Why the Zero-Field Point Lies Outside Opposite Charges

Place a \(+2.0\,\mu\mathrm{C}\) charge on the left and a \(-8.0\,\mu\mathrm{C}\) charge 0.60 m to its right. Where would you look for a point where the electric field is zero?

Between the charges seems reasonable. One is positive, the other is negative, so perhaps their fields cancel somewhere in the gap.

They don’t. Between opposite charges, both electric fields point from the positive charge towards the negative charge. They reinforce each other.

For two unequal, opposite point charges, the zero-field point (also called the neutral point or electric-field null) lies outside the pair, on the side of the charge with the smaller magnitude.

The reason is a combination of direction and distance. Outside the charges, the two fields oppose each other. A point closer to the weaker charge can balance the stronger charge’s field because electric field strength decreases with the square of the distance.

01A field probe can give you the wrong idea about where cancellation occurs

Imagine trying to locate the neutral point experimentally.

You fix two small charged spheres on insulating supports, with their centres 0.60 m apart. The left sphere carries \(+2.0\,\mu\mathrm{C}\), and the right sphere carries \(-8.0\,\mu\mathrm{C}\).

A very small positive test charge is attached to a movable probe. The probe travels along a horizontal ruler through the centres of the two spheres. At each position, an idealised force sensor measures the horizontal electric force on the test charge.

The arrangement is read from left to right:

• The \(+2.0\,\mu\mathrm{C}\) source is fixed at \(x=0\).
• The \(-8.0\,\mu\mathrm{C}\) source is fixed at \(x=0.60\,\mathrm{m}\).
• The positive test charge moves to different positions along the \(x\)-axis.
• Positive \(x\) means rightwards; negative \(x\) means leftwards.

What does the sensor actually measure?

The electric field is a property of the source charges and their arrangement. It exists whether or not you place another charge there.

An ideal test charge allows us to determine that field from the force it would experience:

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

Here:

• \(\vec E\) is the electric field, measured in newtons per coulomb (\(\mathrm{N\,C^{-1}}\)).
• \(\vec F\) is the electric force on the test charge, measured in newtons (N).
• \(q_t\) is the test charge, measured in coulombs (C).

The arrow over a symbol means it is a vector, with both magnitude and direction.

Strictly, the test charge should be small enough that it does not noticeably alter the source-charge arrangement. The field is defined using the limiting behaviour as the positive test charge approaches zero magnitude.

A positive test charge experiences force in the field’s direction. A negative test charge experiences force in the opposite direction.

This gives an important distinction: a neutral object can experience zero electric force without the electric field being zero. A zero-field point requires the electric-field contributions themselves to cancel.

Why testing only the gap would fail

Suppose the first measurement plan checks several positions between the charged spheres.

What would the sensor show?

At every position in the gap, a positive test charge is repelled towards the right by the positive source and attracted towards the right by the negative source.

The sensor therefore predicts a rightwards force everywhere in that interval.

A student might conclude that no zero-field point exists. But the measurement plan has only checked the wrong region.

The revised plan must scan outside the charges as well. On the left of the positive source, the two force contributions point in opposite directions, so cancellation becomes possible.

These are predictions of the ideal point-charge model, not reported experimental measurements. We’ll calculate the expected readings before considering what a real apparatus might change.

02Work out the directions before using an equation

For any point charge, the electric field points:

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

For two charges, apply those rules separately at the same position.

You are not comparing the signs of the charges directly. You are comparing the directions of the electric fields they produce at your chosen point.

Three regions, three different possibilities

Keep the \(+2.0\,\mu\mathrm{C}\) charge at \(x=0\) and the \(-8.0\,\mu\mathrm{C}\) charge at \(x=0.60\,\mathrm{m}\).

The last row needs some explanation. The fields do point in opposite directions on the right. Why can’t they cancel there?

Every point to the right of both sources is closer to the \(-8.0\,\mu\mathrm{C}\) charge than to the \(+2.0\,\mu\mathrm{C}\) charge.

The negative source is already four times stronger at equal distance. Being closer only increases its advantage.

On the left, however, the smaller positive charge is closer. Its distance advantage can compensate for its smaller magnitude.

This is the most reliable first move in a neutral-point question: eliminate regions where the vector directions make cancellation impossible.

For a refresher on reading the arrows, see Electric Field Lines Around a Point Charge.

Can the zero-field point be above or below the two charges?

Not for two non-zero point charges acting alone.

Two vectors can cancel completely only when they have equal magnitudes and exactly opposite directions.

At a point away from the line joining two charges, their electric-field vectors are not directly opposite. Even if one pair of components cancels, another component remains.

For two opposite point charges, any finite zero-field point must therefore lie on the line joining their centres. For unequal magnitudes, it lies outside the pair.

03How the weaker charge can produce an equally strong field

The field magnitude produced by an isolated point charge is

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

where:

• \(E\) is electric field strength in \(\mathrm{N\,C^{-1}}\).
• \(k=8.99\times10^9\,\mathrm{N\,m^2\,C^{-2}}\) is Coulomb’s constant, approximately valid for charges in air.
• \(Q\) is the source charge in coulombs (C).
• \(r\) is the distance from the source charge in metres (m).
• \(|Q|\) means the magnitude of the charge, ignoring its sign.

The equation tells us how strong one contribution is. The sign of the source charge determines its direction separately.

Doubling the distance reduces that source’s field to one-quarter of its original strength.

That gives the weaker charge a way to compete: the neutral point must be sufficiently close to it and sufficiently far from the stronger charge.

The distance relationship

Let two opposite point charges have magnitudes \(a\) and \(b\), where \(b>a\), and let their separation be \(d\).

Measure a distance \(s\) outwards from the smaller charge, away from the larger charge.

At that exterior point:

• Distance from the smaller charge is \(s\).
• Distance from the larger charge is \(s+d\).

Because the field directions oppose, cancellation requires equal magnitudes:

\[ \frac{ka}{s^2} = \frac{kb}{(s+d)^2} \]

Cancel \(k\) and take positive square roots, since distances and charge magnitudes are positive:

\[ \frac{s+d}{s}=\sqrt{\frac{b}{a}} \]

Rearranging gives

\[ s=\frac{d}{\sqrt{b/a}-1} \]

This formula applies to the exterior neutral point for two isolated, unequal, opposite point charges. It is a shortcut after you have identified the correct region, not a replacement for checking directions.

Notice that the denominator is positive because \(b>a\). The position is measured outwards from the smaller-magnitude charge.

What would the revised field scan show?

The following values are illustrative predictions calculated from the point-charge model. They are not experimental observations.

Take rightwards field as positive and leftwards field as negative.

The sign change on either side of \(x=-0.60\,\mathrm{m}\) helps identify the null point.

Notice the large positive field inside the gap. The midpoint doesn’t have a weak field just because it lies between opposite charges.

A real force-probe experiment would need more care than this ideal calculation.

If the test charge is too large, it can disturb the source charges. The charged spheres also have finite size, and nearby conductors or supports may alter the field.

The calculation assumes stationary point charges, electrostatic conditions, negligible probe disturbance, and no other significant electric fields. Those assumptions are what make the predicted cancellation exact.

04Changing the charge arrangement changes the location

There’s a useful contrast that exposes why memorising “the zero point is outside” isn’t enough.

Suppose the charges have the same sign.

Between two positive charges, the field from the left charge points right, while the field from the right charge points left. The fields now oppose each other inside the gap.

For two unequal positive charges, the zero-field point lies between them, closer to the smaller charge.

So the correct location depends on field directions, not merely the relative magnitudes.

05Three limits of the outside-zero-point rule

Before applying the result to a new problem, separate three situations.

Equal and opposite point charges

For \(+Q\) and \(-Q\) with equal magnitudes, there is no finite zero-field point anywhere in space.

Between them, the fields reinforce. Outside them, the nearer charge always produces the stronger field because the magnitudes are equal.

The net field approaches zero as you move infinitely far away, but approaching zero is not the same as having a finite position where it is exactly zero.

You can also see this from the exterior-distance formula. If \(b=a\), then \(\sqrt{b/a}-1=0\), so there is no finite solution.

A uniform field is not an inverse-square field

Two point charges each produce fields whose strengths depend on distance.

Between sufficiently large, closely spaced parallel plates, the field is approximately uniform away from the edges. Moving through the central region does not produce the same inverse-square variation.

If an external uniform field is added to two point-charge fields, the outside-zero-point rule no longer applies without modification. The new field must be included in the vector sum.

For help deciding which model an arrangement requires, see Point Charges vs Parallel Plates: Choosing the Right Field Model.

Zero field does not automatically mean stable equilibrium

At the neutral point, a small test charge experiences zero instantaneous electric force.

That does not guarantee that it will return to the point if displaced. Nor does zero force mean a moving charge must stop there.

Stability requires a separate investigation of the force after a small displacement. The final questions explore why this can be much more subtle than locating the null itself.

06Questions and solutions

Assume ideal point charges in air, treated as vacuum, with no additional fields unless stated. Use \(k=9.00\times10^9\,\mathrm{N\,m^2\,C^{-2}}\). Take rightwards as positive \(x\), and assume the source charges remain fixed unless the question explicitly releases them.

Question 1

Two stationary point charges lie on a horizontal line. A \(+3.0\,\mu\mathrm{C}\) charge is located at \(x=0\), and a \(-12.0\,\mu\mathrm{C}\) charge is located at \(x=0.50\,\mathrm{m}\).

Find the position on the \(x\)-axis where the resultant electric field is zero. Explain why the point cannot be between the charges.

Solution 1

The zero-field point is at \(x=-0.50\,\mathrm{m}\), outside the pair on the side of the smaller positive charge.

Between the sources, the positive charge produces a rightwards field and the negative charge also produces a rightwards field. They cannot cancel.

Let \(s\) be the distance left of the positive charge. For equal field magnitudes,

\[ \frac{s+0.50}{s} = \sqrt{\frac{12.0}{3.0}} \]

Therefore,

\[ \begin{aligned} s+0.50&=2s\\ s&=0.50\,\mathrm{m} \end{aligned} \]

The point is 0.50 m from the positive charge and 1.00 m from the negative charge.

The magnitudes confirm the result:

\[ \begin{aligned} E_+&=\frac{k(3.0\times10^{-6})}{0.50^2}\\ &=1.08\times10^5\,\mathrm{N\,C^{-1}}\\ E_-&=\frac{k(12.0\times10^{-6})}{1.00^2}\\ &=1.08\times10^5\,\mathrm{N\,C^{-1}} \end{aligned} \]

The positive-source field points left, while the negative-source field points right. They cancel.

The common mistake is putting the point between opposite charges because their signs differ. Direction must be established before equating magnitudes.

Question 2

Two equal-magnitude, opposite point charges are fixed in the \(xy\)-plane:

• \(+4.0\,\mu\mathrm{C}\) at \((-0.30\,\mathrm{m},0)\).
• \(-4.0\,\mu\mathrm{C}\) at \((+0.30\,\mathrm{m},0)\).

A field sensor is positioned at \(P=(0,+0.40\,\mathrm{m})\), directly above the midpoint.

A student argues that the field at \(P\) must be zero because the point is equally distant from equal and opposite charges.

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

(b) Explain why the student’s reasoning fails.

(c) Determine whether these two charges have any finite zero-field point elsewhere in space.

Solution 2

The field at \(P\) is \(1.73\times10^5\,\mathrm{N\,C^{-1}}\) to the right, and the pair has no finite zero-field point anywhere.

(a) Calculate the field at \(P\).

The distance from either source to \(P\) is

\[ \begin{aligned} r&=\sqrt{0.30^2+0.40^2}\\ &=0.50\,\mathrm{m} \end{aligned} \]

Each source produces a field of magnitude

\[ \begin{aligned} E&=\frac{k|Q|}{r^2}\\ &=\frac{(9.00\times10^9)(4.0\times10^{-6})}{0.50^2}\\ &=1.44\times10^5\,\mathrm{N\,C^{-1}} \end{aligned} \]

The positive source produces a field directed upwards and rightwards at \(P\).

The negative source produces a field directed downwards and rightwards, towards the negative charge.

The vertical components are equal and opposite, so they cancel.

Each horizontal component is

\[ \begin{aligned} E_x&=E\left(\frac{0.30}{0.50}\right)\\ &=(1.44\times10^5)(0.60)\\ &=8.64\times10^4\,\mathrm{N\,C^{-1}} \end{aligned} \]

Both horizontal components point right, giving

\[ \begin{aligned} E_{\mathrm{net}} &=2(8.64\times10^4)\\ &=1.728\times10^5\,\mathrm{N\,C^{-1}} \end{aligned} \]

Hence the resultant is \(1.73\times10^5\,\mathrm{N\,C^{-1}}\) rightwards, to three significant figures.

(b) Identify the error.

Equal distances and equal charge magnitudes give equal electric-field magnitudes. They do not guarantee opposite vector directions.

At \(P\), the vertical components cancel, but the horizontal components reinforce.

The student has confused cancellation of one component with cancellation of the whole vector.

(c) Check for a neutral point elsewhere.

There is no finite zero-field point.

Off the line joining the charges, the two field vectors are not exactly opposite, so they cannot completely cancel.

Between the charges, both fields point right.

To the left of both charges, the positive source is closer and its leftwards contribution dominates. To the right of both charges, the negative source is closer and its leftwards contribution dominates.

At very large distances, the field becomes weak, but it does not become exactly zero at a finite position.

This shows why symmetry can produce cancellation of selected components without producing a zero resultant field.

Question 3

A \(+4.0\,\mu\mathrm{C}\) charge is fixed at \(x=0\), and a \(-9.0\,\mu\mathrm{C}\) charge is fixed at \(x=0.50\,\mathrm{m}\).

A test bead carrying \(-2.0\,\mathrm{nC}\) can move on a frictionless insulating rail along the \(x\)-axis. Its charge is small enough not to disturb the source charges.

(a) Find the zero-field point.

(b) The negative test bead is placed at \(x=-1.10\,\mathrm{m}\). Calculate the electric force acting on it, including its direction. State the force direction if it is instead placed at \(x=-0.90\,\mathrm{m}\).

(c) Is the neutral point a stable equilibrium for this negative bead when it is constrained to the rail? Would the answer change for a positive test bead of the same magnitude?

(d) If the positive bead is allowed to leave the rail, does a small displacement perpendicular to the \(x\)-axis necessarily produce a restoring force?

Solution 3

The neutral point is at \(x=-1.00\,\mathrm{m}\). It is unstable for the negative bead on the rail, stable against small horizontal displacements for a positive bead on the rail, but not stable in every direction when that positive bead is free to leave the rail.

(a) Locate the neutral point.

The smaller-magnitude source is positive, so the neutral point lies to its left.

Let \(s\) be its distance from \(x=0\).

\[ \begin{aligned} \frac{s+0.50}{s} &=\sqrt{\frac{9.0}{4.0}}\\ &=1.50\\ s+0.50&=1.50s\\ s&=1.00\,\mathrm{m} \end{aligned} \]

Therefore \(x=-1.00\,\mathrm{m}\).

(b) Calculate the force and determine its direction.

At \(x=-1.10\,\mathrm{m}\), the distances to the positive and negative sources are 1.10 m and 1.60 m.

The positive source produces a leftwards field. The negative source produces a rightwards field.

Using signed components,

\[ \begin{aligned} E_x&=\frac{k(9.0\times10^{-6})}{1.60^2}\\ &\quad-\frac{k(4.0\times10^{-6})}{1.10^2}\\ &=+1.89\times10^3\,\mathrm{N\,C^{-1}} \end{aligned} \]

The net field points right.

However, the test bead is negative, so its force points left:

\[ \begin{aligned} F_x&=q_tE_x\\ &=(-2.0\times10^{-9})(1.89\times10^3)\\ &=-3.78\times10^{-6}\,\mathrm{N} \end{aligned} \]

The force is \(3.78\times10^{-6}\,\mathrm{N}\) leftwards.

At \(x=-0.90\,\mathrm{m}\), the positive source is 0.90 m away and the negative source is 1.40 m away.

Comparing field magnitudes, with charge magnitudes in microcoulombs and distances in metres,

\[ \frac{4.0}{0.90^2} > \frac{9.0}{1.40^2} \]

The leftwards field from the positive source dominates. Therefore the negative test bead experiences a rightwards force.

(c) Assess stability along the rail.

The negative bead is pushed left when displaced left of equilibrium, and pushed right when displaced right of equilibrium.

Both forces act away from the neutral point.

Its equilibrium is unstable.

For a positive test bead, the force directions reverse. A small leftward displacement produces a rightwards restoring force, while a small rightward displacement produces a leftwards restoring force.

The positive bead therefore has stable equilibrium for small movements constrained to the \(x\)-axis.

(d) Consider a perpendicular displacement.

The positive bead is not stable against every possible displacement.

Suppose it starts at the neutral point and is moved a tiny distance upwards.

Both sources then produce vertical field components. The positive source produces an upward component, while the negative source produces a downward component.

At the original neutral point, the field magnitudes were equal. But the positive source is closer: 1.00 m compared with 1.50 m.

For a sufficiently small upward displacement, the vertical contributions are proportional to the charge magnitudes divided by the cubes of those original distances.

Compare:

\[ \frac{4.0}{1.00^3} > \frac{9.0}{1.50^3} \]

The upward component from the positive source dominates.

A positive bead is therefore pushed further upwards, away from its original position.

The neutral point is stable along the rail for a positive test bead, but unstable to small transverse displacements when the bead is free to move in space.

A final distinction: if either test bead crosses the neutral point with non-zero speed, it does not automatically stop there. Zero field means zero instantaneous electric acceleration, not zero velocity.

Question 4

Two stationary point charges are separated by 0.40 m:

• \(+1.0\,\mu\mathrm{C}\) at \(x=0\).
• \(-1.0\,\mu\mathrm{C}\) at \(x=0.40\,\mathrm{m}\).

An independently controlled, ideal uniform electric field can be added. When switched on, this background field points to the left and has constant magnitude everywhere on the \(x\)-axis.

Consider three separate conditions:

1. No background field.
2. A background field of \(4.50\times10^5\,\mathrm{N\,C^{-1}}\) leftwards.
3. A background field of \(1.00\times10^6\,\mathrm{N\,C^{-1}}\) leftwards.

For each condition, determine the number and positions of zero-field points on the entire \(x\)-axis.

Explain why adding a uniform field can produce a result that would be impossible for the original pair of opposite charges acting alone.

Solution 4

There are no zero-field points without the background field, one at \(x=0.20\,\mathrm{m}\) when the background strength is \(4.50\times10^5\,\mathrm{N\,C^{-1}}\), and two at \(x=0.10\,\mathrm{m}\) and \(x=0.30\,\mathrm{m}\) when it is \(1.00\times10^6\,\mathrm{N\,C^{-1}}\).

The key is that the added uniform field opposes the point-charge fields inside the gap.

Condition 1: No background field

The opposite sources have equal magnitudes.

Between them, both electric fields point right, so they add.

Outside them, the field from the nearer source always dominates.

There is no finite zero-field point on the \(x\)-axis.

Condition 2: Background field of \(4.50\times10^5\,\mathrm{N\,C^{-1}}\) leftwards

First calculate the field produced by the charges at their midpoint.

Each source is 0.20 m away and produces a rightwards field.

\[ \begin{aligned} E_{\mathrm{pair}} &=2\frac{k(1.0\times10^{-6})}{0.20^2}\\ &=4.50\times10^5\,\mathrm{N\,C^{-1}} \end{aligned} \]

The uniform background field has exactly this magnitude in the opposite direction.

Therefore the net field at \(x=0.20\,\mathrm{m}\) is zero.

Why is that the only zero?

Inside the gap, the combined point-charge field is smallest at the midpoint. Moving towards either charge increases the combined field because of the inverse-square dependence.

At every other interior position, the rightwards field from the charges exceeds \(4.50\times10^5\,\mathrm{N\,C^{-1}}\).

Outside the pair, the field from the two charges already points left, so the added leftwards field reinforces it.

Thus there is exactly one zero-field point, at \(x=0.20\,\mathrm{m}\).

Condition 3: Background field of \(1.00\times10^6\,\mathrm{N\,C^{-1}}\) leftwards

At the midpoint, the rightwards point-charge field is only \(4.50\times10^5\,\mathrm{N\,C^{-1}}\).

The background now dominates there, so the net field points left.

Closer to either charge, however, the rightwards point-charge field becomes very large.

This suggests that cancellation can occur on both sides of the midpoint.

Test \(x=0.10\,\mathrm{m}\).

The distances to the positive and negative charges are 0.10 m and 0.30 m.

\[ \begin{aligned} E_{\mathrm{pair}} &=\frac{k(1.0\times10^{-6})}{0.10^2}\\ &\quad+\frac{k(1.0\times10^{-6})}{0.30^2}\\ &=9.00\times10^5+1.00\times10^5\\ &=1.00\times10^6\,\mathrm{N\,C^{-1}} \end{aligned} \]

The rightwards field from the point charges exactly cancels the leftwards background field.

Therefore \(x=0.10\,\mathrm{m}\) is a zero-field point.

By symmetry, \(x=0.30\,\mathrm{m}\) is another.

These are the only two. The point-charge field decreases from an arbitrarily large value near the positive source to its unique minimum at the midpoint, then rises symmetrically towards the negative source.

Consequently, the chosen background-field magnitude is matched exactly once on each half of the gap.

The surprising result is not a contradiction of the outside-zero-point rule.

That rule assumes only two isolated point charges contribute to the field.

Once an independent uniform field is added, the total field becomes

\[ \vec E_{\mathrm{total}} = \vec E_+ + \vec E_- + \vec E_{\mathrm{uniform}} \]

Cancellation inside the gap becomes possible because the third field opposes the other two.

The same pair of opposite charges can have zero, one, or two zero-field points on the axis under these three different conditions.

Question 5

Three charged beads can move freely along a horizontal insulating track. Treat them as ideal point charges, ignore friction, and neglect every force except their mutual electrostatic forces.

Initially, two beads are held fixed:

• Bead A carries \(+2.0\,\mu\mathrm{C}\) at \(x=0\).
• Bead B carries \(-18.0\,\mu\mathrm{C}\) at \(x=0.40\,\mathrm{m}\).

A third bead C, carrying an unknown charge \(q_C\), is placed at the zero-field point produced by A and B.

The three beads are then to be released simultaneously from rest.

(a) Find the position of C.

(b) Is there a non-zero value of \(q_C\) that makes all three beads experience zero resultant electric force at the instant of release? If so, determine its sign and magnitude.

(c) Verify that your choice balances the forces on all three beads, not just C.

(d) If all three start at those exact positions with zero velocity, describe their motion in the ideal model.

(e) Now suppose A is displaced a tiny distance to the right while B and C remain at their original positions at that instant. Does the resulting force on A tend to restore the original arrangement?

A student argues: “Because C is at the zero-field point, it feels no force. Therefore adding C cannot affect the force on either original bead.”

Explain precisely why that argument fails.

Solution 5

Yes. All three beads can be made force-free at the same instant by placing bead C at \(x=-0.20\,\mathrm{m}\) with charge \(-4.5\,\mu\mathrm{C}\).

The exact configuration remains at rest if released from rest in the ideal model, but its equilibrium is unstable.

The important insight is that a charge experiencing zero net force can still exert non-zero forces on other charges.

(a) Find the neutral point of A and B.

The smaller-magnitude source is A, so the zero-field point lies to its left.

Let \(s\) be the distance left of A.

\[ \begin{aligned} \frac{s+0.40}{s} &=\sqrt{\frac{18.0}{2.0}}\\ &=3.0\\ s+0.40&=3s\\ s&=0.20\,\mathrm{m} \end{aligned} \]

Therefore bead C must be positioned at \(x=-0.20\,\mathrm{m}\).

The distances are:

(b) Determine the charge on C.

Bead A is positive and bead B is negative.

B attracts A to the right.

To balance this, C must attract A towards the left. Since C lies to the left of A, C must be negative.

Now calculate the force on A due to B:

\[ \begin{aligned} F_{AB} &=\frac{k|q_Aq_B|}{r_{AB}^2}\\ &=\frac{(9.00\times10^9)(36.0\times10^{-12})}{0.40^2}\\ &=2.025\,\mathrm{N} \end{aligned} \]

For A to experience zero net force, C must exert a leftwards force of 2.025 N.

Using Coulomb’s law again,

\[ \frac{k|q_Aq_C|}{0.20^2} = 2.025\,\mathrm{N} \]

Rearrange:

\[ \begin{aligned} |q_C| &=\frac{(2.025)(0.20)^2}{k|q_A|}\\ &=\frac{0.0810}{(9.00\times10^9)(2.0\times10^{-6})}\\ &=4.50\times10^{-6}\,\mathrm{C} \end{aligned} \]

Since C must be negative,

\[ q_C=-4.5\,\mu\mathrm{C} \]

This value is fixed by the requirement that A experience zero net force.

(c) Verify all three force balances.

For bead A:

• B attracts A rightwards with a force of 2.025 N.
• C attracts A leftwards with a force of 2.025 N.

The net force on A is zero.

For bead B:

• A attracts B leftwards with a force of 2.025 N.
• C repels B rightwards because both C and B are negative.

Calculate the second force:

\[ \begin{aligned} F_{BC} &=\frac{k|q_Bq_C|}{r_{BC}^2}\\ &=\frac{(9.00\times10^9)(81.0\times10^{-12})}{0.60^2}\\ &=2.025\,\mathrm{N} \end{aligned} \]

The forces on B also cancel.

For bead C:

• A attracts C to the right.
• B repels C to the left.

The distance from C to A is 0.20 m, while the distance from C to B is 0.60 m.

\[ \begin{aligned} F_{CA} &=\frac{k(9.0\times10^{-12})}{0.20^2}\\ &=2.025\,\mathrm{N}\\ F_{CB} &=\frac{k(81.0\times10^{-12})}{0.60^2}\\ &=2.025\,\mathrm{N} \end{aligned} \]

These forces are opposite, so C also experiences zero net force.

All three beads have zero resultant electric force.

(d) Determine their motion when released exactly from rest.

All three remain at rest in the ideal model.

Each has zero initial velocity and zero resultant force. With no disturbance, their positions remain unchanged, so the force balance continues.

This is an equilibrium configuration.

But it is not necessarily stable.

(e) Test a small displacement of A.

Suppose A moves slightly to the right while B and C remain temporarily at their original positions.

A is now closer to B and farther from C.

The rightwards attraction from B becomes stronger, because the A-B distance has decreased.

The leftwards attraction from C becomes weaker, because the A-C distance has increased.

Therefore A experiences a net force to the right.

That force pushes A further from its equilibrium position rather than restoring it.

The configuration is unstable.

Even though the exact arrangement can remain motionless, a small disturbance can make the imbalance grow.

Why the student’s argument fails

Bead C experiences zero net force because two non-zero forces acting on it cancel.

That does not mean C exerts zero force on A or B.

Electrostatic forces occur in interacting pairs. C attracts A and repels B, even while the forces acting on C itself add to zero.

There is also a field-definition subtlety. The field produced by A and B at C’s position is zero. Once the finite charge C is introduced, it creates its own field elsewhere. We should not describe the total field at C’s own position, including its ideal point-charge self-field, as zero.

The distinction is between the field acting on a charge from the other sources and the additional field produced by that charge.

This is why the original neutral point can become part of a three-charge equilibrium arrangement without making the three charges electrically irrelevant to one another.

07What to decide in the next electric-field problem

When two opposite point charges are involved, start with directions. Between the charges, both field contributions point from positive to negative, so cancellation is impossible. Outside, the fields oppose, and the inverse-square relationship allows a finite neutral point on the smaller-magnitude charge’s side, provided the magnitudes are unequal.

Then check what the question is actually asking. Finding where the electric field is zero is different from determining whether a charged particle will stop, whether its equilibrium is stable, or whether an external field changes the result.

A further distinction becomes important when you study electric potential: zero electric potential and zero electric field are not the same condition. Electric potential is a scalar, while electric field is a vector. Keeping those ideas separate prevents the same cancellation misconception from appearing in a different form.