Can a Neutral Object Experience an Electric Force? Explained

A positively charged rod is brought towards an aluminium can. The can rolls towards the rod, even though the can has no net electric charge. The rod never touches it, so no electrons jump across the gap.

Why does the can move?

Yes, a neutral object can experience a net electric force. Neutrality means that an object’s total positive and negative charges balance. It does not mean that the object contains no charges, or that an electric field cannot act on it.

The key is to look at the forces on the positive and negative charges separately. If those forces do not cancel as vectors, the neutral object experiences a net force.

01Neutral does not mean charge-free

An object is electrically neutral when its total positive charge equals its total negative charge.

For example, a neutral object might contain:

• 100 positively charged protons, contributing a total charge of \(+100e\).
• 100 negatively charged electrons, contributing a total charge of \(-100e\).

Its net charge is therefore zero:

\[ Q_{\text{net}}=+100e-100e=0 \]

However, those charges still exist. They can still experience electric forces.

This is where the distinction between neutrality and polarisation becomes important.

Follow the electrons

Return to the positively charged rod and neutral aluminium can.

Suppose the rod became positively charged when electrons transferred from it to a cloth during rubbing.

The charge transfer looks like this:

Now bring the positive rod close to the can.

The rod attracts the can’s mobile electrons. These electrons shift towards the side closest to the rod.

As a result:

• The near side has an excess of electrons and becomes negatively charged.
• The far side has a deficit of electrons and becomes positively charged.

The can has not gained or lost any electrons overall. Its total charge remains zero.

This redistribution of charge is called electrical polarisation.

For a metal, mobile electrons can move through the material. In an insulator, electrons generally cannot move freely through the entire object. Instead, their electron clouds shift slightly relative to the nuclei, or polar molecules rotate.

Both processes can produce polarisation.

The important point is that polarisation separates positive and negative charge without necessarily changing the object’s total charge.

02Why does a neutral object experience a net force?

You might predict that the positive and negative charges should experience equal and opposite forces, leaving the object stationary.

That prediction would be correct if the electric field were uniform across the object.

But near a charged rod, the electric field is not uniform. It is stronger closer to the rod.

Consider a simplified model of the polarised can. Represent it as two equal and opposite charges:

• A negative charge, \(-q\), on the near side.
• A positive charge, \(+q\), on the far side.

Place a positive source charge, \(+Q\), to the left.

There are two electric forces:

Force on the negative charge

The positive source charge attracts the negative charge towards the left.

Force on the positive charge

The positive source charge repels the positive charge towards the right.

The forces act in opposite directions. However, their magnitudes are not equal because the charges are at different distances from the source.

Coulomb’s law gives the magnitude of the force between two point charges:

\[ F=k\frac{|Qq|}{r^2} \]

Here:

• \(F\) is the electric force, measured in newtons (N).
• \(k=8.99\times10^9\ \mathrm{N\,m^2\,C^{-2}}\) is Coulomb’s constant in air or vacuum.
• \(Q\) and \(q\) are the interacting charges, measured in coulombs (C).
• \(r\) is the distance between the charges, measured in metres (m).

Because the negative charge is closer to the positive source charge, it experiences a stronger attractive force.

The weaker repulsive force on the positive charge cannot balance it.

The result is a net force towards the rod.

This is the central idea:

A neutral object can experience a net electric force when the electric field varies across its charge distribution.

The mathematical explanation

The electric force on an individual charge in an electric field is:

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

The direction depends on the sign of the charge:

• A positive charge experiences a force in the direction of the electric field.
• A negative charge experiences a force opposite to the electric field.

For an object containing several charges, the net force is the vector sum:

\[ \vec F_{\text{net}} = \sum_i q_i\vec E(\vec r_i) \]

The position \(\vec r_i\) identifies where each charge is located, and \(\vec E(\vec r_i)\) is the electric field at that position.

If the electric field is uniform, it is the same at every position. We can then write:

\[ \begin{aligned} \vec F_{\text{net}} &=\vec E\sum_i q_i\\ &=Q_{\text{net}}\vec E\\ &=0 \end{aligned} \]

This follows because the object is neutral.

However, in a non-uniform electric field, the field at each charge’s position can be different. We can no longer take the electric field outside the sum.

The individual forces may therefore fail to cancel.

03The important contrast: uniform and non-uniform electric fields

It is worth understanding when a neutral object does not experience a net translational force.

Imagine placing a neutral dipole between two large, oppositely charged parallel plates. In the central region between the plates, the electric field is approximately uniform.

The positive charge experiences a force in the direction of the field.

The negative charge experiences a force in the opposite direction.

Because the field has the same magnitude at both locations, the forces have equal magnitudes:

\[ F_+=qE \]

\[ F_-=qE \]

Their vector sum is zero.

The object therefore experiences no net translational force from the uniform field.

There is one extra detail worth knowing. If the dipole is not aligned with the field, the equal and opposite forces act at different positions. They can produce a torque, causing the dipole to rotate.

So zero net force does not necessarily mean nothing happens to the object.

Compare the two situations

For an ordinary polarizable object near a single charged rod, the net force is generally attractive. However, the direction of the force in more complicated arrangements must be determined by adding the individual force vectors.

For a broader explanation of charge redistribution and induction, see Charged and Neutral Objects: Attraction, Repulsion and Induction.

05When the electric field changes direction as well as magnitude

So far, the examples have involved forces acting along one straight line.

Real electric fields can be more complicated. Their magnitude and direction can both change across an object.

In this situation, subtracting force magnitudes is not enough. You must resolve the forces into components and add them as vectors.

06Questions and solutions

The following questions test whether you can distinguish net charge from the forces acting on individual charges. They progress from basic polarisation to more demanding situations involving electron transfer, vector forces, and unstable equilibrium.

Question 1

A neutral aluminium sphere is placed on an insulating stand. A positively charged rod is brought close to its left side without touching it.

Explain:

a) Which way do the mobile electrons in the sphere move?

b) What is the induced charge on the near and far sides of the sphere?

c) Does the sphere experience a net electric force? If so, state its direction.

Solution 1

The sphere experiences a net attractive force towards the positively charged rod.

a) Electron movement

The positive rod attracts the mobile electrons in the aluminium sphere.

They redistribute towards the left side, which is closest to the rod.

b) Induced charges

The near side develops a negative charge because of its excess electrons.

The far side develops a positive charge because it has a deficit of electrons.

c) Net force

The sphere is attracted towards the rod.

Although the negative and positive regions experience forces in opposite directions, the negative region is closer to the rod. Its attractive force is therefore stronger than the repulsive force on the farther positive region.

The sphere remains neutral because no electrons have entered or left it.

Question 2

A negatively charged rod has a charge of \(-1.5\ \mu\mathrm C\).

It is placed near an isolated neutral conducting sphere. Model the sphere’s polarisation using two effective charges:

• \(+0.30\ \mu\mathrm C\), located \(0.080\ \mathrm m\) from the rod.
• \(-0.30\ \mu\mathrm C\), located \(0.110\ \mathrm m\) from the rod.

Calculate the magnitude and direction of the net electric force on the sphere.

Use:

\[ k=8.99\times10^9\ \mathrm{N\,m^2\,C^{-2}} \]

Assume a point-charge model and take the positive \(x\)-direction as away from the rod.

Solution 2

The sphere experiences a net force of approximately \(0.30\ \mathrm N\) towards the negatively charged rod.

Step 1: Identify the directions of the forces.

The negatively charged rod repels the negative induced charge on the far side of the sphere.

It attracts the positive induced charge on the near side.

Therefore:

• The force on the positive charge acts towards the left.
• The force on the negative charge acts towards the right.

Step 2: Calculate the force on the positive charge.

\[ F_+=k\frac{|Qq|}{r^2} \]

\[ F_+= \frac{ (8.99\times10^9) (1.5\times10^{-6}) (0.30\times10^{-6}) }{(0.080)^2} \]

\[ F_+=0.632\ \mathrm N \]

Since this force acts towards the rod:

\[ F_{+,x}=-0.632\ \mathrm N \]

Step 3: Calculate the force on the negative charge.

\[ F_-=k\frac{|Qq|}{r^2} \]

\[ F_-= \frac{ (8.99\times10^9) (1.5\times10^{-6}) (0.30\times10^{-6}) }{(0.110)^2} \]

\[ F_-=0.334\ \mathrm N \]

This force acts away from the rod:

\[ F_{-,x}=+0.334\ \mathrm N \]

Step 4: Calculate the resultant force.

\[ \begin{aligned} F_{\text{net},x} &=-0.632+0.334\\ &=-0.298\ \mathrm N \end{aligned} \]

Therefore:

\[ \boxed{F_{\text{net}}=0.30\ \mathrm N} \]

The force acts towards the rod.

The tempting mistake is to add the two force magnitudes. They act in opposite directions, so you must subtract them using their signs.

The sphere is still neutral. The non-zero force comes from the unequal distances between the source charge and the two induced charges.

Question 3

An ideal neutral electric dipole is placed between two large parallel plates producing a uniform electric field of magnitude:

\[ E=2.5\times10^4\ \mathrm{N/C} \]

The dipole consists of:

• A positive charge of \(+0.60\ \mu\mathrm C\).
• A negative charge of \(-0.60\ \mu\mathrm C\).

The charges are separated vertically by \(0.040\ \mathrm m\), with the negative charge above the positive charge.

The electric field points horizontally east.

Calculate:

a) The magnitude and direction of the force on each charge.

b) The net translational force on the dipole.

c) Whether the dipole can still experience any motion despite having zero net force.

Solution 3

The net translational force is zero, but the dipole experiences a torque that tends to rotate it into alignment with the electric field.

a) Force on each charge

The force magnitude on an individual charge is:

\[ F=|q|E \]

Substitute:

\[ F=(0.60\times10^{-6})(2.5\times10^4) \]

\[ F=0.015\ \mathrm N \]

The positive charge experiences a force of \(0.015\ \mathrm N\) east.

The negative charge experiences a force of \(0.015\ \mathrm N\) west because its force is opposite to the electric field.

b) Net translational force

Taking east as positive:

\[ \begin{aligned} F_{\text{net}} &=+0.015-0.015\\ &=0\ \mathrm N \end{aligned} \]

The two forces cancel.

Therefore, the dipole experiences no net translational acceleration due to the electric field.

c) Can the dipole still move?

Yes. The forces act at different positions because the charges are separated vertically.

They form a couple, which produces a torque.

The dipole therefore tends to rotate until its dipole moment, which points from the negative charge towards the positive charge, aligns with the electric field.

This is an important distinction:

• Net force determines translational acceleration.
• Net torque can produce rotational acceleration.

A neutral object in a uniform electric field can therefore experience rotational motion even when its net translational force is zero.

Question 4

A neutral conducting sphere is placed near a positively charged rod.

The following sequence occurs:

1. The sphere is initially isolated.
2. A conducting wire connects the sphere to Earth while the positive rod remains nearby.
3. The wire is disconnected while the rod is still nearby.
4. The positive rod is then removed.

Answer the following:

a) Which way do electrons move when the sphere is connected to Earth?

b) What is the net charge on the sphere after the wire is disconnected?

c) What happens to the sphere’s charge distribution when the rod is removed?

d) What would happen if the rod were removed before the Earth connection was disconnected?

Solution 4

The sphere ends up negatively charged because electrons transfer from Earth into the sphere.

a) Electron movement

The positive rod attracts electrons towards the sphere.

When the sphere is connected to Earth, electrons can flow from Earth through the conducting wire into the sphere.

The sphere gains electrons.

b) Net charge after disconnecting the wire

The sphere now has an excess of electrons.

Therefore, it has a net negative charge.

Disconnecting the wire first is essential because it prevents the excess electrons from flowing back to Earth.

c) Removing the rod

When the positive rod is removed, the external electric field responsible for the initial charge separation disappears.

The excess electrons redistribute over the conducting sphere’s surface.

The sphere remains negatively charged because it is now isolated and cannot exchange electrons with Earth.

d) What if the rod is removed first?

If the rod is removed while the sphere is still connected to Earth, the attraction that caused electrons to enter the sphere disappears.

The excess electrons can flow back into Earth.

The sphere returns to a neutral state.

This question demonstrates the difference between polarisation and actual electron transfer.

Initially, the isolated sphere was neutral and polarised. Connecting it to Earth allowed electrons to enter, changing its net charge.

Question 5

Two identical positive point charges, each with charge \(+Q\), are fixed at positions:

\[ x=-0.20\ \mathrm m \]

and

\[ x=+0.20\ \mathrm m \]

A small, neutral, polarizable bead is placed exactly halfway between them, at \(x=0\).

Assume:

• The bead has no permanent electric dipole.
• The bead is much smaller than the separation between the fixed charges.
• Its induced dipole aligns with the local electric field.
• The bead can move only along the \(x\)-axis.
• Gravity and friction are negligible.

A student argues:

“The electric field at the midpoint is zero because the two positive charges produce equal and opposite fields. Therefore, the bead experiences zero force and must be in stable equilibrium.”

Evaluate the student’s reasoning.

In particular:

a) What is the net electric field at the exact midpoint?

b) If the bead is displaced a small distance to the right, which direction does the resultant electric field point?

c) Does the bead experience a force towards the midpoint or further to the right? Explain why the direction of the electric field alone is not enough to predict the bead’s motion.

Solution 5

The student is correct that the net force is zero at the exact midpoint, but incorrect that the equilibrium is stable.

The bead is in an unstable equilibrium.

a) Electric field at the midpoint

At the midpoint, both positive charges are the same distance away.

The electric field produced by the left charge points to the right.

The electric field produced by the right charge points to the left.

Because the charges and distances are identical, the field magnitudes are equal.

Therefore:

\[ \boxed{\vec E_{\text{net}}=0} \]

By symmetry, the net force on the bead is also zero at this exact position.

b) Displace the bead slightly to the right

Now the bead is closer to the positive charge on the right.

The electric field from the right-hand charge is stronger than the field from the left-hand charge.

Both fields still point in opposite directions between the two fixed charges.

However, the stronger field comes from the right-hand charge and points to the left.

Therefore, the resultant electric field at the bead points towards the midpoint.

c) Determine the force on the neutral bead

This is the important part.

The tempting prediction is that the bead must move left because the resultant electric field points left.

That would be correct for a positive test charge.

But the bead is neutral and polarizable. It develops an induced dipole in response to the electric field.

Since the local field points left:

• The positive end of the induced dipole forms on the left.
• The negative end forms on the right.

The negative end is therefore closer to the positive charge on the right.

It experiences a stronger attractive interaction with that charge than the farther positive end experiences through repulsion.

The same unequal-force effect occurs with the left-hand source charge: its stronger interaction also favours a force towards the right.

Consequently, the net force on the bead points further to the right.

The bead therefore moves further away from the midpoint rather than returning to it.

The same reasoning applies if the bead is displaced slightly to the left: it experiences a net force further to the left.

The midpoint is consequently an unstable equilibrium.

The deeper lesson is that the electric field direction tells you the force direction on a positive charge, not necessarily the direction of motion of an entire neutral object. You must account for the object’s internal charge distribution and the variation of the field across it.

07What to remember

When deciding whether a neutral object experiences an electric force, keep these distinctions clear:

A neutral object does not need to gain or lose electrons to experience an electric force.

The most reliable method is to identify the charges, determine the electric field at each charge’s position, calculate the individual forces using \(\vec F=q\vec E\), and add those forces as vectors.

That method is also the starting point for more complex electric-field problems. The next step is understanding how to determine the magnitude and direction of the electric field at different positions, which is covered in Force on a Charge in an Electric Field: HSC Physics Guide.