Why Rubbing Transfers Electrons Rather Than Protons

Rub a dry polythene rod with a piece of wool. After separating them, the rod can become negatively charged while the wool becomes positively charged. But why does rubbing transfer electrons rather than protons?

The short answer is that electrons can be redistributed between the outer regions of atoms, while protons are tightly bound inside atomic nuclei. Ordinary rubbing can change how many electrons an object has, but it cannot pull protons out of nuclei.

In the standard HSC Physics model, electrons move from the wool to the polythene rod. The rod gains negative charge, and the wool becomes positive because it has lost negative charge. Neither object creates charge.

01What actually happens when you rub two materials together?

Imagine that both the polythene rod and the wool are initially electrically neutral.

That means each contains equal amounts of positive and negative charge. Neither has an overall charge.

Now rub the wool against the rod and pull them apart.

Before looking at the explanation, predict what happens if electrons move from the wool onto the rod:

• The rod gains negatively charged particles, so it becomes negative.
• The wool loses negatively charged particles, so it becomes positive.

Notice something important: the wool does not become positive because it gains protons. It becomes positive because it loses electrons.

This process is called contact electrification, commonly known as charging by friction or the triboelectric effect.

Why does rubbing help?

The surfaces of solid materials look smooth, but at a microscopic level they contain irregularities. When two materials are pressed together, only some small regions actually make contact.

At these contact points, the electronic structures of the two materials interact. Depending on the materials and their surface conditions, electrons can transfer from one surface to the other.

When the materials separate, some of this charge remains on each surface.

Rubbing helps because it creates many repeated contacts and separations. It can also expose fresh surface regions.

However, friction is not simply a process of mechanically knocking electrons loose. The materials’ surface properties determine which direction electron transfer favours.

For a dry polythene rod rubbed with wool, we use the standard model in which electrons transfer from wool to polythene.

The direction is not universal. Different material pairs can produce different results, and moisture, contamination, and surface condition can affect the amount and direction of charging.

02Why electrons can move but protons cannot

To understand the difference, look inside an atom.

An atom contains:

• Protons, which have positive charge.
• Neutrons, which have no electric charge.
• Electrons, which have negative charge.

Protons and neutrons are concentrated in a tiny central region called the nucleus. Electrons occupy the surrounding electron structure.

The crucial difference is where these particles are located and how strongly they are bound.

Electrons are accessible at material surfaces

Electrons occupy the outer regions of atoms and are involved in chemical bonding and interactions between materials.

Their energies depend on the material’s electronic structure. When two different surfaces make contact, some electrons can move between them.

These electrons are still associated with the material’s atomic and electronic structure. They are not all freely floating around. However, transferring an outer electron between surfaces is possible without disrupting an atomic nucleus.

Protons are locked inside atomic nuclei

Protons are held inside the nucleus by the strong nuclear force.

This force is enormously stronger than the ordinary electromagnetic interactions involved in surface charging.

Removing a proton from a nucleus would require a nuclear process, not simply rubbing two everyday materials together.

It would also change the identity of the nucleus. For example, removing a proton from a carbon nucleus would turn it into a different element.

The mechanical energy involved in rubbing a plastic rod and a piece of wool is nowhere near what is required to disrupt atomic nuclei.

An electron is also much lighter than a proton, but mass is not the main reason electrons transfer. Their location and binding within the atom are what matter.

But can protons ever be transferred?

Yes. Proton transfer occurs in acid-base chemistry.

For example, an acid can donate a hydrogen ion, \(\mathrm{H^+}\), to another chemical species. This involves changes in chemical bonding, not the mechanical removal of protons from heavier atomic nuclei.

That is a different process from electrostatic charging by rubbing.

For the chemical explanation, see Acid-Base Reactions: Proton Transfer and Neutralisation.

03Follow the electrons: where does the charge go?

Let’s track the charge rather than just memorising which material becomes positive or negative.

An electron has a charge of approximately

\[
-e=-1.60\times10^{-19}\ \mathrm C
\]

where \(e\) is the magnitude of the elementary charge.

If an object gains electrons, its charge becomes negative. If it loses electrons, its charge becomes positive.

For an object that gains or loses \(N\) electrons:

\[
q=\pm Ne
\]

Here:

• \(q\) is the object’s net charge, measured in coulombs (C).
• \(N\) is the number of electrons transferred.
• \(e=1.60\times10^{-19}\ \mathrm C\).
• Use a negative sign when electrons are gained.
• Use a positive sign when electrons are lost.

For an initially neutral pair of insulated objects, the charge transferred from one object must appear as an equal and opposite charge on the other.

Therefore,

\[
q_{\text{rod}}+q_{\text{wool}}=0
\]

This is conservation of electric charge.

The rubbing process does not create extra electrons or protons. It redistributes charge that was already present.

04How can we experimentally test that electrons have transferred?

We cannot watch individual electrons moving between two rubbing surfaces with an ordinary classroom apparatus.

Instead, we can measure the charge left on each object and test whether the results agree with the electron-transfer model.

Consider an investigation using a polythene rod and wool.

The first attempt: measure only the rod

Suppose a student holds a polythene rod in one hand, rubs it with wool, and then tests the rod using an electroscope.

The electroscope indicates that the rod has become negatively charged.

This is useful evidence that the rod has acquired excess negative charge. However, it does not establish where that charge came from.

There are two problems:

1. The student’s hand can provide a conducting path through which charge leaks to or from the Earth.
2. Only the rod has been measured. We do not know whether the wool has acquired an equal positive charge.

The experiment needs to be improved.

The revised experiment: measure both objects

A better investigation uses two insulated objects and two Faraday pails connected to calibrated electrometers.

A Faraday pail is a conducting container that allows the net charge of an object placed inside it to be measured without the object touching the container.

The apparatus is arranged as follows:

1. Place the polythene rod and wool on separate insulating supports.
2. Zero the electrometers and check that both samples initially have approximately zero net charge.
3. Rub the wool against the polythene rod while keeping both objects insulated from the surroundings.
4. Separate the materials.
5. Lower the polythene rod into the first Faraday pail without touching its sides.
6. Record the sign and magnitude of the charge.
7. Repeat the measurement using the wool and the second Faraday pail.

What should the experiment show?

If electrons move from the wool to the polythene rod, we predict:

• The rod should have a negative reading.
• The wool should have a positive reading.
• Their charge magnitudes should be approximately equal if the pair is isolated.

Consider the following illustrative data.

Assume each final reading has an uncertainty of approximately \(0.2\ \mathrm{nC}\).

The measured total of \(-0.1\ \mathrm{nC}\) is consistent with zero within the measurement uncertainty.

This supports the model that negative charge has moved from the wool to the rod.

What does this experiment actually prove?

There is an important limitation.

The electrometers measure net electric charge. They do not directly identify individual particles.

The experiment supports the prediction that the two materials acquire opposite charges, but the identification of electrons as the mobile negative particles relies on our established understanding of atomic structure.

It is also important that the objects remain insulated. If electrons leak to the Earth, the two measured charges might not add to zero.

That would not mean charge conservation had failed. It would mean that the two-object system was not completely isolated.

05What happens to the electric force after rubbing?

Once the rod and wool have separated, they exert an electric force on each other.

Because the rod is negative and the wool is positive, they attract.

This is described by Coulomb’s law:

\[
F=k\frac{|q_1q_2|}{r^2}
\]

where:

• \(F\) is the magnitude of the electric force, measured in newtons (N).
• \(q_1\) and \(q_2\) are the charges, measured in coulombs (C).
• \(r\) is the separation between the charges, measured in metres (m).
• \(k=8.99\times10^9\ \mathrm{N\,m^2\,C^{-2}}\) is Coulomb’s constant.

However, calculating the magnitude is only half the job.

Electric force is a vector. It has both magnitude and direction.

The direction depends on the signs of the charges:

• Like charges repel.
• Opposite charges attract.

For the negatively charged rod and positively charged wool, each force points towards the other object.

The forces have equal magnitudes and opposite directions, in accordance with Newton’s third law.

A useful way to avoid direction errors is to identify the object experiencing the force before drawing the arrow.

For example, the force on the negative rod points towards the positive wool. Do not reverse the arrow simply because the rod has a negative charge.

06A useful contrast: neutral does not mean field-free

There is another misconception worth dealing with before you move on.

A neutral object is not necessarily free from electric forces.

Remember, neutral simply means that the total positive and negative charges balance. The object still contains charged particles.

Consider a negatively charged polythene rod brought close to a neutral aluminium can.

The rod and can do not touch.

What happens?

Because aluminium is a conductor, some of its electrons can move through the material.

The negative rod repels electrons inside the can. These electrons shift towards the far side, leaving the near side with an excess of positive charge.

The can is now polarised.

The can still has zero net charge because no electrons have entered or left it.

However, the positive charge on the near side is closer to the rod than the negative charge on the far side.

The attractive force is therefore stronger than the repulsive force, producing a net attraction towards the rod.

This is called polarisation.

It is different from charging by rubbing:

If the rod is removed without the can touching anything else, the electrons redistribute and the can returns to its original neutral arrangement.

The key distinction is that charge transfer changes an object’s net charge, while polarisation changes the distribution of charge within an object.

For a closer explanation of this effect, see Can a Neutral Object Experience an Electric Force? Explained.

07Questions and solutions

The following questions build from basic electron counting to more demanding combinations of charge conservation, electric fields, and vector forces.

Question 1

A polythene rod and a piece of wool are initially electrically neutral. They are rubbed together while both are insulated from the surroundings.

After rubbing, the polythene rod has a charge of \(-6.4\ \mathrm{nC}\).

Assuming that only electrons are transferred:

(a) Calculate the number of electrons transferred.

(b) Determine the final charge on the wool.

Solution 1

(a) The number of electrons transferred is \(4.0\times10^{10}\). The wool loses these electrons, so its final charge is \(+6.4\ \mathrm{nC}\).

The negative charge on the rod tells us that it has gained electrons.

Use

\[
N=\frac{|q|}{e}
\]

Substitute:

\[
N=\frac{6.4\times10^{-9}}
{1.60\times10^{-19}}
\]

\[
\boxed{N=4.0\times10^{10}\text{ electrons}}
\]

(b) The wool has a positive charge of \(+6.4\ \mathrm{nC}\).

Since the two objects were initially neutral and insulated, the charge gained by the rod must equal the charge lost by the wool.

Therefore,

\[
q_{\text{wool}}=+6.4\ \mathrm{nC}
\]

The total charge remains zero.

The important point is that a positive object does not need to gain protons. Losing electrons is sufficient.

Question 2

A polythene rod A and a piece of wool B are rubbed together while insulated.

After rubbing:

• Rod A has a charge of \(-2.5\ \mathrm{nC}\).
• Wool B has a charge of \(+2.5\ \mathrm{nC}\).

Rod A is positioned at the origin, \((0,0)\), and wool B is positioned at \((0.30,\ 0.40)\ \mathrm m\).

Treat the objects as point charges.

Calculate:

(a) The magnitude of the electric force acting on rod A due to wool B.

(b) The horizontal and vertical components of this force.

(c) The force vector acting on wool B due to rod A.

Solution 2

(a) The force magnitude is \(2.25\times10^{-7}\ \mathrm N\). It acts towards wool B. The force components on A are \(+1.35\times10^{-7}\ \mathrm N\) horizontally and \(+1.80\times10^{-7}\ \mathrm N\) vertically.

Step 1: Calculate the separation.

\[
r=\sqrt{(0.30)^2+(0.40)^2}
\]

\[
r=\sqrt{0.25}
\]

\[
r=0.50\ \mathrm m
\]

Step 2: Calculate the force magnitude.

Using Coulomb’s law:

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

Substitute:

\[
F=
\frac{(8.99\times10^9)
(2.5\times10^{-9})^2}
{(0.50)^2}
\]

\[
F=2.2475\times10^{-7}\ \mathrm N
\]

Therefore,

\[
\boxed{F=2.25\times10^{-7}\ \mathrm N}
\]

Step 3: Determine the direction.

The charges are opposite, so they attract.

The force on A points from the origin towards B.

The direction ratios are

\[
\cos\theta=\frac{0.30}{0.50}=0.6
\]

\[
\sin\theta=\frac{0.40}{0.50}=0.8
\]

Step 4: Calculate the force components.

Horizontal component:

\[
F_x=F\cos\theta
\]

\[
F_x=(2.2475\times10^{-7})(0.6)
\]

\[
F_x=1.35\times10^{-7}\ \mathrm N
\]

Vertical component:

\[
F_y=F\sin\theta
\]

\[
F_y=(2.2475\times10^{-7})(0.8)
\]

\[
F_y=1.80\times10^{-7}\ \mathrm N
\]

Therefore,

\[
\boxed{
\vec F_A=
(1.35\hat{\mathbf i}
+1.80\hat{\mathbf j})
\times10^{-7}\ \mathrm N
}
\]

(c) The force on B is equal and opposite to the force on A.

By Newton’s third law,

\[
\boxed{
\vec F_B=
(-1.35\hat{\mathbf i}
-1.80\hat{\mathbf j})
\times10^{-7}\ \mathrm N
}
\]

Both objects experience forces of equal magnitude, directed towards each other.

A common mistake is to calculate the correct magnitude but forget that the question asks for a vector. Always finish by identifying the direction and signs of the components.

Question 3

A student investigates electron transfer using insulated polythene and wool samples.

Both objects are initially neutral. After rubbing, the student measures their charges using calibrated Faraday pails.

The results are:

The student claims:

“The polythene rod gained more electrons than the wool lost. This means additional negative charge must have been created during rubbing.”

(a) Calculate the total measured charge after rubbing.

(b) Determine whether the results are consistent with an isolated two-object system, taking the stated uncertainties into account.

(c) Explain a possible reason for the discrepancy and suggest one improvement to the experiment.

(d) Does the discrepancy provide evidence that protons were transferred? Explain.

Solution 3

(a) The total measured charge is \(-0.4\ \mathrm{nC}\).

Add the measured charges:

\[
q_{\text{total}}=q_{\text{rod}}+q_{\text{wool}}
\]

\[
q_{\text{total}}=-4.0+3.6
\]

\[
\boxed{q_{\text{total}}=-0.4\ \mathrm{nC}}
\]

(b) The results are not consistent with an isolated two-object system within the stated measurement uncertainty.

The uncertainty in the sum, using a conservative addition of the individual uncertainties, is

\[
\Delta q_{\text{total}}=0.1+0.1
\]

\[
\Delta q_{\text{total}}=0.2\ \mathrm{nC}
\]

Therefore, the measured total is

\[
q_{\text{total}}=-0.4\pm0.2\ \mathrm{nC}
\]

This range does not include zero.

If the two objects were perfectly isolated and initially neutral, conservation of charge predicts

\[
q_{\text{total}}=0
\]

The discrepancy is larger than the stated uncertainty.

(c) Charge may have leaked to the surroundings, or there may be a systematic measurement error.

For example, a damp insulating support could provide a pathway for charge to move between an object and the surroundings.

This means the two objects were not acting as a completely isolated system.

A suitable improvement would be to use clean, dry insulating supports, minimise handling, and check the electrometers against known reference charges before repeating the investigation.

The student should also consider whether charge on the supports needs to be measured.

(d) No, the discrepancy is not evidence that protons were transferred.

The electrometers measure net charge, not the identity of the particles responsible for it.

A mismatch can indicate charge exchange with the surroundings or an experimental error. It does not overturn the established model of electron transfer during ordinary contact electrification.

The tempting conclusion is that charge has been created. The actual issue is that the experiment has not accounted for all the charge in the system.

Question 4

A negatively charged polythene rod is held close to a neutral aluminium can.

The can is resting on an insulating surface. The rod does not touch the can, and the can is not connected to the Earth.

A student argues:

“The can has zero net charge, so the electric force on it must also be zero.”

(a) Explain why the student’s prediction is incorrect.

(b) Describe how the electrons and positive charges in the can are distributed while the rod is nearby.

(c) Determine the direction of the net electric force on the can.

(d) The rod is then removed without the can touching anything else. Explain what happens to the can’s charge distribution and net charge.

Solution 4

(a) The can experiences a net attractive electric force despite having zero net charge.

The student’s mistake is assuming that zero net charge means zero electric force.

The can is neutral overall, but its positive and negative charges can be redistributed.

(b) Electrons move towards the side of the can furthest from the negative rod.

The negative rod repels electrons in the aluminium can.

As a result:

• The side closest to the rod becomes positively charged.
• The far side becomes negatively charged.

The positive nuclei remain in their fixed positions within the metal lattice. The mobile electrons redistribute.

The can remains neutral because the total number of electrons has not changed.

(c) The net force on the can is towards the negatively charged rod.

The positively charged near side is attracted towards the rod.

The negatively charged far side is repelled.

However, the positive charges are closer to the rod, so their attractive force is stronger than the repulsive force on the far side.

The result is a net attraction.

(d) The can returns to a uniform neutral charge distribution, with zero net charge.

When the external rod is removed, the force causing the electrons to redistribute disappears.

The electrons spread back through the conductor, restoring the original distribution.

Because the can was never grounded or touched by the rod, no net charge was transferred to or from it.

This question highlights an important distinction: an object’s net charge determines its overall charge balance, but its internal charge distribution also matters when predicting electric forces.

Question 5

An insulated polythene rod A and a piece of wool B are initially neutral.

They are rubbed together and separated. The resulting charges are:

\[
q_A=-Q,\qquad q_B=+Q
\]

where \(Q\) is a positive constant.

The two objects are then fixed on a horizontal line, with A on the left and B on the right. They are positioned equal distances \(d\) from a midpoint M.

A small, initially neutral conducting sphere is placed exactly at M. It does not touch either charged object.

Ignore gravity and other external forces. Assume the sphere is small compared with the separation between A and B, and use the induced-polarisation model.

A student makes the following argument:

“The charges are equal and opposite, so their electric fields cancel at M. Also, the sphere is neutral, so even if there is an electric field, it cannot experience an electric force.”

Evaluate this argument.

(a) Determine the direction of the electric field at M due to A and B separately. Hence, determine the direction of the resultant electric field.

(b) Explain how the conducting sphere becomes polarised. Identify which side becomes positive and which side becomes negative.

(c) Determine the net electric force on the sphere when it is exactly at M. Explain why this result does not mean the electric field is zero.

(d) The sphere is displaced a small distance towards A and released. Predict the direction of the net electric force. Explain why the sphere’s equilibrium at M is unstable.

Solution 5

The student’s argument is incorrect about the electric field and the reason for the force. The resultant electric field at M is directed towards A. The sphere polarises, but its net force is zero at the exact midpoint because of symmetry. If displaced slightly towards A, it experiences a net force towards A.

(a) The electric fields due to A and B both point towards A.

Consider the two contributions separately.

• A is negatively charged. The electric field due to a negative charge points towards that charge. Therefore, the field at M due to A points left.
• B is positively charged. The electric field due to a positive charge points away from that charge. Since M is to the left of B, the field at M due to B also points left.

The two fields therefore add rather than cancel.

The resultant field is

\[
\boxed{\vec E_{\text{net}}\text{ points left}}
\]

Equal and opposite charges do not necessarily produce cancelling electric fields.

The direction of each electric field must be determined separately before adding the vectors.

(b) The conducting sphere becomes polarised, with positive charge on its left side and negative charge on its right side.

The resultant electric field points left.

Electrons in the conducting sphere experience an electric force in the opposite direction to the electric field.

They therefore shift towards the right, leaving the left side with an excess of positive charge.

The sphere develops a separated charge distribution:

• Left side: positive.
• Right side: negative.

Its net charge remains zero because no electrons have entered or left the sphere.

(c) The net electric force on the sphere is zero at the exact midpoint, even though the electric field is non-zero.

This is the subtle part.

The electric field is directed left, so the sphere polarises. However, the arrangement of A, B, and the sphere is perfectly symmetric about M.

The forces on the induced positive and negative regions balance at this exact position.

Therefore,

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

This is not because the sphere is neutral and therefore incapable of experiencing an electric force.

It is because the forces balance due to the symmetric arrangement.

A neutral conducting sphere can experience a net force in a non-uniform electric field. Its separated charges experience different forces because they occupy different positions.

(d) The sphere accelerates towards A, moving further away from the midpoint.

Once the sphere moves slightly towards A, the arrangement is no longer symmetric.

The electric field becomes stronger towards the nearer charged object. The induced positive charge on the sphere’s left side is closer to the negatively charged rod, so the attraction towards A becomes stronger.

The net force is therefore towards A.

The sphere moves further away from M rather than returning to its original position.

The midpoint is an unstable equilibrium.

This is the important distinction between electric field and electric force:

• The electric field describes the electric influence at a position.
• The electric force depends on the charge distribution of the object placed in that field.
• A neutral object can polarise and experience a net force when the field is non-uniform.

The student’s conclusion that the sphere experiences no force at the midpoint happens to be correct, but the reasoning is wrong. The field does not cancel, and the sphere is not immune to electric forces simply because its net charge is zero.

08Where this leads next

You now have three separate ideas to keep track of:

1. Rubbing can transfer electrons between materials, changing their net charges.
2. Separated charged objects exert electric forces whose directions depend on their charge signs and positions.
3. Neutral objects can still polarise and experience electric forces without gaining or losing any net charge.

The next useful step is charging by induction and grounding. These processes take the same electron-movement model and apply it to conductors, where electrons can redistribute internally or move between an object and the Earth.

Once you can distinguish electron transfer, polarisation, and grounding, predicting the final charge of an object becomes much more straightforward.