Electric charge and Coulomb's law

Highlights only0 highlighted

Everything in this unit, and most of the course, grows from one fact: objects can carry electric charge, and charged objects push or pull on each other without touching.

Rub a balloon on your hair and it lifts the hair toward it from a distance.

A comb pulls a thin stream of tap water sideways.

Lightning, a clinging sock from the dryer and the sting from a doorknob in a dry Minnesota winter are all the same interaction.

This section gives you the exact rule for that push or pull, Coulomb's law, and shows you how to add several such forces together.

You need to be able to describe the electric force in three ways: in words (which way it points and why), as an equation (Coulomb's law, used with vector addition), and as a picture (a force diagram drawn as arrows from a dot).

The exam uses all three.

How the points work

A free-response question is scored in points, each awarded on its own. There are no levels: every point you earn counts.

The first point of a Derive part is for starting from the fundamental equation (Coulomb's law, $\vec{E} = \vec{F}_E/q$ , the integral for $\vec{E}$ , or Gauss's law) inside a multistep derivation.

Each substitution (the charge element $dq$ , its distance, the area of the Gaussian surface, the enclosed charge) and the limits of integration earn their own points, so write each on its own line.

If an earlier answer is wrong but you use it correctly later, you still earn the later points.

A claim point (Indicate) is earned only with an attempt at a relevant justification; the reasoning points need named physics (each field's direction, each distance, what stays the same).

A Sketch earns points for shape features (where it is zero, which way it curves, where it is largest), not for exact values.

A value you calculate from a graph in the lab question is credited inside a range of accepted answers.

Final numerical answers carry units.

No point values are printed on the parts. The question type tells you the split: Mathematical Routines (MR) 10, Translation Between Representations (TBR) 12, Experimental Design and Analysis (LAB) 10, Qualitative/Quantitative Translation (QQT) 8.

Section I has no penalty for a wrong answer, so answer every multiple-choice question.

What charge is

Charge is a property that matter has, in the same way it has mass.

Unlike mass, it comes in two kinds, which we call positive and negative.

The names are only labels (Benjamin Franklin chose them in the 1740s), but the signs are useful: add the charges in an object with their signs and you get its net charge.

An object with equal amounts of each is neutral, even though it is packed with charged particles.

Charge is a scalar: it has a size and a sign, but no direction.

That is why you can add charges as plain signed numbers, while the forces they produce must be added as vectors.

Ordinary matter is built from protons, neutrons and electrons.

The proton carries charge $+e$ , the electron $-e$ , and the neutron none, where

e = 1.60\times10^{-19}\ \text{C}

is the elementary charge.

The SI unit of charge is the coulomb (C).

One coulomb is a huge amount of charge in electrostatics: the charges on rubbed balloons and laboratory spheres are nanocoulombs (nC, $10^{-9}$ C) to microcoulombs (μC, $10^{-6}$ C).

Because every free charge you meet is made of whole electrons and protons, a net charge is always a whole-number multiple of $e$ , so for this course $e$ is the smallest amount of charge there is.

Finally, a point charge is a model, not a kind of particle.

A charged sphere 2 mm across, 50 cm from another charge, behaves as if all its charge sat at one point, because its size is negligible compared with the distances in the problem.

Two such spheres 3 mm apart are not point charges: the charges on each would shift toward or away from each other, and the distance $r$ would be poorly defined.

Key terms

Charge: Charge is a fundamental property of all matter. (i) Charge is a scalar quantity and is described as positive or negative; (ii) The magnitude of the charge of a single electron or proton, the elementary charge e, can be considered to be the smallest indivisible amount of charge; (iii) The charge of an electron is −e and the charge of a proton is +e, and a neutron has no electric charge; (iv) A point charge is a model in which the physical size of a charged object or system is negligible in the context of the situation being analyzed.

Coulomb's law

In 1785 Charles-Augustin de Coulomb used a torsion balance to measure how the force between two small charged spheres depends on their charges and separation.

The result is

|\vec{F}_E| = \frac{1}{4\pi\varepsilon_0}\frac{|q_1 q_2|}{r^2} = k\frac{|q_1 q_2|}{r^2}

where $r$ is the distance between the centers of the two point charges, $k = 9.0\times10^{9}\ \text{N·m}^2/\text{C}^2$ and $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/(\text{N·m}^2)$ is the electric permittivity of free space (the next section explains what permittivity means).

The two forms are the same law: $k = 1/(4\pi\varepsilon_0)$ .

Use whichever makes the algebra cleaner.

With $\varepsilon_0$ the factor $4\pi$ is the area of a unit sphere, which is why Gauss's law, later in this unit, comes out simplest in that form.

The drawing below shows the kind of torsion balance Coulomb built.

A charged ball on a light needle hangs from a very fine wire; a second charged ball repels it, the wire twists, and the angle of twist, read from the scale, measures the force, because a wire's twist is proportional to the twisting force on it.

Illustration of a torsion balance of the kind Coulomb used: a glass cylinder with a tall tube on its lid, a very fine wire hanging from a dial at the top of the tube, a light horizontal needle on the wire carrying a small ball at one end and a paper counterweight at the other, a second small ball fixed on an insulating rod beside the first, and a scale around the cylinder for reading the angle of twist.

Read the law as two proportions.

$F_E \propto |q_1|$ and $F_E \propto |q_2|$ : double either charge and the force doubles; double both and it is four times as large.

$F_E \propto 1/r^2$ : double the distance and the force falls to a quarter; a third of the distance makes it nine times larger.

The inverse square is the same geometry that makes light dimmer with distance: whatever spreads out from a point is shared over a sphere whose area grows as $r^2$ .

That link becomes Gauss's law in the last sections of this unit.

Is it a sensible size? Two charges of 1.0 μC each, 10 cm apart, push on each other with $F_E = (9.0\times10^9)(1.0\times10^{-6})^2/(0.10)^2 = 0.90\ \text{N}$ , about the weight of a medium apple.

Two charges of 1 C at 1 m would push with $9\times10^9$ N, the weight of about a million cars, which is why you never see a coulomb of excess charge sitting on anything.

Key terms

Coulomb's law: Coulomb’s law describes the electrostatic force between two charged objects as directly proportional to the magnitude of each of the charges and inversely proportional to the square of the distance between the objects. Relevant equation: |F⃗_E| = (1/(4πε₀))·|q₁q₂|/r² = k|q₁q₂|/r²

Direction: along the line, like repel, unlike attract

Coulomb's law with absolute values gives only the magnitude.

You find the direction separately, from the signs:

the force on each charge acts along the line joining the two charges;

charges of the same sign repel (each is pushed directly away from the other);

charges of opposite sign attract (each is pulled directly toward the other).

The diagram below shows the three cases.

In every pair the two forces are equal in size and opposite in direction, as Newton's third law requires, even when one charge is much larger than the other: a 50 nC sphere pushes on a 1 nC bead exactly as hard as the bead pushes back.

Three pairs of point charges labeled 1 and 2. Two positive charges push each other apart, two negative charges push each other apart, and a positive and a negative charge pull toward each other. In every pair the two force arrows are equal in length, opposite in direction and along the line joining the charges.

Keeping magnitude and direction apart is the safest method.

If you put the signs into $kq_1q_2/r^2$ and get a negative number, it tells you "attractive", but it does not tell you whether that is $+x$ or $-y$ .

Find the size from the magnitudes, then draw the arrow from the rule.

Key terms

Direction of the electrostatic force: The direction of the electrostatic force depends on the signs of the charges of the interacting objects and is along the line of separation between the objects. (i) Two objects with charges of the same sign exert repulsive forces on each other; (ii) Two objects with charges of opposite signs exert attractive forces on each other.

More than two charges: superposition

When several charges act on one charge, each pair interacts as if the others were not there, and the net force is the vector sum of the separate forces.

This is the principle of superposition.

You need to be able to find the net force from up to four charged objects (more only when the arrangement is highly symmetric, such as charges equally spaced around a circle).

The method is always:

1. Draw the charge you want the force on, and one arrow for each other charge, pointing along the line to that charge (toward it for opposite signs, away for like signs).

2. Find the magnitude of each force from Coulomb's law with magnitudes only.

3. Resolve each force into $x$ - and $y$ -components, using the geometry, with signs from your arrows.

4. Add the components, then find the magnitude $\sqrt{F_x^2 + F_y^2}$ and the direction $\tan^{-1}(F_y/F_x)$ .

Symmetry can save you work: in a symmetric arrangement, components that are equal and opposite cancel, and you can see which ones before you calculate anything.

Contact forces are electric forces in disguise

The normal force from a table, the friction on a sliding box and the tension in a rope are all, at the level of atoms, electric forces: the electron clouds of neighboring atoms repel when they are pushed together and attract when slightly pulled apart.

A book resting on a table is held up by trillions of such interactions.

Nobody adds them one by one.

Instead we treat their combined effect as a single contact force with simple rules (a normal force perpendicular to the surface, a tension along the string).

In this unit you will use both kinds side by side: Coulomb's law for the force between charged objects that are apart, and the familiar contact forces for strings, ramps and springs that hold them.

Key terms

Electric forces and contact forces: Electric forces are responsible for some of the macroscopic properties of objects in everyday experiences. However, the large number of particle interactions that occur make it more convenient to treat everyday forces in terms of nonfundamental forces called contact forces, such as normal force, friction, and tension.

Charged objects in equilibrium: force diagrams

A charged object held still by a string, a ramp or a spring is a mechanics problem with one extra force.

Draw a force diagram: a dot for the object and one arrow per force, each starting on the dot and pointing the way the force acts, labeled with the usual symbols: $F_E$ (electric), $F_T$ (tension), $F_g$ (gravitational), $F_N$ (normal).

Draw forces, not components, and draw only forces that act: an extra force (a "net force" arrow, or a separate "centripetal" force) costs points.

Then apply Newton's second law in two perpendicular directions with $a = 0$ and eliminate the force you do not know.

The justify paragraph

Many points in this course come from Indicate ... Justify items.

A full justification is two to four sentences that name, in order: the relation that controls the quantity (here Coulomb's law), the quantity that changes and which way, the quantities that stay the same, and then the claim.

Write the dependence as a proportion when you can ("the force is inversely proportional to the square of the distance").

A claim restated as its own reason ("the force is larger because the charges push harder") earns nothing.

You will practice this in every section of the unit.

On the Table of Information

Coulomb's law, $|\vec{F}_E| = \dfrac{1}{4\pi\varepsilon_0}\dfrac{|q_1q_2|}{r^2} = k\dfrac{|q_1q_2|}{r^2}$ , is printed there.

So are the constants this section uses: $e = 1.60\times10^{-19}$ C, $k = 9.0\times10^{9}\ \text{N·m}^2/\text{C}^2$ , $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/(\text{N·m}^2)$ , $g = 9.8\ \text{m/s}^2$ , and Newton's second law on the mechanics page.

Derive or recall: not on the Table

The direction rules: along the line joining the charges, like charges repel, unlike attract.

The superposition method: add forces as vectors, by components.

How to set up equilibrium: $\sum F_x = 0$ and $\sum F_y = 0$ from a force diagram.

The worked examples climb from a one-line use of Coulomb's law to an exam-length derivation, then model two of the answers the exam asks you to draw or write in words.

Worked example 1: the force between two small spheres

A small sphere carries $+3.0\ \mu\text{C}$ and another carries $-5.0\ \mu\text{C}$ .

Their centers are 12 cm apart.

Calculate the magnitude of the electric force on each sphere, and give its direction.

1. Convert: $q_1 = 3.0\times10^{-6}$ C, $q_2 = 5.0\times10^{-6}$ C (magnitudes), $r = 0.12$ m.

2. $|\vec{F}_E| = k\dfrac{|q_1q_2|}{r^2} = (9.0\times10^9\ \text{N·m}^2/\text{C}^2)\dfrac{(3.0\times10^{-6}\ \text{C})(5.0\times10^{-6}\ \text{C})}{(0.12\ \text{m})^2}$ .

3. $= \dfrac{0.135}{0.0144}\ \text{N} = 9.4\ \text{N}$ .

Answer: 9.4 N on each sphere, attractive: each is pulled toward the other along the line joining them.

Check: both spheres feel the same 9.4 N even though their charges differ (Newton's third law).

Forgetting to convert centimeters gives $9.4\times10^{-4}$ N, ten thousand times too small, a common slip.

The next example needs a force diagram of positions, drawn below.

Three small charged spheres. A, carrying +2.0 nC, is at the corner of a right angle; B, carrying −4.0 nC, is 0.30 m to its right; C, carrying +3.0 nC, is 0.40 m above it. On A, the force from B points right, toward B; the force from C points down, away from C; a dashed arrow shows their vector sum pointing right and slightly down. A key shows +x to the right and +y up.

Worked example 2: net force from two charges, by components

Three small charged spheres are fixed in the $xy$ -plane.

Sphere A ( $+2.0$ nC) is at the origin, sphere B ( $-4.0$ nC) is 0.30 m away along $+x$ , and sphere C ( $+3.0$ nC) is 0.40 m away along $+y$ , as shown in Figure 1 (the diagram above).

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

1. Force from B (opposite signs, attractive, so toward B, the $+x$ -direction): $F_B = (9.0\times10^9)\dfrac{(2.0\times10^{-9})(4.0\times10^{-9})}{(0.30)^2} = 8.0\times10^{-7}$ N.

2. Force from C (same signs, repulsive, so away from C, the $-y$ -direction): $F_C = (9.0\times10^9)\dfrac{(2.0\times10^{-9})(3.0\times10^{-9})}{(0.40)^2} = 3.4\times10^{-7}$ N.

3. Components: $F_x = +8.0\times10^{-7}$ N, $F_y = -3.4\times10^{-7}$ N.

4. Magnitude: $F = \sqrt{(8.0\times10^{-7})^2 + (3.4\times10^{-7})^2} = 8.7\times10^{-7}$ N.

5. Direction: $\tan^{-1}(3.4/8.0) = 23°$ below the $+x$ -axis.

Answer: $8.7\times10^{-7}$ N at $23°$ below the $+x$ -direction.

Check: the net force must lie between "toward B" and "away from C", and be closer to the larger force, from B; it does.

Adding the two magnitudes ( $1.1\times10^{-6}$ N) would be wrong: these forces are perpendicular.

Recent exams have asked you to draw a free-body diagram for a charged object held still by contact forces and to use Newton's second law along two axes.

The next example does that for a sphere on a ramp, shown below with its force diagram.

Left: a small sphere with charge plus q rests on a frictionless insulating ramp inclined at angle θ to the horizontal. A fixed sphere with charge plus Q on an insulating stand sits a horizontal distance d to its left, at the same height. Right: the free-body diagram of the small sphere, drawn as three arrows starting on a dot: the electric force F_E horizontal to the right, the gravitational force F_g straight down, and the normal force F_N perpendicular to the ramp, pointing up and to the left.

Worked example 3: a charged sphere held on a ramp (derive, then calculate)

A small sphere of mass $m$ and charge $+q$ rests on a frictionless insulating ramp inclined at angle $\theta$ above the horizontal.

A second sphere with charge $+Q$ is fixed on an insulating stand at the same height, a horizontal distance $d$ to the left, as shown in Figure 1 (the diagram above).

Both spheres can be modeled as point charges.

Part A. Draw the forces exerted on the small sphere, as arrows starting on the dot.

Label each force with a distinct symbol.

(2 points)

Part B. Derive an expression for $d$ .

Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Express your answer in terms of $m$ , $q$ , $Q$ , $\theta$ , and physical constants, as appropriate.

(4 points)

Part C. The small sphere has $m = 2.0$ g, $\theta = 30°$ , and $q = Q = 0.20\ \mu\text{C}$ .

Calculate $d$ .

(2 points)

Example Response, Part A. The force diagram above, right: $F_E$ horizontal, away from the fixed sphere (like charges repel); $F_g$ vertically down; $F_N$ perpendicular to the ramp surface, away from it.

Point A1: For an electric force directed horizontally away from the fixed sphere, and a downward gravitational force.

Point A2: For a normal force perpendicular to the ramp, and no other forces.

Scoring Note: an extra arrow (a "net force", or the components of $F_g$ drawn as separate forces) limits Part A to one point.

Example Response, Part B.

1. Newton's second law with $a = 0$ : $\sum \vec{F} = m\vec{a} = 0$ , in two directions.

2. Horizontal: $F_E - F_N\sin\theta = 0$ .

3. Vertical: $F_N\cos\theta - mg = 0$ , so $F_N = \dfrac{mg}{\cos\theta}$ .

4. Eliminate $F_N$ : $F_E = F_N\sin\theta = mg\tan\theta$ .

5. Coulomb's law: $F_E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{qQ}{d^2}$ , so $\dfrac{qQ}{4\pi\varepsilon_0 d^2} = mg\tan\theta$ .

6. $d = \sqrt{\dfrac{qQ}{4\pi\varepsilon_0\,mg\tan\theta}}$ .

Point B1: For a multistep derivation that starts from Newton's second law applied in two directions (or along and perpendicular to the ramp).

Point B2: For the correct components of $F_N$ (or of $F_E$ and $F_g$ along the ramp).

Point B3: For Coulomb's law with $d$ as the separation.

Point B4: For the correct expression for $d$ .

Example Response, Part C. $mg\tan\theta = (2.0\times10^{-3}\ \text{kg})(9.8\ \text{m/s}^2)\tan 30° = 1.13\times10^{-2}$ N. Then $d = \sqrt{\dfrac{(9.0\times10^9)(0.20\times10^{-6})^2}{1.13\times10^{-2}}} = 0.18$ m.

Point C1: For substituting values consistent with Part B, with grams converted to kilograms and microcoulombs to coulombs.

Point C2: For $d = 0.18$ m, with units.

Check: if the ramp were steeper ( $\theta$ larger), $\tan\theta$ grows, the sphere needs a bigger push, so $d$ must be smaller: the expression agrees.

Working along and perpendicular to the ramp gives the same result: along the ramp, $F_E\cos\theta = mg\sin\theta$ .

Worked example 4: four equal charges at the corners of a square (a symmetric case)

Four particles, each of charge $+q$ , sit at the corners of a square of side $a$ .

Derive an expression for the magnitude of the net electric force on one of them, in terms of $q$ , $a$ , and physical constants, as appropriate.

1. Coulomb's law for each pair: $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{r^2}$ .

2. The two neighbors are at distance $a$ , along two sides at $90°$ to each other: each pushes with $F_1 = \dfrac{q^2}{4\pi\varepsilon_0 a^2}$ .

3. Their sum points along the diagonal, away from the square: $\sqrt{2}\,F_1$ .

4. The opposite corner is at distance $\sqrt{2}a$ , also along the diagonal: $F_2 = \dfrac{q^2}{4\pi\varepsilon_0 (2a^2)} = \tfrac{1}{2}F_1$ .

5. All three now point the same way, so they add: $F = \left(\sqrt{2} + \tfrac{1}{2}\right)\dfrac{q^2}{4\pi\varepsilon_0 a^2}$ .

Answer: $F = \left(\sqrt{2} + \tfrac{1}{2}\right)\dfrac{q^2}{4\pi\varepsilon_0 a^2} \approx 1.91\dfrac{q^2}{4\pi\varepsilon_0 a^2}$ , directed along the diagonal, away from the center.

Check: symmetry says the force must lie along the diagonal (the square looks the same reflected in it), and it does.

Worked example 5: model answer, Indicate and Justify (3 points)

Two small spheres attract each other with a force of magnitude $F_0$ .

The charge on one sphere is doubled, the charge on the other is unchanged, and their separation is doubled.

Indicate whether the new force is greater than, less than, or equal to $F_0$ by writing one of the following.

$F_{\text{new}} > F_0$

$F_{\text{new}} < F_0$

$F_{\text{new}} = F_0$

Justify your answer.

Full-credit answer: " $F_{\text{new}} < F_0$ .

By Coulomb's law the force is proportional to the product of the charges and inversely proportional to the square of the separation.

Doubling one charge doubles the force, but doubling the separation divides it by $2^2 = 4$ .

The other charge is unchanged, so $F_{\text{new}} = 2F_0/4 = F_0/2$ ."

How it scores: Point 1 for "less than" with an attempt at a relevant justification; Point 2 for the effect of the charge (proportional, factor 2); Point 3 for the effect of the separation (inverse square, factor 1/4).

Weaker answer: "Less than, because the spheres are further apart."

It earns the claim point only: the charge also changed, and the answer never shows that the distance effect wins.

Worked example 6: model answer, Sketch (2 points)

Two small charged spheres are moved apart along the line joining them.

Sketch the magnitude of the electric force between them as a function of their separation $r$ , from $r_0$ to $4r_0$ .

Full-credit answer: a curve that starts high at $r_0$ and decreases all the way, steeply at first and then more gently (concave up), reaching one-sixteenth of its starting value at $4r_0$ , never touching the axis.

How it scores: Point 1 for a curve that decreases throughout and is concave up; Point 2 for a value at $4r_0$ much smaller than at $r_0$ (about $1/16$ , and well below a quarter) with the curve not reaching zero.

Weaker answer: a straight line falling to zero at $4r_0$ .

It earns no points: a $1/r^2$ curve is never straight and never reaches zero at a finite distance.

Connect it

Used again in: electric fields (later in this unit), where the field of a point charge comes straight from Coulomb's law and $\vec{E} = \vec{F}_E/q$ ; the potential energy of point charges (Unit 9, Electric Potential); the force between polarized conductors (Unit 10, Conductors and Capacitors).

You also need: vector addition by components and Newton's second law (AP Physics C: Mechanics or your mechanics course).

Exam tip

When a charged object is held still, start your derivation from Newton's second law written in two perpendicular directions, eliminate the force you do not know, and only then put in Coulomb's law.

The first point of the derivation goes to that start.

Do not begin from the answer you expect and work backwards: a derivation that only reaches the printed result by rearranging it earns little.

Common mistakes

Forgetting to convert. Charges in nC or μC and distances in cm or mm must be in C and m before you substitute. It is tempting to plug in the numbers as given because Coulomb's law looks simple; one missed conversion of cm is a factor of $10^4$ in the force.

Adding force magnitudes. Forces in different directions add as vectors. Writing $F_{\text{net}} = F_1 + F_2$ feels natural because charges add as plain numbers, but forces do not.

Using $r$ instead of $r^2$ , or the edge-to-edge gap. $r$ is center to center, and it is squared. Students reach for the visible gap between two spheres, which is smaller than $r$ .

Thinking the larger charge exerts the larger force. The forces on the two charges are equal and opposite (Newton's third law). The smaller object may accelerate more, because its mass may be smaller, which is what makes the idea tempting.

Drawing components as extra forces. In a force diagram, draw $F_g$ , not " $mg\sin\theta$ " and " $mg\cos\theta$ " as two more arrows. Components belong in the equations.

Blank axes for a sketch: the magnitude of the electric force on the vertical axis against the separation r on the horizontal axis, with reference marks at r zero and four r zero and no numerical scale.

The photograph below shows the apparatus for question 9.

Photograph of the apparatus: a small metal sphere on an insulating rod attached to a force sensor held in a clamp, and an identical sphere on an insulating stand that slides along a track beside a meterstick, at the same height; a data logger and a pair of calipers are on the bench.

Check your understanding

1. Two small spheres carry charges of $+4.0$ nC and $+6.0$ nC, with their centers 3.0 cm apart. What is the magnitude of the electric force each exerts on the other?

(A) $2.4\times10^{-8}$ N

(B) $7.2\times10^{-6}$ N

(C) $2.4\times10^{-4}$ N

(D) $2.4\times10^{2}$ N

Total [1]

2. Particle 1 has charge $+q$ and particle 2 has charge $+4q$ ; they are fixed a distance $L$ apart on the $x$ -axis, with particle 1 at the origin. A third particle, of charge $-Q$ , is placed on the $x$ -axis between them. At what position is the net electric force on the third particle zero?

(A) $x = L/5$

(B) $x = L/3$

(C) $x = L/2$

(D) $x = 2L/3$

Total [1]

3. Sphere X has charge $+2q$ and sphere Y has charge $-q$ . Which row correctly compares the magnitude of the force on each sphere and gives the direction of the force on Y?

Magnitudes

Force on Y

(A)

$F_X = 2F_Y$

toward X

(B)

$F_X = F_Y$

toward X

(C)

$F_X = F_Y$

away from X

(D)

$F_Y = 2F_X$

away from X

(A) Row (A)

(B) Row (B)

(C) Row (C)

(D) Row (D)

Total [1]

4. Two small spheres exert an electric force of magnitude $F_0$ on each other. Both charges are tripled and the separation is changed until the force is again $F_0$ . By what factor was the separation multiplied?

(A) $\sqrt{3}$

(B) 3

(C) 9

(D) 1/3

Total [1]

5. A small sphere of mass $m$ and charge $-q$ hangs at rest from an insulating thread. A large horizontal sheet below it produces a uniform upward electric force on the sphere of magnitude $F_E$ , smaller than $mg$ . A student draws the forces on the sphere.

Draw the forces exerted on the sphere, as arrows starting on a dot.

Label each force with a distinct symbol, and draw the relative lengths of the arrows to be reasonable.

Total [2]

6. Particles of charge $+q$ are fixed at $(-a, 0)$ and at $(+a, 0)$ . A particle of charge $+Q$ is at $(0, y)$ with $y > 0$ .

Derive an expression for the magnitude of the net electric force on the particle at $(0, y)$ .

Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Express your answer in terms of $q$ , $Q$ , $a$ , $y$ , and physical constants, as appropriate.

Total [4]

7. A small charged bead and a much larger charged sphere repel each other. The bead's charge is 0.50 nC and the sphere's charge is 40 nC. A student claims the sphere exerts a larger force on the bead than the bead exerts on the sphere.

Indicate whether the force exerted on the bead by the sphere is greater than, less than, or equal to the force exerted on the sphere by the bead by writing one of the following.

$F_{\text{on bead}} > F_{\text{on sphere}}$

$F_{\text{on bead}} < F_{\text{on sphere}}$

$F_{\text{on bead}} = F_{\text{on sphere}}$

Justify your answer.

Total [3]

8. On axes like those shown in Figure 1 (the blank axes just above this list), sketch a graph of the magnitude of the electric force between two point charges as a function of their separation $r$ , from $r = r_0$ to $r = 4r_0$ .

Total [2]

9. Experimental Design and Analysis (LAB)

Two identical small conducting spheres, A and B, are mounted on insulating rods.

Sphere A is attached to a force sensor that reads the horizontal force on it.

Sphere B sits on an insulating stand that slides along a track with a meterstick.

Both spheres are given the same charge $q$ by touching them together after one has been charged, and the spheres are then separated.

In Experiment 1, students are asked to determine $q$ .

They have the apparatus shown in Figure 1 (the photograph above this list), a set of calipers, and no other equipment.

Part A

Describe a procedure the students could use to collect data that would allow them to determine $q$ .

Include a step that would reduce experimental uncertainty.

Part B

Describe how the students could use a graph of their data to determine $q$ .

Include which quantities they should graph to produce a straight line, and how $q$ can be found from that line.

In Experiment 2, the separation of the centers of the spheres is kept at 0.100 m.

Sphere A keeps a fixed charge $q_A$ .

The charge $q_B$ on sphere B is set to five different values, measured each time with a charge sensor, and the force on A is recorded, as shown in Table 1.

Table 1

$q_B$ (nC)

$F_E$ (mN)

0.37

0.70

1.10

1.42

1.81

Part C

i. Indicate which quantities the students should plot to produce a linear graph.

ii. On a grid, label the axes with the quantities and units, choose a linear scale, and plot the data.

iii. Draw a best-fit line for the data.

Part D

Using the best-fit line, calculate the charge $q_A$ .

Total [10]

Answers

1. (C). $F = (9.0\times10^9)\dfrac{(4.0\times10^{-9})(6.0\times10^{-9})}{(0.030)^2} = 2.4\times10^{-4}$ N.

(A) leaves the separation in centimeters ( $r = 3.0$ ).
(B) uses $r$ in place of $r^2$ .
(D) treats the charges as microcoulombs.

2. (B). For zero net force the two attractions balance: $\dfrac{kqQ}{x^2} = \dfrac{k(4q)Q}{(L - x)^2}$ , so $L - x = 2x$ and $x = L/3$ . The point is closer to the smaller charge, as it must be.

(A) comes from $L - x = 4x$ , forgetting the square root.
(C) assumes the point is midway, ignoring the unequal charges.
(D) is the right distance measured from the wrong particle: it puts the point closer to the larger charge.

3. (B). The forces on the two spheres are a Newton's third law pair: equal in magnitude. Opposite charges attract, so the force on Y is toward X.

(A) and (D) give the larger force to one sphere.
(C) has the right magnitudes but treats opposite charges as repelling.

4. (B). $F \propto q_1q_2/r^2$ . Tripling both charges multiplies the numerator by 9; to keep $F$ the same, $r^2$ must be multiplied by 9, so $r$ by 3.

(A) triples only one of the charges.
(C) forgets the square root.
(D) moves the spheres closer, which would make the force larger still.

5. Point 1: For a downward gravitational force ( $F_g$ ) and an upward electric force ( $F_E$ ), each drawn from the dot. Point 2: For an upward tension ( $F_T$ ) and no other forces, with $F_T + F_E = F_g$ shown by $F_g$ as the longest arrow.

Example Response: three vertical arrows from the dot: $F_g$ down (longest), $F_T$ up, $F_E$ up, with $F_T$ and $F_E$ together as long as $F_g$ .

Scoring Note: an extra force (a "net force" or a "force of the sheet" drawn twice) limits the response to Point 1.

6. Point A1: For a multistep derivation that starts from Coulomb's law. Point A2: For the distance from each fixed particle, $r = \sqrt{a^2 + y^2}$ , substituted. Point A3: For recognizing that the $x$ -components cancel and the $y$ -components add, with the component factor $y/\sqrt{a^2 + y^2}$ . Point A4: For the correct expression.

Example Response: each force has magnitude $F_1 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{qQ}{a^2 + y^2}$ , directed away from its source.

By symmetry the $x$ -components are equal and opposite.

Each $y$ -component is $F_1\dfrac{y}{\sqrt{a^2 + y^2}}$ .

So $F = \dfrac{2qQy}{4\pi\varepsilon_0(a^2 + y^2)^{3/2}} = \dfrac{qQy}{2\pi\varepsilon_0(a^2 + y^2)^{3/2}}$ , in the $+y$ -direction.

Checks: $F = 0$ at $y = 0$ (the pushes balance); for $y \gg a$ , $F \to \dfrac{2qQ}{4\pi\varepsilon_0 y^2}$ , the force from a single charge $2q$ .

7. Point 1: For indicating $F_{\text{on bead}} = F_{\text{on sphere}}$ with an attempt at a relevant justification. Point 2: For identifying the two forces as an interaction pair (Newton's third law), equal in magnitude and opposite in direction. Point 3: For using Coulomb's law: the magnitude depends on the product of the two charges, $|q_1q_2|$ , which is the same number for both forces, and on the same separation.

Scoring Note: "equal, because the bead accelerates more" does not earn Point 2 or 3; a larger acceleration comes from a smaller mass, not a larger force.

8. Point 1: For a curve that decreases throughout and is concave up. Point 2: For a curve that does not reach zero, with its value at $4r_0$ about one-sixteenth of its value at $r_0$ (clearly less than a quarter).

Example Response: a smooth curve falling steeply just after $r_0$ and flattening toward the axis by $4r_0$ .

9. Part A. Point A1: For a procedure that measures the center-to-center separation $r$ and the force $F_E$ for at least five different separations, with the charges kept the same. (The calipers measure the spheres' diameter; $r$ is the gap read from the meterstick plus one diameter, or the distance between matching edges.) Point A2: For a step that reduces uncertainty: repeat each force reading and average, use many separations over a wide range, or zero the force sensor with sphere B far away before each run.

Example Response: "Measure the diameter of each sphere with calipers.

Move B to a separation $r$ , read $r$ from the meterstick (center to center), and record the force sensor reading.

Repeat for six separations from 5 cm to 20 cm.

Take three readings at each separation and average them."

Part B. Point B1: For plotting $F_E$ against $1/r^2$ (or an equivalent linearizing pair, such as $1/\sqrt{F_E}$ against $r$ ).

Point B2: For relating the slope to $q$ : $F_E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{r^2}$ , so the slope is $\dfrac{q^2}{4\pi\varepsilon_0}$ and $q = \sqrt{4\pi\varepsilon_0 \times \text{slope}}$ .

Scoring Note: if the charge were known instead, the same slope would give $\varepsilon_0 = \dfrac{q^2}{4\pi \times \text{slope}}$ .

Part C. i.

Point C1: For $F_E$ against $q_B$ (either axis choice).

$F_E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_A}{r^2}q_B$ , so with $r$ fixed the force is proportional to $q_B$ .

ii.

Point C2: For axes labeled with quantity and unit (" $q_B$ (nC)", " $F_E$ (mN)") and a linear scale with equal steps (for example 0 to 60 nC in steps of 10 nC, and 0 to 2.0 mN in steps of 0.2 mN).

Point C3: For the five points plotted correctly, consistent with the axes.

iii.

Point C4: For a straight best-fit line drawn with a ruler, following the trend of the points, with points on both sides.

Part D. Point D1: For relating the slope to $q_A$ : slope $= \dfrac{q_A}{4\pi\varepsilon_0 r^2}$ , so $q_A = 4\pi\varepsilon_0 r^2 \times \text{slope}$ .

Point D2: For a value of $q_A$ from 36 nC to 44 nC.

Example Response: two points read from the line, $(15\ \text{nC},\ 0.54\ \text{mN})$ and $(45\ \text{nC},\ 1.62\ \text{mN})$ : slope $= \dfrac{(1.62 - 0.54)\times10^{-3}\ \text{N}}{(45 - 15)\times10^{-9}\ \text{C}} = 3.6\times10^{4}\ \text{N/C}$ .

Then $q_A = \dfrac{r^2 \times \text{slope}}{k} = \dfrac{(0.100\ \text{m})^2(3.6\times10^4\ \text{N/C})}{9.0\times10^9\ \text{N·m}^2/\text{C}^2} = 4.0\times10^{-8}$ C $= 40$ nC.

Scoring Note: a slope from two data points that do not lie on the best-fit line, or from one point and the origin, does not earn Point D1.

Electric charge and Coulomb's law

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