E_{3x}=E_{3y}
=-\frac{2.88\times10^5}{\sqrt2}
\]

\[
E_{3x}=E_{3y}
=-2.04\times10^5\text{ N/C}
\]

Step 4: Calculate the net components.

Horizontal:

\[
\begin{aligned}
E_x&=-2.88\times10^5-2.04\times10^5\\
&=-4.92\times10^5\text{ N/C}
\end{aligned}
\]

Vertical:

\[
\begin{aligned}
E_y&=+4.32\times10^5-2.04\times10^5\\
&=+2.28\times10^5\text{ N/C}
\end{aligned}
\]

Step 5: Calculate the resultant magnitude.

\[
E_{\text{net}}=
\sqrt{(4.92\times10^5)^2+
(2.28\times10^5)^2}
\]

\[
\boxed{
E_{\text{net}}=5.42\times10^5\text{ N/C}
}
\]

Step 6: Calculate the angle.

\[
\theta=
\tan^{-1}\left(
\frac{2.28\times10^5}
{4.92\times10^5}
\right)
\]

\[
\boxed{\theta=24.9^\circ}
\]

The horizontal component is negative, and the vertical component is positive. Therefore, the resultant is north of west.

The tempting mistake here is to treat the negative source charge as if it were positive. That would reverse the vertical contribution and change the final direction.

Question 3

A square has side length \(0.20\text{ m}\).

Two known source charges are positioned at its corners:

• Bottom-right: \(+2.0\,\mu\text{C}\)
• Top-left: \(-3.0\,\mu\text{C}\)

An unknown positive charge \(Q\) is placed at the top-right corner.

Determine the value of \(Q\) required for the net electric field at the bottom-left corner to be exactly horizontal.

Calculate the magnitude and direction of the resulting field.

Solution 3

The required diagonal charge is \(+6\sqrt2\,\mu\text{C}\), approximately \(+8.49\,\mu\text{C}\). The resulting field is \(1.125\times10^6\text{ N/C}\) west.

The key is that a horizontal resultant must have a vertical component of exactly zero.

Step 1: Calculate the field from the bottom-right charge.

\[
E_1=
\frac{(9.0\times10^9)(2.0\times10^{-6})}
{(0.20)^2}
\]

\[
E_1=4.50\times10^5\text{ N/C}
\]

It points left.

Step 2: Calculate the field from the top-left charge.

\[
E_2=
\frac{(9.0\times10^9)(3.0\times10^{-6})}
{(0.20)^2}
\]

\[
E_2=6.75\times10^5\text{ N/C}
\]

It points upwards because the source charge is negative.

Step 3: Determine what the diagonal charge must do.

The resultant must be horizontal, so the vertical components must cancel.

The diagonal charge is positive, so its field at \(P\) points down-left.

Its downward component must equal the upward field from the top-left charge.

Therefore:

\[
\frac{E_3}{\sqrt2}=6.75\times10^5
\]

Rearranging:

\[
E_3=6.75\times10^5\sqrt2
\]

Step 4: Use the inverse-square relationship to calculate \(Q\).

The diagonal distance is:

\[
r=\sqrt2(0.20)
\]

Using:

\[
E_3=\frac{kQ}{2(0.20)^2}
\]

Substitute:

\[
6.75\times10^5\sqrt2
=
\frac{(9.0\times10^9)Q}{0.08}
\]

Rearranging:

\[
Q=
\frac{(6.75\times10^5\sqrt2)(0.08)}
{9.0\times10^9}
\]

\[
\boxed{Q=6\sqrt2\times10^{-6}\text{ C}}
\]

Therefore:

\[
\boxed{Q=+8.49\,\mu\text{C}}
\]

The positive sign is essential. A negative diagonal charge would produce an upward component, making the vertical fields reinforce rather than cancel.

Step 5: Calculate the resulting horizontal field.

The bottom-right charge produces a leftward field of:

\[
4.50\times10^5\text{ N/C}
\]

The diagonal charge produces a leftward component of:

\[
6.75\times10^5\text{ N/C}
\]

Therefore:

\[
E_x=-4.50\times10^5-6.75\times10^5
\]

\[
E_x=-1.125\times10^6\text{ N/C}
\]

The vertical component is zero.

Final answer:

\[
\boxed{
E_{\text{net}}=1.125\times10^6\text{ N/C west}
}
\]

The important idea is that we used the required direction of the resultant to establish a condition on the components before calculating the unknown charge.

Question 4

A square has side length \(0.30\text{ m}\), with the following charges:

• Bottom-right: \(+2.0\,\mu\text{C}\)
• Top-left: \(+2.0\,\mu\text{C}\)
• Top-right: \(-2\sqrt2\,\mu\text{C}\)

A uniform electric field of \(1.0\times10^5\text{ N/C}\) acts horizontally towards the east.

Part A: Calculate the net electric field at the bottom-left corner.

Part B: The square is then enlarged so that its side length doubles to \(0.60\text{ m}\). All source charges remain unchanged, and the uniform electric field remains \(1.0\times10^5\text{ N/C}\) east.

Calculate the new net electric field at the bottom-left corner.

Explain why the two contributions change differently.

Solution 4

Initially, the net field is \(1.0\times10^5\text{ N/C}\) south.

After doubling the square’s side length, the net field becomes approximately \(7.91\times10^4\text{ N/C}\), directed \(18.4^\circ\) south of east.

The key is that the point-charge fields decrease with distance, whereas the uniform field remains unchanged under the stated conditions.

Part A: Original square

Step 1: Calculate the fields from the two adjacent positive charges.

\[
E_1=E_2=
\frac{(9.0\times10^9)(2.0\times10^{-6})}
{(0.30)^2}
\]

\[
E_1=E_2=2.0\times10^5\text{ N/C}
\]

The bottom-right field points west, and the top-left field points south.

Step 2: Calculate the diagonal field.

\[
E_3=
\frac{(9.0\times10^9)(2\sqrt2\times10^{-6})}
{2(0.30)^2}
\]

\[
E_3=1.414\times10^5\text{ N/C}
\]

Since the diagonal source charge is negative, its field points towards the source, which is northeast.

Its components are:

\[
E_{3x}=E_{3y}=1.0\times10^5\text{ N/C}
\]

Step 3: Add the point-charge contributions.

Horizontal:

\[
E_{\text{charges},x}
=-2.0\times10^5+1.0\times10^5
\]

\[
E_{\text{charges},x}=-1.0\times10^5
\]

Vertical:

\[
E_{\text{charges},y}
=-2.0\times10^5+1.0\times10^5
\]

\[
E_{\text{charges},y}=-1.0\times10^5
\]

Both values are in N/C.

Step 4: Add the uniform field.

The uniform field points east:

\[
\vec E_u=(1.0\times10^5,0)
\]

Therefore:

\[
E_{\text{net},x}
=-1.0\times10^5+1.0\times10^5=0
\]

\[
E_{\text{net},y}=-1.0\times10^5
\]

The resultant is:

\[
\boxed{
E_{\text{net}}=1.0\times10^5\text{ N/C south}
}
\]

Part B: Enlarged square

Step 5: Determine how the point-charge fields change.

The side length doubles:

\[
0.30\text{ m}\rightarrow0.60\text{ m}
\]

The distance from each adjacent source charge doubles.

The diagonal distance also doubles.

Since:

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

every point-charge contribution becomes one-quarter of its original value.

The original point-charge components were:

\[
E_{\text{charges},x}
=-1.0\times10^5
\]

\[
E_{\text{charges},y}
=-1.0\times10^5
\]

The new components are:

\[
E’_{\text{charges},x}
=-2.5\times10^4
\]

\[
E’_{\text{charges},y}
=-2.5\times10^4
\]

Step 6: Add the unchanged uniform field.

The uniform field remains:

\[
\vec E_u=(1.0\times10^5,0)
\]

Therefore:

\[
E’_{\text{net},x}
=1.0\times10^5-2.5\times10^4
\]

\[
E’_{\text{net},x}=7.5\times10^4
\]

And:

\[
E’_{\text{net},y}=-2.5\times10^4
\]

Step 7: Calculate the new resultant magnitude.

\[
E’_{\text{net}}=
\sqrt{(7.5\times10^4)^2+
(2.5\times10^4)^2}
\]

\[
\boxed{
E’_{\text{net}}=7.91\times10^4\text{ N/C}
}
\]

Step 8: Find the direction.

\[
\theta=
\tan^{-1}\left(
\frac{2.5\times10^4}{7.5\times10^4}
\right)
\]

\[
\theta=18.4^\circ
\]

The horizontal component is positive, and the vertical component is negative.

Therefore, the field points south of east.

Final answer:

\[
\boxed{
E’_{\text{net}}=
7.91\times10^4\text{ N/C},
\quad18.4^\circ\text{ south of east}
}
\]

The important distinction is that enlarging the square changes the point-charge fields because their distances from \(P\) change. It does not change the separately specified uniform field.

Question 5

This final problem combines vector superposition, inverse-square fields, uniform fields, and the independence of electric field from a test charge.

A square has side length \(0.20\text{ m}\).

Three positive source charges are positioned at its corners:

• Bottom-right: \(+1.0\,\mu\text{C}\)
• Top-left: \(+1.0\,\mu\text{C}\)
• Top-right: \(+2\sqrt2\,\mu\text{C}\)

A uniform electric field also acts throughout the square, with components:

\[
\vec E_u=
(4.5\times10^5\hat i+
4.5\times10^5\hat j)\text{ N/C}
\]

A small positive test charge is placed at the bottom-left corner \(P\). It experiences zero net electric force.

A student concludes:

“The electric field is zero at \(P\), so the entire square must be field-free.”

Test this conclusion.

Part A: Verify that the net electric field at \(P\) is zero.

Part B: Calculate the net electric field at the centre of the square, labelled \(M\).

Part C: A positive test charge of \(+2.0\,\mu\text{C}\) is placed at \(M\). Calculate the electric force acting on it.

Part D: Explain what happens to the electric field and force if the positive test charge is replaced by a negative test charge of the same magnitude.

Solution 5

The student is incorrect. The field is zero at \(P\), but at the centre \(M\), the net field is \(6.36\times10^5\text{ N/C}\) southwest. A zero field at one location does not mean the entire region has zero field.

The important detail is that the point-charge fields change as we move around the square, while the uniform field stays constant.

Part A: Verify that the field at P is zero

Step 1: Calculate the fields from the two adjacent positive charges.

The side length is:

\[
a=0.20\text{ m}
\]

For either adjacent charge:

\[
E_1=E_2=
\frac{(9.0\times10^9)(1.0\times10^{-6})}
{(0.20)^2}
\]

\[
E_1=E_2=2.25\times10^5\text{ N/C}
\]

The bottom-right charge produces a leftward field.

The top-left charge produces a downward field.

Step 2: Calculate the diagonal contribution.

The diagonal source charge is:

\[
Q_3=2\sqrt2\times10^{-6}\text{ C}
\]

Its distance from \(P\) is:

\[
r=\sqrt2(0.20)
\]

Therefore:

\[
E_3=
\frac{(9.0\times10^9)(2\sqrt2\times10^{-6})}
{2(0.20)^2}
\]

\[
E_3=3.18\times10^5\text{ N/C}
\]

It points down-left.

Its horizontal and vertical components are:

\[
E_{3x}=E_{3y}
=-\frac{3.18\times10^5}{\sqrt2}
\]

\[
E_{3x}=E_{3y}=-2.25\times10^5\text{ N/C}
\]

Step 3: Add the point-charge fields.

The horizontal component is:

\[
E_{\text{charges},x}
=-2.25\times10^5-2.25\times10^5
\]

\[
E_{\text{charges},x}=-4.50\times10^5
\]

The vertical component is:

\[
E_{\text{charges},y}
=-2.25\times10^5-2.25\times10^5
\]

\[
E_{\text{charges},y}=-4.50\times10^5
\]

Both are in N/C.

Step 4: Include the uniform field.

The uniform field has components:

\[
E_{u,x}=+4.50\times10^5
\]

\[
E_{u,y}=+4.50\times10^5
\]

Therefore:

\[
E_{\text{net},x}
=-4.50\times10^5+4.50\times10^5=0
\]

\[
E_{\text{net},y}
=-4.50\times10^5+4.50\times10^5=0
\]

Hence:

\[
\boxed{\vec E_P=0}
\]

The student’s observation at \(P\) is correct. The conclusion about the entire square is not.

Part B: Calculate the electric field at the centre M

This is where the problem becomes interesting.

The centre of a square is a distance of \(a/\sqrt2\) from every corner.

The distance from each source charge to \(M\) is therefore:

\[
r_M=\frac{0.20}{\sqrt2}
\]

\[
r_M=0.141\text{ m}
\]

Step 5: Calculate the contributions from the two adjacent charges.

Each adjacent positive charge produces a field of magnitude:

\[
E_M=
\frac{(9.0\times10^9)(1.0\times10^{-6})}
{(0.20/\sqrt2)^2}
\]

\[
E_M=4.50\times10^5\text{ N/C}
\]

However, their directions are opposite.

The bottom-right charge produces a field towards the northwest at \(M\).

The top-left charge produces a field towards the southeast at \(M\).

They have equal magnitudes and opposite directions.

Therefore, they cancel:

\[
\boxed{\vec E_{1,M}+\vec E_{2,M}=0}
\]

This is a crucial observation. The two adjacent charges contributed to the field at \(P\), but their contributions cancel at the centre.

Step 6: Calculate the field from the diagonal charge at M.

The diagonal charge is now only \(0.141\text{ m}\) away.

Its field magnitude is:

\[
E_{3,M}=
\frac{(9.0\times10^9)(2\sqrt2\times10^{-6})}
{(0.20/\sqrt2)^2}
\]

\[
E_{3,M}=1.27\times10^6\text{ N/C}
\]

The charge is positive, so its field points away from it.

At the centre, this means southwest.

Its components are:

\[
E_{3,M,x}=E_{3,M,y}
=-\frac{1.27\times10^6}{\sqrt2}
\]

\[
E_{3,M,x}=E_{3,M,y}
=-9.0\times10^5\text{ N/C}
\]

Step 7: Add the uniform field.

The uniform field is unchanged at \(M\):

\[
E_{u,x}=E_{u,y}=+4.50\times10^5
\]

The net horizontal component is:

\[
E_{\text{net},x}
=-9.0\times10^5+4.50\times10^5
\]

\[
E_{\text{net},x}=-4.50\times10^5
\]

The net vertical component is:

\[
E_{\text{net},y}
=-9.0\times10^5+4.50\times10^5
\]

\[
E_{\text{net},y}=-4.50\times10^5
\]

Step 8: Calculate the resultant field at M.

\[
E_M=
\sqrt{(4.50\times10^5)^2+
(4.50\times10^5)^2}
\]

\[
\boxed{
E_M=6.36\times10^5\text{ N/C}
}
\]

Both components are negative, so the field points southwest.

\[
\boxed{
\vec E_M=
6.36\times10^5\text{ N/C southwest}
}
\]

The reason the result differs from \(P\) is that the diagonal charge is closer to the centre than to \(P\). Its field becomes stronger, while the two adjacent charges now cancel one another.

The uniform field, however, has not changed.

Part C: Calculate the force on a positive test charge at M

The positive test charge is:

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

Using:

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

The force magnitude is:

\[
F=|q|E
\]

\[
F=(2.0\times10^{-6})(6.36\times10^5)
\]

\[
\boxed{F=1.27\text{ N}}
\]

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

Therefore:

\[
\boxed{F=1.27\text{ N southwest}}
\]

Part D: Replace the positive test charge with a negative test charge

The electric field does not change.

It is still:

\[
\boxed{
E_M=6.36\times10^5\text{ N/C southwest}
}
\]

However, the electric force reverses direction because the test charge is now negative.

For a charge of \(-2.0\,\mu\text{C}\):

\[
F=|q|E=1.27\text{ N}
\]

The force points northeast.

\[
\boxed{F=1.27\text{ N northeast}}
\]

This final distinction is important: changing the test charge changes the force, not the electric field produced by the source charges.

The field at \(P\) was genuinely zero, but that cancellation was specific to that location. Moving to the centre changed the distances and directions of the individual point-charge contributions.

A field can cancel at one point without cancelling everywhere around it.

10Where to go next

Once you can calculate the net electric field at a square’s corner, the next step is to use that field to determine the force on a particular charge.

The relationship is:

\[
\boxed{\vec F=q\vec E}
\]

The calculation is straightforward once the resultant field is known. The sign of the charge determines whether the force follows or opposes the field direction.

See Force on a Charge in an Electric Field: HSC Physics Guide for the next step.