
Physics II course sequence · Interactive lesson
Charging by Friction, Contact & Induction
Three ways to leave an object charged, and one rule under all of them: electrons change hands. Learn to say which way they went, how many went, and what sign each object is left holding — then check that the books balance.
UPII-01.03 · Lesson 3 of 8 in this unit. Position is not a completion score.Build the model
Nothing is created. Electrons change owner, and the books must balance.
The key idea
Charging by friction, contact and induction are three ways of moving electrons between objects: rubbing transfers them at a surface, touching shares them between conductors, and induction uses a nearby charge plus a path to earth to strand a charge of the opposite sign.
Example. A polythene rod rubbed with wool ends at −2.0 × 10⁻⁸ C: 1.2 × 10¹¹ electrons have crossed onto the rod, the wool is left at +2.0 × 10⁻⁸ C, and the pair together is still exactly neutral.
The complete physical model and assumptions
No process in this section creates charge; each one moves electrons and leaves a deficit behind. Rubbing two dissimilar insulators presses their surfaces into intimate contact at countless points, and because the two materials hold their outer electrons with different tenacity a thin layer of electrons crosses and stays where it lands, so the two objects separate with equal and opposite charges. Touching a charged conductor to a neutral one is a different mechanism: the mobile electrons of a metal redistribute until both bodies sit at one common potential, so the charge is shared and both end with the same sign. Induction is different again. A charge held nearby polarizes a conductor without touching it; opening a path to earth lets the unwanted sign leave, and breaking that path before withdrawing the inducing charge strands a charge opposite in sign to it. Throughout, two ledgers must balance: the total charge of the isolated system, and the whole number of electrons N = |Δq|/e. The model runs out where charge stops staying put — humid air leaks it away in seconds, and no surface holds more than about 2.7 × 10⁻⁵ C m⁻² before the air beside it breaks down.
1Rubbing hands electrons over, it does not make them
Press two dissimilar materials together and their outermost electrons feel two different pulls; at every point of genuine contact a few electrons cross to the material that grips them harder and stay there. Rubbing does not create that transfer, it multiplies it, because two rough surfaces sliding over one another meet at far more points than two surfaces merely laid together. The triboelectric series lists materials in the order they surrender electrons — hair and wool near the giving end, polythene and PTFE near the taking end — so wool rubbed on polythene leaves the polythene negative and the wool equally positive. The amounts are tiny in electron terms and large in electrostatic ones. Take the rod above at −2.0 × 10⁻⁸ C: a 25 g polythene rod contains about 8.6 × 10²⁴ electrons, so the 1.2 × 10¹¹ that crossed are about 1.5 parts in 10¹⁴ of them. Because polythene is an insulator, those electrons stay in the patch where they landed instead of spreading over the rod.
2Contact charging shares, and size sets the share
A metal behaves differently, because its conduction electrons roam the whole body. Touch a charged conductor to a neutral one and the pair becomes a single conductor; electrons flow until it reaches one common potential, and both objects come away carrying the same sign as the charged one. For two identical spheres symmetry does the work — whatever one has the other must have — so the total divides in half, and every further touch halves again. For unequal spheres size sets the split. Keep them far apart, join them with a thin wire, and each sits at V = kq/R, so equal potentials give q ∝ R. A sphere of radius 5.0 cm carrying +14 nC, wired to a neutral sphere of radius 2.0 cm, ends at +10 nC while the small one takes +4.0 nC. Notice that the larger share went to the larger sphere, not to the one that started charged: the split is decided by radius alone. Watch what the electrons did — the big sphere fell from +14 nC to +10 nC, so 2.5 × 10¹⁰ electrons ran from the small neutral sphere onto it. Why both must reach one potential is proved in the unit on electric potential; the ratio is usable before then.
3Induction: the order of operations decides the sign
Charging by induction never touches the object being charged. Hold a rod at −6.0 nC beside a metal sphere on an insulating stand. Step one: the rod's field pushes the sphere's conduction electrons to the far side, leaving the near side short of them. The net charge is still zero, so nothing has been won yet. Step two: run a wire from the sphere to earth. The electrons crowded on the far side are now free to leave, and they go, because earth accepts them. Step three: remove the wire while the rod is still in place. The sphere is now short of electrons with no route back to earth. Step four: withdraw the rod, and the positive charge left behind spreads over the whole surface. The sphere is charged, its sign is opposite to the rod's, and the rod is untouched. Swap steps three and four and you get nothing: withdraw the rod first and the electrons simply climb back up the wire, leaving a neutral sphere. The sequence is the physics.
4What the earth actually does
Earth is not a drain that empties things. It is a reservoir so large that charge entering or leaving it does not shift its potential. Which way charge moves through a grounding wire is decided by the field at the grounded object, never by the wire itself. A negative rod nearby drives electrons down into the earth and leaves the object positive; a positive rod draws electrons up out of the earth and leaves it negative. This is also why a grounded conductor sitting near a charge is not neutral. Grounding does not make the field inside the metal vanish — that is already true of any conductor in electrostatic equilibrium, charged or not. What the wire fixes is the potential: the conductor takes on exactly the charge that holds it at the earth's potential, and how much that is depends on the geometry. For a point charge q at distance d from the centre of a grounded sphere of radius R, the charge drawn from earth is −qR/d, a result the method of images will prove in a later unit. With q = −6.0 nC, R = 4.0 cm and d = 12 cm the sphere takes +2.0 nC, and at d = 24 cm only +1.0 nC. The nearer the inducing charge when you cut the wire, the more you keep.
5Predict the sign before you compute anything
Three questions in order settle the sign of every problem in this topic. Did the charged object touch the target? If it did, this is contact charging: the target ends with the same sign as the charger, and the charger itself ends with less charge than it started with. If it did not, was there a path to earth that was broken while the charged object was still present? If so, this is induction: the target ends with the opposite sign, the charger is unchanged, and the charge was traded with the earth. If neither, the target was only polarized, and it ends neutral however dramatic it looked while the rod was near. Run the four-step sequence above through these questions, at the 12 cm spacing used earlier, and the answers appear before any arithmetic — sphere +2.0 nC, rod still −6.0 nC, earth −2.0 nC, the three summing to −6.0 nC, which is what the system held before anything was done to it. When a calculation disagrees with that ledger, the calculation is wrong.
6Where the charging model runs out
Four limits matter. Charge does not stay put: on a humid day a film of water on an insulator carries it away in seconds, which is why these demonstrations fail in August and work in January. There is also a hard ceiling, because a surface cannot exceed σₘₐₓ = ε₀Eb, and with Eb = 3.0 × 10⁶ V m⁻¹ for dry air that is 2.7 × 10⁻⁵ C m⁻²; a charged face of 1.0 × 10⁻² m² therefore carries at most 2.7 × 10⁻⁷ C, which is why these demonstrations deal in nanocoulombs. The triboelectric series is a rule of thumb rather than a law — its order shifts with roughness, contamination and humidity, and some pairs reverse outright. And the sharing results assume ideal conductors held far apart; two spheres in contact are close enough to distort each other's charge, so q ∝ R is a sound estimate rather than an exact answer. What survives all four is the bookkeeping: whatever leaves one object arrives somewhere else, in whole electrons.
Key equations
Read the units, assumptions and limits before substituting.
Conservation in a transfer
q₁′ + q₂′ = q₁ + q₂
Turns the question of how much moved into one subtraction: a change Δq on one body is −Δq on the other.
Charges in coulombs (C) for an isolated pair; the primes mark the values after the process.Electrons behind the charge
N = |Δq| / e, with e = 1.602 × 10⁻¹⁹ C
Converts a measured charge into the number of electrons that actually changed owner.
Δq in coulombs, N a whole number of electrons — a transfer is never a fraction of one.Contact sharing, identical spheres
q₁′ = q₂′ = (q₁ + q₂) / 2
Fixes both sign and size after a touch, and halves what is left with every repeat.
Charges in coulombs, for two identical isolated conducting spheres touched once and separated.Contact sharing, unequal spheres
qᵢ = (q₁ + q₂) Rᵢ / (R₁ + R₂)
Equal potentials, not equal charges: the larger sphere always takes the larger share.
Radii in metres; spheres far apart on a thin wire, so each sits at V = kq/R with k = 1/(4πε₀).Polarization moves nothing net
qₙₑₐᵣ + qfar = 0
Says why induction needs a ground: with no path out, nothing leaves and the net stays zero.
Induced charges in coulombs on an isolated, ungrounded conductor of any shape.Ceiling on surface charge
σₘₐₓ = ε₀ Eb ≈ 2.7 × 10⁻⁵ C m⁻²
Caps what any charging process can leave behind: past it the air sparks the charge away.
ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²; dry air breaks down at Eb ≈ 3.0 × 10⁶ V m⁻¹.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Set d = 8 cm and walk step 1 → 2 → 3 → 4: the sphere ends at +{q/2|2} nC although the bead never touched it. Repeat at d = 24 cm and the keeper is a third as large — what you strand is fixed by the gap at the instant the wire is cut.
Live results
- NET ON SPHERE
- 2.00 nC
- NET ON EARTH
- -2.00 nC
- ELECTRONS TO EARTH
- 12.48 ×10⁹
- BEAD + SPHERE + EARTH
- -6.0 nC
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Step in the sequence: 2; Bead charge magnitude: 6.0 nC; Bead distance from centre: 12 cm
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The commonest error is to treat the inducing rod as a supply. In charging by induction the rod never touches the sphere and never loses a single electron: test it on an electroscope afterwards and it reads exactly what it read before. The electrons that move come from the earth, through the grounding wire, and they move in the direction the rod's field pushes them — away from a negative rod, towards a positive one. That is why the sphere is left with the sign opposite to the rod's, and why induction is the one process here whose final sign is not the charger's. Contact charging is the mirror image: the rod does give up charge, and the object it touched ends with the rod's own sign. Decide which process happened before you assign any sign.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A glass rod is rubbed with a silk cloth and afterwards carries +4.8 nC. Say which way the electrons moved, how many of them moved, and what charge the silk is left with. Take e = 1.602 × 10⁻¹⁹ C.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Positive means a shortage of electrons, so the glass lost them. Charge is never created, so the electrons the glass lost are now on the silk, and the silk must carry −4.8 nC.
- Charge is quantised, so the transfer is a whole number of electrons: N = |Δq|/e = (4.8 × 10⁻⁹ C) ÷ (1.602 × 10⁻¹⁹ C) = 3.0 × 10¹⁰ electrons.
- Check the ledger: qglass + qₛᵢₗₖ = +4.8 nC + (−4.8 nC) = 0, the same total the neutral pair started with.
- Check the size: 3.0 × 10¹⁰ sounds vast, but a 30 g glass rod holds about 9.0 × 10²⁴ electrons, so about three parts in 10¹⁵ have moved. Rubbing skims the surface and nothing more.
Answer. 3.0 × 10¹⁰ electrons crossed from the glass to the silk, leaving the silk at −4.8 nC and the pair still neutral overall.
MediumExample 2
Two isolated conducting spheres stand far apart on insulating posts. Sphere A has radius 3.0 cm and carries +12 nC; sphere B has radius 9.0 cm and is neutral. A long thin wire joins them briefly and is then removed. Find the final charge on each, and the number of electrons that crossed the wire and in which direction.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Joined by the wire the spheres act as one conductor, so charge moves until they share one potential. Far apart, each sphere's potential is V = kq/R, so equal potentials mean qA/RA = qB/RB.
- The pair is isolated, so the total is fixed at qA + qB = +12 nC. The charge therefore splits in the ratio of the radii, 3.0 : 9.0, which is 1 : 3.
- qA = 12 nC × 3.0/12.0 = +3.0 nC and qB = 12 nC × 9.0/12.0 = +9.0 nC. Both are positive, as they must be, since sharing cannot invent a negative charge.
- Sphere A fell from +12 nC to +3.0 nC, so it gained 9.0 nC of negative charge: N = (9.0 × 10⁻⁹ C) ÷ (1.602 × 10⁻¹⁹ C) = 5.6 × 10¹⁰ electrons.
- Those electrons came off sphere B. Electrons drift from the lower potential towards the higher one, so they run from the neutral large sphere onto the charged small one, even though the small one is the one that started charged.
Answer. qA = +3.0 nC and qB = +9.0 nC, with 5.6 × 10¹⁰ electrons crossing from B to A. The total is still +12 nC.
HardExample 3
A metal sphere of radius 2.0 cm on an insulating stand is neutral. A rod at −6.0 nC is brought near, the sphere is earthed for a moment while 2.5 × 10¹⁰ electrons leave it, the wire is removed, and then the rod is withdrawn. (a) Find the final charge on the sphere and on the rod. (b) It is then touched to an identical neutral sphere and separated: find each charge. (c) Compare (a) with the most it could hold in dry air, where σₘₐₓ = ε₀Eb and Eb = 3.0 × 10⁶ V m⁻¹.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- (a) Losing electrons leaves the sphere positive: q = Ne = 2.5 × 10¹⁰ × 1.602 × 10⁻¹⁹ C = 4.005 × 10⁻⁹ C, so +4.0 nC. Removing the wire before the rod is what strands it; had the rod gone first the electrons would have flowed straight back.
- The rod touched nothing, so it is unchanged at −6.0 nC. The charge the sphere gained came from the earth, which is left at −4.0 nC, and rod + sphere + earth still totals −6.0 nC.
- (b) The spheres are identical, so a touch splits the charge evenly: q′ = (4.005 nC + 0)/2 = 2.0 nC on each. Contact charging gives both the sign of the charged one.
- (c) σₘₐₓ = ε₀Eb = 8.85 × 10⁻¹² × 3.0 × 10⁶ = 2.66 × 10⁻⁵ C m⁻², carrying a guard digit through the working, and the sphere's area is A = 4πR² = 4π(0.020 m)² = 5.03 × 10⁻³ m², so qₘₐₓ = σₘₐₓ A = 1.34 × 10⁻⁷ C, which is 0.13 μC to the two figures the data support.
- 4.005 nC is 3.0% of that ceiling, so the sphere carries this charge comfortably. Push it past 0.13 μC and the field just outside the surface reaches 3.0 × 10⁶ V m⁻¹ and the surrounding air breaks down.
Answer. (a) sphere +4.0 nC, rod still −6.0 nC. (b) +2.0 nC on each sphere. (c) 4.0 nC is 3.0% of the 0.13 μC dry-air ceiling.
Quick check
Select an answer and explain it before checking.
Review & continue
Charging by friction, contact and induction are three ways of moving electrons between objects: rubbing transfers them at a surface, touching shares them between conductors, and induction uses a nearby charge plus a path to earth to strand a charge of the opposite sign.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics II completion tasks.