Charging by Induction: Electron Transfer and Electric Force

A neutral metal sphere can be left with a net electric charge even though a charged rod never touches it. Bring the rod close, connect the sphere to Earth, disconnect the Earth connection while the rod is still nearby, and then remove the rod. This process is called charging by induction.

The important detail is that the rod’s electric field causes electrons to move, but the electrons do not jump across the air gap. They travel through a separate conducting path, such as a grounding wire.

The final charge depends on which direction the electrons move. A negatively charged rod can leave the sphere positively charged, while a positively charged rod can leave it negatively charged.

01Neutral does not mean field-free

Start with a neutral metal sphere and a negatively charged rod.

Bring the rod close to the sphere without touching it. The rod repels electrons in the metal, causing them to move towards the side farther from the rod.

The sphere now has:

• A relative shortage of electrons on the side nearest the rod.
• An excess of electrons on the far side.
• Zero net charge overall.

Why doesn’t the sphere become positively charged?

Because the electrons have only moved within the sphere. None have entered or left it.

This process is called induced charge separation. It changes where charge is distributed, not the total amount of charge on an isolated object.

This gives us an important distinction:

A neutral object can still be affected by an electric field. In fact, the charged rod creates a field in the space around it, including the air gap between the rod and sphere.

Inside a conductor at electrostatic equilibrium, the electric field within the metal is zero. This happens because mobile electrons have redistributed until their electric field cancels the field inside the conducting material.

It does not mean that the space around the conductor is field-free.

For a closer look at how this temporary charge separation produces attraction, see Charged and Neutral Objects: Attraction, Repulsion and Induction.

02The experiment: why the Earth connection matters

We now need to solve a different problem.

We want the sphere to retain a net charge after the rod is removed. Simply moving electrons around inside it won’t achieve that.

A standard induction demonstration uses:

• A neutral conducting metal sphere.
• An insulating stand.
• A negatively charged rod held on an insulating handle.
• A conducting wire with a switch connecting the sphere to Earth.
• A leaf electroscope to test whether the sphere retains a net charge.

The insulating stand prevents charge from leaking through the bench. The rod remains separated from the sphere by an air gap throughout the experiment.

First attempt: bring the rod close

Start with the sphere isolated from Earth.

Bring the negatively charged rod close to it without touching.

Electrons move towards the far side of the sphere, creating induced charge separation.

Now remove the rod and test the sphere with the electroscope.

The sphere shows no persistent net charge.

This is the first important experimental observation. The rod has changed the distribution of charge, but it has not transferred charge into or out of the isolated sphere.

If we want the sphere to retain a net charge, we need to give electrons somewhere to go.

Revised experiment: introduce a grounding wire

Repeat the experiment, but this time connect the sphere to Earth while the negatively charged rod is still nearby.

Earth acts as an enormous reservoir of charge. It can accept or supply electrons without its electric potential changing appreciably.

Follow the sequence carefully.

Step 1: Bring the negatively charged rod close.

Electrons in the sphere are repelled towards its far side.

Step 2: Connect the sphere to Earth.

The negative rod continues repelling electrons.

Because there is now a conducting path to Earth, electrons flow out of the sphere and into the ground.

Step 3: Disconnect the Earth wire while the rod is still nearby.

This is the crucial step.

The sphere now has fewer electrons than it had initially. It has a net positive charge.

Disconnecting the wire prevents electrons from returning from Earth.

Step 4: Remove the charged rod.

The remaining positive charge redistributes across the sphere’s surface.

The sphere retains its positive net charge. An electroscope test now shows a persistent charge.

The rod never touches the sphere. Its electric field causes the electron movement, while the grounding wire provides the route for actual electron transfer.

Why must you disconnect Earth before removing the rod?

Suppose you reverse the final two steps.

You remove the charged rod while the sphere is still grounded.

The electric field that was pushing electrons away disappears. Electrons can now flow back from Earth into the sphere, restoring its electrical neutrality.

If you then disconnect the Earth wire, the sphere remains neutral.

This is a common HSC examination trap. The order of the steps determines whether the sphere retains a net charge.

What if the rod is positively charged?

The electron movement reverses.

A positive rod attracts electrons towards the sphere. When the grounding wire is connected, electrons flow from Earth into the sphere.

Disconnect the Earth wire first, then remove the rod.

The sphere retains an excess of electrons and becomes negatively charged.

The general rule is:

Charging by induction leaves the conductor with a net charge opposite in sign to the inducing rod.

03Can you charge two objects without connecting either to Earth?

Yes. There is another useful induction method involving two conducting spheres.

This time, we don’t need a grounding wire.

Place two initially neutral metal spheres, A and B, in contact with each other. Both spheres are supported by insulating stands.

Bring a negatively charged rod close to sphere A.

The negative rod repels electrons away from A. Because the spheres are touching, electrons can travel from A into B.

The result is:

• Sphere A loses electrons and becomes relatively positive.
• Sphere B gains electrons and becomes relatively negative.

Now separate the spheres while the charged rod is still nearby.

Finally, remove the rod.

Sphere A retains its positive charge, and sphere B retains its negative charge.

Notice something important.

The combined system started neutral and remains neutral. Electrons have simply moved from one sphere to the other.

No charge has been created, and the charged rod has not touched either sphere.

If the rod were positively charged instead, electrons would move from B towards A. Sphere A would become negative, and sphere B would become positive.

This gives us three distinct situations:

The third arrangement is particularly useful when a question asks how two objects can acquire opposite charges without either touching the charged rod.

04Determining the electric force after induction

Knowing the final charge is only half the problem. You also need to determine the direction of the electric force.

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

For a small charged object in a uniform electric field:

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

where:

• \(\vec F\) is the electric force in newtons (N).
• \(q\) is the signed charge in coulombs (C).
• \(\vec E\) is the electric field strength in newtons per coulomb (N C\(^{-1}\)).

The sign of the charge determines the force direction.

For example, consider a negatively charged rod positioned to the left of a sphere.

The electric field produced by the rod points towards the rod.

If the sphere has been positively charged by induction, its electric force is towards the left, in the same direction as the field.

If the rod were positive instead, the field at the sphere would point away from the rod. The sphere would become negative, so its force would again be towards the rod.

The force and field directions are not interchangeable. Always account for the sign of the charge.

One limitation is worth remembering. The equation \(\vec F=q\vec E\) is directly useful for a small charged object in a uniform external field. The actual rod-and-sphere arrangement has a non-uniform field and a changing surface charge distribution, so you should not automatically model the entire conductor as a point charge at its centre.

05A reliable method for HSC induction questions

When you encounter an unfamiliar induction arrangement, work through these decisions in order.

1. Identify the charged object.

Is the inducing rod positive or negative?

2. Track the electrons.

Electrons are mobile in a metal. Positive ions remain fixed in the lattice.

A negative rod repels electrons. A positive rod attracts them.

3. Identify the available conducting paths.

Can electrons move to Earth, from Earth, or between two touching conductors?

If there is no conducting path out of an isolated sphere, it cannot acquire a net charge through induction alone.

4. Check when the conducting path is disconnected.

For a grounded sphere, disconnect Earth before removing the rod.

For two touching spheres, separate them before removing the rod.

5. Determine the final charge.

An electron deficit means positive charge. An electron surplus means negative charge.

6. Determine the force direction.

Draw the electric field direction first. Then use the sign of the charge to establish the force direction.

Keep electron movement, charge distribution, and electric force as three separate ideas. They are related, but they describe different parts of the physics.

08Questions and solutions

Question 1

A neutral conducting sphere is placed on an insulating stand.

A positively charged rod is brought close to the sphere without touching it. While the rod remains nearby, the sphere is connected to Earth.

The Earth connection is then disconnected, and finally the rod is removed.

Explain:

• The direction of electron movement during grounding.
• The final net charge on the sphere.
• The direction of the electric force on the sphere while the rod remains nearby.

Solution 1

The sphere becomes negatively charged, and the electric force on it is directed towards the positively charged rod.

The positive rod attracts electrons towards the sphere. When the grounding wire is connected, electrons flow from Earth into the sphere.

The Earth connection is then disconnected, trapping the additional electrons on the sphere. When the rod is removed, these electrons redistribute across the sphere’s surface.

The sphere therefore has an excess of electrons and a negative net charge.

To determine the force direction, consider the electric field produced by the positive rod.

The electric field points away from the positive rod. Because the sphere is negatively charged, its electric force acts opposite to the field.

Therefore, the sphere experiences a force towards the rod.

The key distinction is that the electric field direction is defined using a positive test charge. A negative charge experiences a force in the opposite direction.

Question 2

A conducting sphere A is charged by induction and ends up with a net charge of \(+2.4\ \mathrm{nC}\).

It is positioned at the origin of a coordinate system.

Two other small charged spheres are placed nearby:

• Sphere B has a charge of \(-1.5\ \mathrm{nC}\) and is positioned \(0.20\ \mathrm m\) due east of A.
• Sphere C has a charge of \(+3.0\ \mathrm{nC}\) and is positioned \(0.30\ \mathrm m\) due north of A.

Calculate:

(a) The number of electrons that must have left sphere A during grounding to produce its final charge.

(b) The magnitude and direction of the resultant electrostatic force on A due to B and C.

Use:

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

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

Assume the spheres can be treated as point charges.

Solution 2

(a) \(1.5\times10^{10}\) electrons left sphere A. (b) The resultant force is \(1.08\times10^{-6}\ \mathrm N\), directed \(41.6^\circ\) south of east.

Part (a): Number of electrons transferred

Sphere A has a positive charge, meaning it has lost electrons.

Use:

\[
q=Ne
\]

Rearrange:

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

Substitute:

\[
N=\frac{2.4\times10^{-9}}
{1.60\times10^{-19}}
\]

\[
\boxed{N=1.5\times10^{10}\text{ electrons}}
\]

Part (b): Resultant electrostatic force

First calculate the force on A due to B.

Both charges have opposite signs, so the force is attractive. It acts east, towards B.

Using Coulomb’s law:

\[
F_{AB}=k\frac{|q_Aq_B|}{r_{AB}^2}
\]

Substitute:

\[
F_{AB}=
\frac{
(8.99\times10^9)
(2.4\times10^{-9})(1.5\times10^{-9})
}{
(0.20)^2
}
\]

\[
\boxed{F_{AB}=8.09\times10^{-7}\ \mathrm N}
\]

Now calculate the force on A due to C.

Both charges are positive, so the force is repulsive.

Since C is north of A, the force on A acts south.

\[
F_{AC}=k\frac{|q_Aq_C|}{r_{AC}^2}
\]

Substitute:

\[
F_{AC}=
\frac{
(8.99\times10^9)
(2.4\times10^{-9})(3.0\times10^{-9})
}{
(0.30)^2
}
\]

\[
\boxed{F_{AC}=7.19\times10^{-7}\ \mathrm N}
\]

Now resolve the forces into components.

Take east as positive \(x\) and north as positive \(y\).

The horizontal component is:

\[
F_x=+8.09\times10^{-7}\ \mathrm N
\]

The vertical component is:

\[
F_y=-7.19\times10^{-7}\ \mathrm N
\]

The negative sign indicates south.

The resultant force is:

\[
\vec F_{\text{net}}
=F_x\hat{\mathbf i}+F_y\hat{\mathbf j}
\]

\[
\vec F_{\text{net}}
=(8.09\hat{\mathbf i}-7.19\hat{\mathbf j})
\times10^{-7}\ \mathrm N
\]

Its magnitude is:

\[
\begin{aligned}
F_{\text{net}}
&=\sqrt{F_x^2+F_y^2}\\
&=\sqrt{(8.09\times10^{-7})^2
+(7.19\times10^{-7})^2}\\
&=1.08\times10^{-6}\ \mathrm N
\end{aligned}
\]

The direction is:

\[
\tan\theta=\frac{7.19}{8.09}
\]

\[
\theta=41.6^\circ
\]

Therefore:

\[
\boxed{
F_{\text{net}}=
1.08\times10^{-6}\ \mathrm N
\text{ at }41.6^\circ
\text{ south of east}
}
\]

The important step was determining each force direction before adding the vectors. Simply adding the two force magnitudes would have been incorrect because the forces act at right angles.

Question 3

A student performs an induction experiment using a neutral conducting sphere and a negatively charged rod.

They follow this procedure:

1. Bring the negative rod close to the sphere.
2. Connect the sphere to Earth.
3. Remove the rod while the Earth wire is still connected.
4. Disconnect the Earth wire.

The student expects the sphere to retain a positive charge but finds that it is neutral when tested with an electroscope.

Explain why the experiment produces this result.

Describe how the procedure should be changed to leave the sphere positively charged.

Solution 3

The sphere remains neutral because the student disconnects the Earth wire too late.

While the negative rod is close and the sphere is grounded, electrons are repelled from the sphere and flow into Earth.

At this stage, the sphere has an electron deficit and would retain a positive charge if the grounding path were removed.

However, the student removes the rod while the Earth wire is still connected.

This removes the external electric field that was causing the electron redistribution. Because the sphere is still connected to Earth, electrons can now flow back into it.

The sphere returns to electrical neutrality.

Disconnecting the Earth wire after this has happened cannot preserve the positive charge.

The correct sequence is:

1. Bring the negative rod close to the neutral sphere.
2. Connect the sphere to Earth.
3. Disconnect the Earth wire while the rod is still nearby.
4. Remove the rod.

This time, the electrons cannot return from Earth after the grounding connection is broken.

The sphere retains its positive net charge.

The mistake is confusing the order of the physical operations. Both arrangements involve the same rod and grounding wire, but only one leaves the sphere electrically charged.

Question 4

An isolated neutral copper sphere is placed near a positively charged rod.

After the charges have reached electrostatic equilibrium, two measurements are made:

• A field probe positioned inside the conducting material reads zero electric field.
• A field probe positioned in the air gap between the rod and sphere reads a non-zero electric field.

A student makes two claims:

Claim A: The sphere is neutral, so there cannot be an electric field around it.

Claim B: The electric field inside the metal is zero, so electrons must not have moved.

Evaluate both claims.

Also explain what happens to the sphere’s net charge when the rod is removed.

Solution 4

Both claims are incorrect. The sphere remains neutral overall, but its electrons redistribute, and the electric field in the surrounding space is not necessarily zero.

Claim A: Neutral does not mean field-free.

The positive rod creates an electric field around itself.

This field acts on the mobile electrons in the neutral sphere, attracting them towards the side nearest the rod.

The near side becomes relatively negative, while the far side becomes relatively positive.

The total positive and negative charges still balance, so the sphere remains neutral.

The field in the air gap can therefore be non-zero even though the sphere has zero net charge.

Claim B: Zero internal field does not mean electrons have not moved.

In a conductor at electrostatic equilibrium, mobile electrons redistribute until the electric field within the conducting material becomes zero.

The induced surface charges create an electric field that cancels the internal field.

The zero reading is actually consistent with the electrons having moved and reached equilibrium.

What happens when the rod is removed?

The external electric field disappears.

The induced charge separation relaxes, and the electrons redistribute across the sphere.

Because the sphere was never connected to Earth or another conductor, no net charge was transferred into or out of it.

It remains neutral.

This question tests an important distinction: zero net charge, zero internal electric field, and zero electric field everywhere are three different statements.

Question 5

Two identical neutral conducting spheres, A and B, are initially touching on insulating supports.

The combined system is isolated from Earth, and there are no other charged objects nearby.

A negatively charged rod is brought close to sphere A. The spheres are allowed to reach electrostatic equilibrium.

The following sequence then occurs:

1. The spheres are separated while the negative rod remains near A.
2. The rod is moved to the opposite side, close to sphere B, while the spheres remain separated.
3. The spheres are reconnected while the rod remains near B.
4. After the spheres reach electrostatic equilibrium, the rod is removed while the spheres are still touching.
5. The spheres are separated again.

Assume the spheres are identical, there is no charge leakage, and electrostatic equilibrium is reached at each stage.

Determine the final net charge on each sphere.

Explain the direction of electron movement during the sequence.

Would the final result change if the spheres were separated immediately before the rod was removed, rather than after it?

Solution 5

Both spheres finish electrically neutral. The crucial detail is that the rod is removed while the spheres are connected, allowing the induced charge separation to disappear before they are separated for the final time.

Follow the electron movement through each stage.

Stage 1: Negative rod approaches sphere A.

The negative rod repels electrons away from A.

Because the spheres are touching, electrons can move through their conducting contact from A into B.

Sphere A develops an electron deficit, while sphere B develops an electron surplus.

Stage 2: Separate the spheres while the rod remains near A.

This disconnects the conducting path between the spheres.

Their net charges are now fixed because neither sphere is connected to Earth or another conductor.

Therefore:

\[
q_A>0,\qquad q_B<0
\]

Sphere A is positive, and sphere B is negative.

Stage 3: Move the rod towards B.

The rod is now close to B rather than A.

However, the spheres are still separated.

The electric field can redistribute charge across the surfaces of each sphere, but it cannot change their individual net charges because there is no conducting path between them.

Their net charges remain positive on A and negative on B.

This distinction between surface redistribution and net charge transfer is important.

Stage 4: Reconnect the spheres while the negative rod is near B.

The spheres are now electrically connected again.

The negative rod repels electrons away from B and towards A.

Electrons therefore flow from B back towards A.

The induced charge distribution reverses: A becomes relatively negative, while B becomes relatively positive.

The total charge of the combined system remains zero throughout.

Stage 5: Remove the rod while the spheres are still touching.

This is the decisive step.

Once the rod is removed, the external electric field that was maintaining the charge separation disappears.

Because the spheres are still connected, electrons can redistribute between them.

The combined system has zero net charge. Since the spheres are identical and there are no other external influences, the final equilibrium has no net charge on either sphere.

Thus:

\[
\boxed{q_A=0}
\]

\[
\boxed{q_B=0}
\]

Separating the spheres after this point does not change their net charges.

What if the spheres were separated before the rod was removed?

The result would be different.

With the negative rod still near B, electrons have been pushed towards A through the connecting contact.

If the spheres are separated at this stage:

• A retains an excess of electrons and becomes negative.
• B retains an electron deficit and becomes positive.

Removing the rod afterwards cannot undo this transfer because the spheres are no longer connected.

The final charges would then be:

\[
q_A<0,\qquad q_B>0
\]

The tempting approach is to look only at which sphere was initially nearest the rod and assume those original charges must remain.

But the spheres were reconnected, creating another opportunity for electrons to move. Removing the rod while they were connected allowed the induced separation to disappear.

The deciding factor is not simply where the charged rod is positioned. It is whether the conductors are electrically connected when the external field is removed.

09Where this leads next

Charging by induction is fundamentally about controlling where electrons can move. Once you can track that movement, you can determine the final charge without guessing.

The next useful step is applying the electric field to predict the motion of charged particles. The same relationship, \(\vec F=q\vec E\), determines whether a particle accelerates with or against the field, and how its trajectory changes.

See Charged-Particle Trajectories in Electric Fields: HSC Physics for the next stage of that reasoning.