High school · ScienceAbout 180 minutes
ScienceHigh schoolAbout 180 minutes

Trace electron movement in charging by friction, contact and induction, and calculate electric forces with Coulomb's law

Several sessions: a laboratory investigation with low-voltage or mechanical apparatus only, modeled problems checked in SI units, independent practice and a written explanation. g = 9.8 m/s² throughout.

Ohio learning standards this draft targets

science-2018:P.EM.1science-2018:P.EM.2

Read the standard text

science-2018:P.EM.1 Charging objects

science-2018:P.EM.2 Coulomb's law

Provisional targets. Clause-by-clause educational review is still pending.

Section 1 · LearnLearn the idea
For two 1.0 µC charges, F = (8.99 × 10⁹)(1.0 × 10⁻⁶)² ÷ r². At 0.10 m the force is 0.90 N; doubling the separation to 0.20 m cuts it to a quarter, 0.22 N; tripling it to 0.30 m cuts it to a ninth, 0.10 N. The curve is an inverse-square curve, falling much faster than a simple 1/r line would.For two 1.0 µC charges, F = (8.99 × 10⁹)(1.0 × 10⁻⁶)² ÷ r². At 0.10 m the force is 0.90 N; doubling the separation to 0.20 m cuts it to a quarter, 0.22 N; tripling it to 0.30 m cuts it to a ninth, 0.10 N. The curve is an inverse-square curve, falling much faster than a simple 1/r line would.Coulomb force between two 1.0 µC charges vs.separation00.10.20.30.40.500.20.40.60.810.90 N0.22 N0.10 Nseparation r (m)force F (N)
For two 1.0 µC charges, F = (8.99 × 10⁹)(1.0 × 10⁻⁶)² ÷ r². At 0.10 m the force is 0.90 N; doubling the separation to 0.20 m cuts it to a quarter, 0.22 N; tripling it to 0.30 m cuts it to a ninth, 0.10 N. The curve is an inverse-square curve, falling much faster than a simple 1/r line would.

Learning goal: Explain charging by friction, contact and induction by tracing the movement of electrons, distinguish conductors from insulators and explain polarization and the attraction between charged and neutral objects, describe how an electroscope indicates charge, and use Coulomb's law to calculate forces between point charges, including finding where the net force on a charge in a line of charges is zero, and, as an extension, the net force from two charges that are not in a line.

Conservation of charge. Charge is carried by electrons (negative) and protons (positive); ordinary objects are neutral because they contain equal numbers. Whichever way a neutral object is charged, the bookkeeping is the same: whatever excess of positive charge appears on one object or system, an equal excess of negative charge appears on the other. Charge is never created or destroyed, only moved: this is the law of conservation of charge. Since protons are locked in nuclei, it is electrons that move, and tracing them step by step is what explains the difference between the three methods.

Three ways to charge, traced electron by electron. Friction: rub two neutral insulators together (a rubber rod on wool) and electrons transfer from the material that holds them less tightly to the one that holds them more tightly. The rubber gains electrons and becomes negative; the wool loses the same number and becomes equally positive. Contact: touch a charged object to a neutral one. If the rod is negative, some of its excess electrons move onto the neutral object, which becomes negative with the same sign as the rod; if the charged object is positive, electrons move from the neutral object onto it, leaving the neutral object positive. Contact always leaves both objects with the same sign. Induction: bring a charged object near a neutral conductor without touching it. A negative rod repels the conductor's free electrons to the far side, leaving the near side positive. Now connect the far side briefly to the ground (touch it with a finger): the repelled electrons leave through the ground path. Remove the ground connection first and then the rod, and the conductor is left with a net positive charge, opposite to the rod's sign, having never touched it. The order matters: if the rod leaves first, the electrons flow back and nothing is charged.

Conductors, insulators and where charge sits. Metal conductors have a lattice of fixed positive metal ions surrounded by a "sea" of electrons that move freely through the lattice; charge placed on a conductor spreads out over its surface as the like charges repel one another. In an insulator, electrons are bound to their atoms and cannot move through the material; an excess or deficit of electrons stays localized on the small area of the surface where it was put. This is why you can charge one end of a plastic rod and not the other, and why a charged metal sphere cannot hold charge on only one side.

Polarization: why a charged object attracts a neutral one. A charged object can pull on a neutral one. In a metal, the free electrons drift toward a nearby positive object or away from a negative one, so electrons crowd onto one side of the conductor and the other side is left short of them. The side nearer the charged object now carries the opposite charge, and because it is closer, its attraction outweighs the repulsion of the far side: a net attractive force results. If the neutral object is an insulator, its electrons cannot travel across it, but the electron cloud of each atom shifts slightly so that it is no longer centered on the nucleus. The separation is very small, far less than the diameter of an atom, yet across billions of atoms these tiny shifts add up to a net attractive force. That is how a rubbed balloon sticks to a wall and how a charged comb lifts scraps of paper, and it is why attraction alone never proves that the second object is charged: only repulsion does. The same effect lets you balance a meter stick on an upturned cup and turn it with a charged balloon held near one end.

The electroscope. An electroscope is a metal rod with a knob on top and two thin metal leaves at the bottom, insulated from its case. Touch the knob with a negative rod and electrons flow down the conductor to the leaves; both leaves become negative, repel and spread apart. Merely bringing a charged object near the knob polarizes the electroscope, electrons moving toward or away from the knob, so the leaves diverge without any charge being transferred and fall back when the object is removed. The leaves show that charge is present and how much (larger divergence, more charge), but not its sign unless you already know the sign of the electroscope's charge: approaching slowly from a distance, a like charge widens the gap and an unlike charge at first narrows it (a strong enough unlike charge brought very close can reverse the leaves' charge and spread them again). Static cling in a clothes dryer is charging by friction between different fabrics; solutions such as dryer sheets, higher humidity or a conductive strip work by letting the separated charge leak back rather than build up.

Coulomb's law. Two charged objects that are small compared with the distance between them can be modeled as point charges. The electric force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them: F = k|q₁q₂|/r² for the size of the force, with k = 8.99 × 10⁹ N·m²/C² (its value in a vacuum, and very nearly the same in air), charges in coulombs and r in meters. The signs of the charges are not put into this equation; they are used separately to decide the direction. Like charges repel, unlike charges attract, and the force on each charge is equal in magnitude and opposite in direction. The equation can be solved for the force, for one of the charges or for the separation. Doubling the distance cuts the force to a quarter; tripling it cuts the force to a ninth. Plotting F against r gives the steeply falling inverse-square curve in the diagram. Worked model: charges of +2.0 µC and −3.0 µC are 0.30 m apart. F = (8.99 × 10⁹)(2.0 × 10⁻⁶)(3.0 × 10⁻⁶)/(0.30)² = 0.60 N, attractive because the signs differ. Each charge feels 0.60 N toward the other.

Gravity and electricity compared. Newton's law of gravitation, F = G m₁ m₂/r², has the same inverse-square form with masses in place of charges. The differences matter. Gravity always attracts, so in something as massive as a planet the pulls of all its parts add up to a force we notice; electric forces are both attractive and repulsive and, in ordinary matter, tend to cancel because positive and negative charges are mixed. Within and between atoms the electric force is vastly stronger: for an electron and a proton the ratio of electric to gravitational force is about 10³⁹.

Net force from several charges. When more than two charges are present, the force on any one of them is the vector sum of the Coulomb forces from each of the others, calculated pairwise. The problems in this course keep the charges in a line, where the forces simply add or subtract with their signs; when charges are not in a line (an extension), resolve each force into components and add them. For three or four charges in a line, the problem may ask where the net force on a charge is zero.

Investigation: charge and distance. Materials: two strips of clear tape, a balloon, wool, small paper scraps, a plastic rod, a ruler, an electroscope if available. Steps: press two tape strips onto a table and pull them off quickly; hang them and bring them together (they repel, having been charged the same way by friction); bring a charged balloon near a strip and record whether it attracts or repels at 10, 5 and 2 cm by the strip's deflection; test the balloon near paper scraps and near the electroscope's knob without touching, then touching. Record every observation with the distance. A result that supports the claims: like-charged strips repel; deflection increases sharply as the distance falls; neutral paper is attracted; the leaves diverge on approach and relax when the balloon leaves, but stay apart after contact. What this qualitative test cannot show: the inverse-square law itself, which needs a simulation or a force sensor. Safety: no high-voltage sources; keep balloons away from eyes when they burst.

Section 2 · ModelWork through the examples

Worked example 1 — Coulomb's law for two charges. +2.0 µC and −3.0 µC separated by 0.30 m: F = k|q₁q₂|/r² = (8.99 × 10⁹)(2.0 × 10⁻⁶)(3.0 × 10⁻⁶)/0.090 = 0.60 N, attractive. At 0.60 m the force is 0.15 N, one quarter; at 0.10 m it is 5.4 N, nine times larger. To find the separation at which the force would be 1.0 N: r² = k|q₁q₂|/F = 0.0539 m², r = 0.23 m.

Worked example 2 (extension) — net force from two non-collinear charges. This course requires only charges in a line; the same vector addition extends the idea to charges that are not. Charges A = +4.0 µC at (0, 0) and B = +4.0 µC at (0.40 m, 0); a charge C = +1.0 µC sits at (0.20 m, 0.15 m). Distance from A to C = √(0.20² + 0.15²) = 0.25 m; the same from B. Each force on C has magnitude (8.99 × 10⁹)(4.0 × 10⁻⁶)(1.0 × 10⁻⁶)/0.0625 = 0.575 N, directed away from the like charge. Force from A points along (0.20, 0.15)/0.25 = (0.80, 0.60); force from B along (−0.80, 0.60). Sum: x-components cancel (0.46 − 0.46 = 0); y-components add: 2(0.575)(0.60) = 0.69 N. The net force on C is 0.69 N straight up, away from the pair, by symmetry.

Worked example 3 — where the net force is zero. +4.0 µC at x = 0 and +1.0 µC at x = 0.30 m. A third charge placed between them feels opposite pushes; the net force is zero where k(4.0)/x² = k(1.0)/(0.30 − x)², so 2/x = 1/(0.30 − x), 0.60 − 2x = x, x = 0.20 m from the larger charge. The result does not depend on the sign or size of the third charge, because both forces scale with it.

Worked example 4 — induction, step by step. A negative rod approaches a neutral metal sphere on an insulating stand. (1) Free electrons in the sphere are repelled to the far side; the near side is positive. (2) A finger touches the far side: electrons flow through the finger to the ground. (3) The finger is removed while the rod is still near. (4) The rod is removed; the remaining electrons redistribute over the surface, and the sphere carries a net positive charge, opposite in sign to the rod. The rod lost none of its charge.

Section 3 · DoPractice, demonstrate, and explain

Representation task: three charging diagrams and a force graph. Draw before-and-after diagrams for charging by friction, by contact and by induction, showing the electrons that move with arrows and writing the final sign of each object, with a sentence for each on why conservation of charge holds. Then plot Coulomb force against separation for two 1.0 µC charges at 0.10, 0.20, 0.30, 0.40 and 0.50 m (0.90, 0.22, 0.10, 0.056 and 0.036 N), and describe in one sentence how the graph shows an inverse-square relationship rather than a simple inverse one.

Independent performance: procedures and calculations. Part 1: devise two procedures, one for testing how the force depends on the amount of charge and one for how it depends on distance, using either the tape-strip apparatus or a charge-force simulation; state the variable you change, the variables you hold constant, what you measure and what result would support Coulomb's law. Part 2: solve and present, with diagrams, (a) the force between +5.0 µC and +2.0 µC at 0.50 m (0.36 N, repulsive), (b, extension) the net force on a −1.0 µC charge placed 0.30 m from each of two +3.0 µC charges that are 0.36 m apart (magnitude 0.48 N toward the midpoint), and (c) the point between +9.0 µC and +1.0 µC charges 0.40 m apart where a third charge feels no net force (0.30 m from the larger). Success criteria: every force has a magnitude, a direction and units; vector components are shown for (b); a comparison of Coulomb's law with Newton's law of gravitation names two similarities and two differences.

Section 4 · Check7-question lesson practice test

An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.

Problem 1

A glass rod is rubbed with silk and becomes positive. Trace the electrons and explain what the silk's charge must be and why.

Show worked solution

Rubbing transfers electrons from the material that holds them less tightly (the glass) to the one that holds them more tightly (the silk). The glass loses electrons and is left with a surplus of positive charge; the silk gains exactly those electrons and carries an equal surplus of negative charge. Charge is conserved: the total is still zero, it has only been separated, so the silk must be negative by the same amount that the glass is positive.

Problem 2

Describe, step by step, how to give a neutral metal sphere a negative charge by induction using a positively charged rod, and explain why the order of the steps matters.

Show worked solution

Bring the positive rod near (not touching): free electrons in the sphere are attracted to the near side, leaving the far side positive. Touch the far side with a finger: electrons flow up from the ground into the sphere to neutralize the positive far side. Remove the finger first, then the rod; the extra electrons spread over the surface and the sphere is negative, opposite to the rod. If the rod were removed before the finger, the extra electrons would flow back to the ground and the sphere would end up neutral.

Problem 3

Explain why a charged balloon sticks to an uncharged wall, using the movement or shift of electrons, and say why this attraction does not prove the wall is charged.

Show worked solution

The wall is an insulator, so its electrons cannot travel through it, but near the negative balloon the electron cloud of each atom in the wall shifts slightly away from the balloon, leaving the side facing the balloon slightly positive. The shift is tiny, less than an atomic diameter, but summed over billions of atoms it produces a net attraction, because the nearer positive side attracts more strongly than the farther negative side repels. A neutral object is attracted to either sign of charge, so attraction cannot reveal charge; only repulsion proves an object is charged.

Problem 4

Contrast how charge is distributed on a charged metal sphere and on a charged plastic rod, and explain the difference in terms of electron movement.

Show worked solution

On the metal sphere the free electrons of the conductor move through the lattice of fixed positive ions, and because like charges repel, any excess charge spreads out over the whole surface. On the plastic rod the electrons are bound to their atoms and cannot move through the material, so the excess or deficit of electrons stays localized in the small area that was rubbed; one end of a rod can be charged while the other end is not.

Problem 5

Two point charges of +6.0 µC and +2.0 µC are 0.20 m apart. Calculate the force on each and state its direction; then find the force if the separation is increased to 0.60 m.

Show worked solution

F = k|q₁q₂|/r² = (8.99 × 10⁹)(6.0 × 10⁻⁶)(2.0 × 10⁻⁶)/(0.20)² = 0.108/0.040 = 2.7 N. Both charges are positive, so each is pushed directly away from the other with 2.7 N; the forces are equal and opposite. At 0.60 m, three times the distance, the force is 2.7/9 = 0.30 N, because the force is inversely proportional to the square of the separation.

Problem 6

A +2.0 µC charge is fixed at x = 0 and a +8.0 µC charge at x = 0.60 m. (a) Find the net force on a +1.0 µC charge placed at x = 0.30 m. (b) Where between the two fixed charges would the net force on it be zero?

Show worked solution

(a) From the +2.0 µC charge: F = (8.99 × 10⁹)(2.0 × 10⁻⁶)(1.0 × 10⁻⁶)/(0.30)² = 0.20 N, pushing in the +x direction (away from it). From the +8.0 µC charge: F = (8.99 × 10⁹)(8.0 × 10⁻⁶)(1.0 × 10⁻⁶)/(0.30)² = 0.80 N, pushing in the −x direction. The charges are in a line, so the forces subtract: net 0.80 − 0.20 = 0.60 N in the −x direction, toward the smaller charge. (b) The forces balance where k(2.0)/x² = k(8.0)/(0.60 − x)², so (0.60 − x)/x = √4 = 2 and x = 0.20 m, one third of the way from the smaller charge. Check: (8.99 × 10⁹)(2.0 × 10⁻⁶)(1.0 × 10⁻⁶)/(0.20)² = 0.45 N and (8.99 × 10⁹)(8.0 × 10⁻⁶)(1.0 × 10⁻⁶)/(0.40)² = 0.45 N, equal and opposite. The zero point lies closer to the smaller charge, and its position does not depend on the size or sign of the third charge.

Problem 7

Compare Coulomb's law with Newton's law of gravitation: give two similarities and two differences, and explain why gravity dominates between planets while electric forces dominate inside atoms.

Show worked solution

Similarities: both are inverse-square laws (F ∝ 1/r²), and both are proportional to the product of the two interacting quantities (charges or masses), acting along the line joining them with equal and opposite forces on each object. Differences: gravity is only attractive while the electric force can attract or repel; and the electric force between elementary particles is enormously stronger (about 10³⁹ times for a proton and an electron). Inside atoms the charges are unbalanced at short range, so the electric force dominates. Planets are electrically neutral, their positive and negative charges canceling, while their masses only add, so the always-attractive gravitational force accumulates into a large, noticeable effect.