Answer key · Unit 12

Electricity.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 6

Lesson 1 of 6 · About 20 minutes

Electric charge

Rubbing moves electrons: positive and negative charge, conservation of charge, counting the electrons in a charge, and why a charged balloon pulls on neutral paper.

Open the lesson

The 4 steps and their answers

Step 1 of 4
Open this step
Rubbing moves electrons
Practice · Choose
Question
After being rubbed on hair, the balloon is negatively charged. What happened during the rubbing?

Correct answer
- Correct Electrons moved from the hair to the balloon
Setup, choices and feedback
The Scene at this step
Charge rubbed on
0 nC
Gap above the paper
3 cm
Show charges and field
Balloon
Choices
Shown in a random order.
- Correct Electrons moved from the hair to the balloon
- Wrong The friction made new negative charge on the balloon
  Misconception: Thinks rubbing creates new charge rather than moving electrons.
  Feedback from the catalogue
  Rubbing does not create charge. Every material already holds positive and negative charges in equal numbers, and rubbing only moves some of the electrons from one body to the other. The balloon gains extra, and the hair is left short by the same amount.
- Wrong Protons moved from the balloon to the hair
  Misconception: Thinks protons (positive charges) move when objects are charged by rubbing.
  Feedback from the catalogue
  Protons sit in the nuclei, bound tightly inside the atoms, and do not move when things are rubbed. Only the outer electrons move. A body becomes positive when it loses electrons, and negative when it gains them.
- Wrong Electrons moved to the balloon, and the hair stayed neutral
  Misconception: Thinks one body can become charged while the other stays neutral.
  Feedback from the catalogue
  Charge is conserved: it is neither made nor destroyed. Every electron that reached the balloon left the hair, so the hair carries exactly as much positive charge as the balloon carries negative. That is why the hairs stand up and push each other apart.
Feedback for a right answer
Rubbing only moves electrons from one place to another. The balloon gained extra ones, and the hair is missing exactly as many, so the hair is as positive as the balloon is negative.

Hint
What can move between two materials that touch: the nuclei, or the outermost electrons?

Worked solution
1. Every material has positive protons in its nuclei and negative electrons around them, and in a neutral material there are as many of each.
2. The protons are bound in the nuclei and do not move when rubbed. Only outer electrons can pass from one material to another.
3. A negative charge on the balloon means extra electrons, so electrons came to it from the hair.
4. Charge is conserved: the hair is missing exactly those electrons, and carries a positive charge of the same size.
Rubbing does not make charge: it only separates charges that were there all along, and moves some of them from one body to another.
Step 2 of 4
Open this step
Counting electrons
Practice · Calculate
Question
The balloon carries a charge of 250 nC. How many extra electrons are on it? Use $e = 1.6 \times 10^{-19}\ \mathrm{C}$ ⁠. You can type a large number like this: 2.5e12 is $2.5 \times 10^{12}$ ⁠.

Correct answer
1560000000000 ± 3%
Setup, choices and feedback
The Scene at this step
Charge rubbed on
250 nCfixed
Gap above the paper
3 cm
Show charges and field
0.4 (Balloon → Charges and field)
Accepted within
± 3%
Computed from the Scene at this step’s setting
Extra electrons on the balloon, charge / e/nodes/balloon/outputs/electrons
Wrong answers it recognises
- 4 × 10⁻²⁶ ± 5%
  Misconception: Multiplies the charge by e instead of dividing to count electrons.
  Feedback from the catalogue
  Each electron carries $1.6 \times 10^{-19}\ \mathrm{C}$ ⁠, a tiny amount, so any charge you can measure holds a vast number of electrons. From $Q = Ne$ ⁠ you get $N = \dfrac{Q}{e}$ ⁠: you divide by the charge of one electron. An answer smaller than $1$ ⁠ says something is upside down.
- 1.562 × 10²¹ ± 5%
  Misconception: Uses a charge in nanocoulombs as if it were in coulombs.
  Feedback from the catalogue
  Nano is $10^{-9}$ ⁠: $10\ \mathrm{nC} = 10^{-8}\ \mathrm{C}$ ⁠. The formulas of electricity, such as Coulomb’s law and $Q = Ne$ ⁠, work in SI units, so convert to coulombs before substituting. Without that the answer comes out a billion times too big, or a billion squared.
Feedback for a right answer
$N = \dfrac{2.5 \times 10^{-7}}{1.6 \times 10^{-19}} \approx 1.56 \times 10^{12}$ ⁠. Each − in the Scene, 5 nC, is about 31 billion electrons.

Hint
From $Q = Ne$ ⁠ you get $N = \dfrac{Q}{e}$ ⁠. Convert the charge to coulombs first.

Worked solution
1. The givens: $Q = 250\ \mathrm{nC} = 250 \times 10^{-9}\ \mathrm{C} = 2.5 \times 10^{-7}\ \mathrm{C}$ ⁠, and $e = 1.6 \times 10^{-19}\ \mathrm{C}$ ⁠.
2. The formula: $Q = Ne$ ⁠, so $N = \dfrac{Q}{e}$ ⁠.
3. Substituting: $N = \dfrac{2.5 \times 10^{-7}}{1.6 \times 10^{-19}}$ ⁠.
4. The result: $N \approx 1.56 \times 10^{12}$ ⁠, more than one and a half trillion electrons.
If you got a tiny number such as $4 \times 10^{-26}$ ⁠, you multiplied by $e$ ⁠ instead of dividing by it. If you got $1.56 \times 10^{21}$ ⁠, you forgot that the charge was given in nanocoulombs.
Step 3 of 4
Open this step
Neutral paper
Practice · Predict, then look
Question
The balloon hangs 3 cm above neutral scraps of paper. It is rubbed to 250 nC. What will happen to the scraps?

Correct answer
- Correct They will be pulled towards the balloon, and some will jump up to it
Setup, choices and feedback
The Scene at this step
Charge rubbed on
0 nC
Gap above the paper
3 cm
Show charges and field
Balloon
Choices
- Correct They will be pulled towards the balloon, and some will jump up to it
- Wrong Nothing: the paper has no charge, so the balloon exerts no force on it
  Misconception: Thinks a neutral object feels no force from a charged one.
  Feedback from the catalogue
  A neutral body can be pulled. Its charges shift in the charged body’s field: the opposite ones move closer and the like ones move away. That is polarisation. The force weakens with distance, so the pull on the near side beats the push on the far side.
- Wrong They will be pushed away from the balloon
  Feedback
  Repulsion happens only between two bodies charged with the same sign. The paper has no net charge, so there is nothing to push it away.
Once the prediction is locked in, the Scene changes
Charge rubbed on
250 nC
Show charges and field
0.4 (Balloon → Charges and field)
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The scraps are neutral, but their charges have moved: the + is near the balloon and the − far from it. The pull on the near side beats the push on the far side.

Worked solution
1. Each scrap is neutral: it has as many positive charges as negative ones.
2. The charges in paper can move a little. The negative balloon pushes the scrap’s electrons to its far side, and the near side is left slightly positive: the scrap is polarised.
3. The positive side is pulled towards the balloon and the negative side pushed away, but the positive side is closer, and the electric force is stronger at a smaller distance.
4. So the pull wins, and the whole scrap is drawn in. When the upward part of the pull beats its weight, it jumps.
If you answered “nothing”, you thought a neutral body feels no electric force. It does, when its charges can move.
Step 4 of 4
Open this step
Polarisation
Practice · Choose
Question
A glass rod rubbed with silk carries a positive charge. It is brought close to the same neutral scraps of paper. What will happen?

Correct answer
- Correct It will pull them too
Setup, choices and feedback
The Scene at this step
Charge rubbed on
250 nC
Gap above the paper
3 cm
Show charges and field
0.4 (Balloon → Charges and field)
Choices
Shown in a random order.
- Correct It will pull them too
- Wrong It will push them away, because they were pulled by the negative balloon
  Misconception: Thinks a polarised neutral object carries a net charge of its own.
  Feedback from the catalogue
  A scrap of paper pulled to a balloon did not gain a charge: it stayed neutral, and its charges only shifted. So it is not “positive” and will not be pushed away by a positive rod. A positive rod polarises it the other way round and pulls on it too. Any charged body attracts neutral ones.
- Wrong Nothing: only negative charge pulls on paper
  Feedback
  A positive charge polarises the paper too, only the other way round: it draws the scrap’s electrons towards itself.
Feedback for a right answer
The polarisation flips: now the scrap’s electrons gather on the side near the rod. The near side again has the opposite charge, so the pull wins either way.

Worked solution
1. The scrap is neutral and has no charge of its own: what it feels depends on the charges that move in it.
2. Near a positive rod, the scrap’s electrons are drawn to the near side, and the far side is left positive.
3. Now the near side is negative and pulled towards the rod, and the far side is positive and pushed away. The near side is closer, so the pull wins again.
A charged body of either sign attracts neutral bodies. So attraction tells you nothing about the sign; only repulsion shows that both bodies are charged, and with the same sign.

Lesson 2 of 6 · About 30 minutes

Coulomb's law, field and potential

The force between two charges and how it falls with the square of the distance, the electric field as force per coulomb, fields that add as vectors, and potential: energy per coulomb.

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The 6 steps and their answers

Step 1 of 6
Open this step
Coulomb's law
Practice · Calculate
Question
The left sphere carries +10 nC, the right one −10 nC, and their centres are 10 cm apart. What is the size of the force each sphere exerts on the other? Use $k = 8.99 \times 10^9\ \mathrm{N \cdot m^2/C^2}$ ⁠.

Correct answer
0.00008987 N ± 3%
Setup, choices and feedback
The Scene at this step
Left charge
10 nCfixed
Right charge
-10 nCfixed
Spacing
10 cmfixed
Show the potential
Field lines
Accepted within
± 3%
Computed from the Scene at this step’s setting
Size of the push or pull each sphere feels from the other, k·|q₁q₂|/r²/nodes/pair/outputs/force
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.00000899 N ± 3%
  Misconception: Divides by r instead of r² in Coulomb's law (force halves when distance doubles).
  Feedback from the catalogue
  In Coulomb’s law the distance is squared: $F = k\dfrac{|q_1 q_2|}{r^2}$ ⁠. Twice the distance is a quarter of the force, and three times is a ninth. It is an inverse-square law, just like gravity.
- 8.99 × 10⁻⁹ N ± 3%
  Misconception: Puts centimetres (or square centimetres) into a formula that needs metres.
  Feedback from the catalogue
  The constants $k$ ⁠ and $\varepsilon_0$ ⁠ are given in SI units, so the distances must be in metres too: $5\ \mathrm{cm} = 0.05\ \mathrm{m}$ ⁠. An area converts with the square: $1\ \mathrm{cm^2} = 10^{-4}\ \mathrm{m^2}$ ⁠, so $400\ \mathrm{cm^2} = 0.04\ \mathrm{m^2}$ ⁠.
- 89900000000000 N ± 3%
  Misconception: Uses a charge in nanocoulombs as if it were in coulombs.
  Feedback from the catalogue
  Nano is $10^{-9}$ ⁠: $10\ \mathrm{nC} = 10^{-8}\ \mathrm{C}$ ⁠. The formulas of electricity, such as Coulomb’s law and $Q = Ne$ ⁠, work in SI units, so convert to coulombs before substituting. Without that the answer comes out a billion times too big, or a billion squared.
Feedback for a right answer
$F = 8.99 \times 10^9 \cdot \dfrac{10^{-8} \cdot 10^{-8}}{0.1^2} \approx 8.99 \times 10^{-5}\ \mathrm{N}$ ⁠: a small force, but the spheres are light, and it is billions of times stronger than the gravity between them.

Hint
Convert to SI units: $10\ \mathrm{nC} = 10^{-8}\ \mathrm{C}$ ⁠ and $10\ \mathrm{cm} = 0.1\ \mathrm{m}$ ⁠. Then $F = k\dfrac{|q_1 q_2|}{r^2}$ ⁠.

Worked solution
1. The givens in SI units: $q_1 = 10^{-8}\ \mathrm{C}$ ⁠, $|q_2| = 10^{-8}\ \mathrm{C}$ ⁠, $r = 0.1\ \mathrm{m}$ ⁠.
2. The formula: $F = k\dfrac{|q_1 q_2|}{r^2}$ ⁠.
3. Substituting: $F = 8.99 \times 10^9 \cdot \dfrac{10^{-8} \cdot 10^{-8}}{0.1^2} = 8.99 \times 10^9 \cdot \dfrac{10^{-16}}{10^{-2}}$ ⁠.
4. The result: $F \approx 8.99 \times 10^{-5}\ \mathrm{N}$ ⁠, about nine hundredths of a millinewton. It is a pull, because the charges are opposite.
If you got $8.99 \times 10^{-6}\ \mathrm{N}$ ⁠, you divided by $r$ ⁠ instead of $r^2$ ⁠. If you got $8.99 \times 10^{-9}\ \mathrm{N}$ ⁠, you put the distance in centimetres, $r = 10$ ⁠.
Step 2 of 6
Open this step
The square of the distance
Practice · Predict, then look
Question
The spheres are 5 cm apart. The distance is increased to 10 cm, and the charges stay the same. What will happen to the force between them?

Correct answer
- Correct It will become 4 times smaller
Setup, choices and feedback
The Scene at this step
Left charge
10 nCfixed
Right charge
-10 nCfixed
Spacing
5 cm
Show the potential
Field lines
Choices
- Correct It will become 4 times smaller
- Wrong It will become 2 times smaller
  Misconception: Divides by r instead of r² in Coulomb's law (force halves when distance doubles).
  Feedback from the catalogue
  In Coulomb’s law the distance is squared: $F = k\dfrac{|q_1 q_2|}{r^2}$ ⁠. Twice the distance is a quarter of the force, and three times is a ninth. It is an inverse-square law, just like gravity.
- Wrong It will not change, because the charges have not changed
  Feedback
  The charges set part of the force, but the distance is in Coulomb’s law too: $r^2$ ⁠ in the denominator.
- Wrong It will become 2 times larger
  Feedback
  An electric force, like gravity, weakens with distance; it does not grow.
Once the prediction is locked in, the Scene changes
Spacing
10 cm
The Scene then shows: 0.08987 mN · Size of the push or pull each sphere feels from the other, k·|q₁q₂|/r²/nodes/pair/outputs/force
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The force goes as $\dfrac{1}{r^2}$ ⁠: twice the distance is a quarter of the force. Gravity behaves the same way.

Worked solution
1. In Coulomb’s law $F = k\dfrac{|q_1 q_2|}{r^2}$ ⁠, the charges are fixed, so $F$ ⁠ is proportional to $\dfrac{1}{r^2}$ ⁠.
2. The distance grows 2 times, so $r^2$ ⁠ grows $2^2 = 4$ ⁠ times.
3. The force becomes 4 times smaller: from about $0.36\ \mathrm{mN}$ ⁠ at 5 cm to about $0.09\ \mathrm{mN}$ ⁠ at 10 cm.
If you answered “2 times”, you forgot the square. It is an inverse-square law, like gravity.
Step 3 of 6
Open this step
The electric field
Practice · Calculate
Question
The spheres carry +10 nC and −10 nC, 10 cm apart. What is the strength of the electric field at the point halfway between them? Use $k = 8.99 \times 10^9\ \mathrm{N \cdot m^2/C^2}$ ⁠.

Correct answer
71900 V/m ± 3%
Setup, choices and feedback
The Scene at this step
Left charge
10 nCfixed
Right charge
-10 nCfixed
Spacing
10 cmfixed
Show the potential
Field lines
Accepted within
± 3%
Computed from the Scene at this step’s setting
Field strength halfway between the spheres/nodes/pair/outputs/midField
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0 V/m ± 1 V/m
  Misconception: Thinks the fields of opposite charges cancel between them.
  Feedback from the catalogue
  The field points away from a positive charge and towards a negative one. Between a positive sphere and a negative sphere those two directions are the same: from the positive towards the negative. So the fields there add up rather than cancel. They cancel between two like charges.
- 35950 V/m ± 3%
  Misconception: Counts the field of only one of two charges.
  Feedback from the catalogue
  At every point the fields of all the charges act together, and the total field is their vector sum. Halfway between the two spheres, each contributes about 35,950 V/m, both in the same direction, so the field is double.
Feedback for a right answer
Each sphere gives about 35,950 V/m, and both fields point from the positive towards the negative, so they add to about 71,900 V/m. The field lines in the middle are dense and straight.

Hint
Work out the field of one sphere at $0.05\ \mathrm{m}$ ⁠, $E = k\dfrac{|Q|}{r^2}$ ⁠. Then ask: which way does each of the two point?

Worked solution
1. Each sphere is $r = 0.05\ \mathrm{m}$ ⁠ from the point, with $|Q| = 10^{-8}\ \mathrm{C}$ ⁠.
2. The field of one sphere: $E_1 = k\dfrac{|Q|}{r^2} = 8.99 \times 10^9 \cdot \dfrac{10^{-8}}{0.05^2} \approx 35{,}950\ \mathrm{V/m}$ ⁠.
3. The directions: the positive sphere’s field points away from it, to the right; the negative sphere’s field points towards it, also to the right. Both point the same way.
4. So they add: $E = 2 \cdot 35{,}950 \approx 71{,}900\ \mathrm{V/m}$ ⁠.
If you answered $0$ ⁠, you thought opposite charges cancel in the field too. If you answered about 36,000 V/m, you counted only one sphere.
Step 4 of 6
Open this step
Fields add as vectors
Practice · Predict, then look
Question
Now the right sphere will be positive too, +10 nC, at the same distance. What will the field be at the point halfway?

Correct answer
- Correct Zero
Setup, choices and feedback
The Scene at this step
Left charge
10 nCfixed
Right charge
-10 nC
Spacing
10 cmfixed
Show the potential
Field lines
Choices
- Correct Zero
- Wrong Twice as before, about 144,000 V/m
  Misconception: Adds electric fields as numbers, ignoring their directions.
  Feedback from the catalogue
  An electric field is a vector. Between two positive charges, each one’s field points away from it, so in the middle they point against each other. Two vectors of the same size in opposite directions cancel, so the field in the middle is zero.
- Wrong The same as before, about 71,900 V/m
  Feedback
  The size of each sphere’s field has not changed, but the right sphere’s field has turned round.
Once the prediction is locked in, the Scene changes
Right charge
10 nC
The Scene then shows: 0 V/m · Field strength halfway between the spheres/nodes/pair/outputs/midField
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The two fields are the same size and point against each other, so in the middle they cancel. The field lines swerve aside there and leave an empty point.

Worked solution
1. Each sphere still gives about 35,950 V/m in the middle, because the charge and the distance have not changed.
2. A positive sphere’s field points away from it. In the middle, the left sphere’s field points right, and the right sphere’s field points left.
3. Two vectors of the same size in opposite directions cancel: $E = 35{,}950 - 35{,}950 = 0$ ⁠.
This is the null point between like charges: a test charge placed exactly there feels no force. Adding the fields as numbers, without direction, gives twice as much.
Step 5 of 6
Open this step
Potential
Practice · Calculate
Question
Only the left sphere is charged, with +10 nC, and the right one is uncharged. What is the potential at the point halfway, 5 cm from the centre of the left sphere? Use $k = 8.99 \times 10^9\ \mathrm{N \cdot m^2/C^2}$ ⁠.

Correct answer
1798 V ± 3%
Setup, choices and feedback
The Scene at this step
Left charge
10 nCfixed
Right charge
0 nCfixed
Spacing
10 cmfixed
Show the potential
Potential landscape
Accepted within
± 3%
Computed from the Scene at this step’s setting
Electric potential halfway between the spheres/nodes/pair/outputs/midPotential
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 35950 V ± 3%
  Misconception: Uses kQ/r² (the field) for the potential.
  Feedback from the catalogue
  The potential of a point charge is $V = k\dfrac{Q}{r}$ ⁠, with $r$ ⁠ to the first power. $k\dfrac{Q}{r^2}$ ⁠ is the field. The units tell them apart too: potential in volts, J/C, and field in volts per metre.
- 0.018 V ± 5%
  Misconception: Puts centimetres (or square centimetres) into a formula that needs metres.
  Feedback from the catalogue
  The constants $k$ ⁠ and $\varepsilon_0$ ⁠ are given in SI units, so the distances must be in metres too: $5\ \mathrm{cm} = 0.05\ \mathrm{m}$ ⁠. An area converts with the square: $1\ \mathrm{cm^2} = 10^{-4}\ \mathrm{m^2}$ ⁠, so $400\ \mathrm{cm^2} = 0.04\ \mathrm{m^2}$ ⁠.
Feedback for a right answer
$V = 8.99 \times 10^9 \cdot \dfrac{10^{-8}}{0.05} \approx 1798\ \mathrm{V}$ ⁠. On the landscape, that point is on the slope of the hill around the positive sphere.

Hint
$V = k\dfrac{Q}{r}$ ⁠, with $r$ ⁠ in metres, and no square.

Worked solution
1. The givens: $Q = +10^{-8}\ \mathrm{C}$ ⁠ and $r = 0.05\ \mathrm{m}$ ⁠. The uncharged sphere adds no potential of its own.
2. The formula: $V = k\dfrac{Q}{r}$ ⁠.
3. Substituting: $V = 8.99 \times 10^9 \cdot \dfrac{10^{-8}}{0.05}$ ⁠.
4. The result: $V \approx 1798\ \mathrm{V}$ ⁠. Each coulomb placed there would have about 1800 J of potential energy.
If you got about 36,000, you divided by $r^2$ ⁠ and found the field, not the potential. If you got about $0.018$ ⁠, you put the distance in centimetres.
Step 6 of 6
Open this step
Zero potential
Practice · Choose
Question
Between +10 nC and −10 nC, the potential halfway is 0 V. Is the electric field there zero too?

Correct answer
- Correct No: the field there is about 71,900 V/m, from the positive to the negative
Setup, choices and feedback
The Scene at this step
Left charge
10 nCfixed
Right charge
-10 nCfixed
Spacing
10 cmfixed
Show the potential
Potential landscape
Choices
Shown in a random order.
- Correct No: the field there is about 71,900 V/m, from the positive to the negative
- Wrong Yes: where there is no potential, there is no field either
  Misconception: Thinks the field is zero wherever the potential is zero.
  Feedback from the catalogue
  The field is set by the slope of the potential, not by its height. Halfway between $+Q$ ⁠ and $-Q$ ⁠ the potential is zero, but it falls steeply there from the positive side to the negative, so the field is strong. And the other way round: between two like charges the field in the middle is zero, but the potential there is high.
- Wrong There is no way to tell without measuring
  Feedback
  You can: the field comes from how the potential changes, and the slope shows on the landscape.
Feedback for a right answer
The potential is zero, but the landscape there slopes steeply from the hill to the pit, and a positive charge placed there would roll straight downhill. The field is the slope, not the height.

Worked solution
1. Potential is a scalar: $V = k\dfrac{q_1}{r} + k\dfrac{q_2}{r}$ ⁠, and in the middle $q_1 = -q_2$ ⁠, so $V = 0$ ⁠.
2. The field is a vector, and both fields in the middle point the same way, from the positive to the negative. We saw that they add to about 71,900 V/m.
3. The field depends not on the height of the potential but on its slope: how fast $V$ ⁠ changes along the way. On the landscape, the middle point is at height zero, but halfway down a steep slope from the positive hill to the negative pit.
It is the same with gravity: being at sea level does not mean the ground there is flat.

Lesson 3 of 6 · About 25 minutes

Current, voltage and resistance

What flows in a wire, voltage as energy per coulomb, Ohm's law, why a battery does not give a fixed current, and the power a bulb takes.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
What flows
Practice · Predict, then look
Question
The voltage is turned down to zero. What will happen to the electrons in the wire?

Correct answer
- Correct They will stay in the wire and keep darting about, but without drifting round the loop
Setup, choices and feedback
The Scene at this step
Voltage
6 V
Resistance
6 Ω
Look inside the wire
Inside the wire
Choices
- Correct They will stay in the wire and keep darting about, but without drifting round the loop
- Wrong They will disappear, because they come from the battery
  Misconception: Thinks the electrons in a current come out of the battery.
  Feedback from the catalogue
  The free electrons belong to the metal of the wire, and they are there even with no battery. The battery does not supply them; it gives them energy, the push that drives the whole crowd round the circuit. With no voltage they stay in the wire and dart about at random.
- Wrong They will all go back into the battery
  Feedback
  The battery does not store the wire’s electrons. They belong to the metal itself, and stay in it.
Once the prediction is locked in, the Scene changes
Voltage
0 V
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The electrons did not come from the battery: they were in the wire all along. The battery only pushes them, and with no voltage they dart about at random and do not drift.

Worked solution
1. The free electrons belong to the metal: every atom in the wire gives up one or two, and they are there even when the wire lies in a drawer.
2. The battery does not supply electrons. It gives them energy, a voltage, that pushes the whole crowd the same way.
3. With no voltage there is no push: the electrons keep darting about at random, but on average go nowhere, and the current is zero.
That is why the bulb lights the moment the circuit is closed: the wire is already full of electrons, and they all start moving at once.
Step 2 of 5
Open this step
Ohm's law
Practice · Calculate
Question
The voltage is 12 V and the resistance 8 Ω. What is the current in the circuit?

Correct answer
1.5 A ± 2%
Setup, choices and feedback
The Scene at this step
Voltage
12 Vfixed
Resistance
8 Ωfixed
Look inside the wire
Circuit
Accepted within
± 2%
Computed from the Scene at this step’s setting
Current round the loop, V / R/nodes/loop/outputs/current
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 96 A ± 2%
  Misconception: Rearranges Ohm's law wrongly (I = V·R or I = R/V).
  Feedback from the catalogue
  Ohm’s law is $I = \dfrac{V}{R}$ ⁠: more voltage, more current; more resistance, less current. Check with the units: an ohm is a volt per ampere, so volts divided by ohms are amperes. Multiplying, or dividing the other way, does not give amperes.
- 0.6667 A ± 3%
  Misconception: Rearranges Ohm's law wrongly (I = V·R or I = R/V).
  Feedback from the catalogue
  Ohm’s law is $I = \dfrac{V}{R}$ ⁠: more voltage, more current; more resistance, less current. Check with the units: an ohm is a volt per ampere, so volts divided by ohms are amperes. Multiplying, or dividing the other way, does not give amperes.
Feedback for a right answer
$I = \dfrac{12}{8} = 1.5\ \mathrm{A}$ ⁠: a coulomb and a half passes through the bulb each second.

Hint
$I = \dfrac{V}{R}$ ⁠.

Worked solution
1. The givens: $V = 12\ \mathrm{V}$ ⁠ and $R = 8\ \Omega$ ⁠.
2. Ohm’s law: $I = \dfrac{V}{R}$ ⁠.
3. Substituting: $I = \dfrac{12}{8}$ ⁠.
4. The result: $I = 1.5\ \mathrm{A}$ ⁠.
Check by units: volts divided by ohms are amperes. If you got $96$ ⁠, you multiplied instead of dividing; if you got about $0.67$ ⁠, you divided the resistance by the voltage.
Step 3 of 5
Open this step
Setting the current
Practice · Reach the target
Question
The voltage is fixed at 12 V. Change Resistance so that 0.5 A flows in the circuit, and press Check.

Goal
- 0.5 A ± 0.01 A
  At the start of the step: 2 A
Setup, choices and feedback
The Scene at this step
Voltage
12 Vfixed
Resistance
6 Ω
Look inside the wire
Circuit
Read from the Scene at
Current round the loop, V / R/nodes/loop/outputs/current
One setting that reaches it
Resistance
24 Ω
Feedback for a right answer
At 24 Ω, $\dfrac{12}{24} = 0.5\ \mathrm{A}$ ⁠ flows. The bulb is dim, and the drift is slow.

Feedback for a wrong answer
Not yet. If the current is too large, raise the resistance; if it is too small, lower it.

Hint
Solve Ohm’s law for $R$ ⁠: $R = \dfrac{V}{I}$ ⁠.

Worked solution
1. The givens: $V = 12\ \mathrm{V}$ ⁠, and the current wanted is $I = 0.5\ \mathrm{A}$ ⁠.
2. From $I = \dfrac{V}{R}$ ⁠, solve for $R = \dfrac{V}{I}$ ⁠.
3. Substituting: $R = \dfrac{12}{0.5} = 24\ \Omega$ ⁠.
A quarter of the current needs four times the resistance: from 6 Ω to 24 Ω. Current and resistance are inversely proportional.
Step 4 of 5
Open this step
A battery does not give a fixed current
Practice · Predict, then look
Question
The voltage is 6 V and the resistance 6 Ω, so the current is 1 A. The resistance is doubled to 12 Ω. What will the current be?

Correct answer
- Correct 0.5 A
Setup, choices and feedback
The Scene at this step
Voltage
6 Vfixed
Resistance
6 Ω
Look inside the wire
Circuit
Choices
- Correct 0.5 A
- Wrong 1 A, because the battery always gives the same current
  Misconception: Thinks a battery always supplies the same current, whatever the circuit.
  Feedback from the catalogue
  A battery gives a fixed voltage, a fixed energy per coulomb. How many coulombs pass each second depends on the circuit: $I = \dfrac{V}{R}$ ⁠. More resistance, less current; another branch in parallel, more current.
- Wrong 2 A
  Feedback
  More resistance hinders the current more, so it shrinks; it does not grow.
Once the prediction is locked in, the Scene changes
Resistance
12 Ω
The Scene then shows: 0.5 A · Current round the loop, V / R/nodes/loop/outputs/current
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The same voltage, twice the resistance: half the current. The battery pushes just as hard, but the way through is twice as hard.

Worked solution
1. The battery sets the voltage, $V = 6\ \mathrm{V}$ ⁠, and it does not change.
2. Ohm’s law: $I = \dfrac{V}{R} = \dfrac{6}{12}$ ⁠.
3. The result: $I = 0.5\ \mathrm{A}$ ⁠, half of before.
A battery gives a fixed voltage, not a fixed current. The current is set by the resistance of the whole circuit.
Step 5 of 5
Open this step
Power
Practice · Calculate
Question
The voltage is 12 V and the resistance 6 Ω. What power does the bulb take?

Correct answer
24 W ± 2%
Setup, choices and feedback
The Scene at this step
Voltage
12 Vfixed
Resistance
6 Ωfixed
Look inside the wire
Circuit
Accepted within
± 2%
Computed from the Scene at this step’s setting
Power the bulb takes, V² / R/nodes/loop/outputs/power
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 72 W ± 2%
  Misconception: Uses the wrong formula for electrical power (V·R, or stops at the current).
  Feedback from the catalogue
  Power is energy per second: $P = VI$ ⁠, volts (joules per coulomb) times amperes (coulombs per second) gives joules per second, watts. With Ohm’s law: $P = I^2 R = \dfrac{V^2}{R}$ ⁠. The product $V R$ ⁠ is not a power.
- 2 W ± 2%
  Misconception: Uses the wrong formula for electrical power (V·R, or stops at the current).
  Feedback from the catalogue
  Power is energy per second: $P = VI$ ⁠, volts (joules per coulomb) times amperes (coulombs per second) gives joules per second, watts. With Ohm’s law: $P = I^2 R = \dfrac{V^2}{R}$ ⁠. The product $V R$ ⁠ is not a power.
Feedback for a right answer
$I = 2\ \mathrm{A}$ ⁠, so $P = 12 \cdot 2 = 24\ \mathrm{W}$ ⁠: each second the bulb turns 24 J into heat and light.

Hint
First the current, $I = \dfrac{V}{R}$ ⁠, then $P = VI$ ⁠.

Worked solution
1. The givens: $V = 12\ \mathrm{V}$ ⁠ and $R = 6\ \Omega$ ⁠.
2. The current, from Ohm’s law: $I = \dfrac{12}{6} = 2\ \mathrm{A}$ ⁠.
3. The power: $P = VI = 12 \cdot 2 = 24\ \mathrm{W}$ ⁠. Or in one go: $P = \dfrac{V^2}{R} = \dfrac{144}{6} = 24\ \mathrm{W}$ ⁠.
If you got $2$ ⁠, you stopped at the current. If you got $72$ ⁠, you multiplied the voltage by the resistance, and $V R$ ⁠ is not a power.

Lesson 4 of 6 · About 30 minutes

Circuits in series and parallel

Three bulbs on a 12 V battery: the total resistance in series and in parallel, current that is not used up, voltage that is shared, and what happens to one bulb when its neighbour is switched off.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
In series
Practice · Calculate
Question
Three 6 Ω bulbs in series on a 12 V battery. What is the current from the battery? Give two decimal places.

Correct answer
0.6667 A ± 3%
Setup, choices and feedback
The Scene at this step
Wiring
Seriesfixed
Switch for bulb C
onfixed
Resistance of bulb C
6 Ωfixed
Measure the circuit
Circuitfixed
Accepted within
± 3%
Computed from the Scene at this step’s setting
Current from the battery/nodes/board/outputs/totalCurrent
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 2 A ± 3%
  Misconception: Works out a series circuit as if each bulb were alone on the battery.
  Feedback from the catalogue
  In series the current has to pass through every resistor, one after another, so the total resistance is the sum: $R = R_A + R_B + R_C$ ⁠. The current is the battery’s voltage divided by the total resistance, not by one bulb.
Feedback for a right answer
$R = 18\ \Omega$ ⁠, so $I = \dfrac{12}{18} \approx 0.67\ \mathrm{A}$ ⁠: a third of the current of one bulb on its own, so the bulbs are dim.

Hint
First the total resistance, $R = R_A + R_B + R_C$ ⁠, then Ohm’s law.

Worked solution
1. The givens: $V = 12\ \mathrm{V}$ ⁠, and three resistances of $6\ \Omega$ ⁠ in series.
2. The total resistance: $R = 6 + 6 + 6 = 18\ \Omega$ ⁠.
3. Ohm’s law: $I = \dfrac{V}{R} = \dfrac{12}{18}$ ⁠.
4. The result: $I \approx 0.67\ \mathrm{A}$ ⁠.
If you got $2\ \mathrm{A}$ ⁠, you worked as if each bulb were on the battery alone. In series the current has to pass through all three, and each one adds resistance.
Step 2 of 6
Open this step
The current is not used up
Practice · Predict, then look
Question
In series, the current passes through the bulbs one after another. What happens to it on the way?

Correct answer
- Correct It is the same in every bulb and every wire
Setup, choices and feedback
The Scene at this step
Wiring
Seriesfixed
Switch for bulb C
onfixed
Resistance of bulb C
6 Ωfixed
Measure the circuit
Circuit
Choices
- Correct It is the same in every bulb and every wire
- Wrong It shrinks after each bulb, because each bulb uses some of it up
  Misconception: Thinks current is used up in a bulb, so less flows after it.
  Feedback from the catalogue
  Current is the rate at which charge flows, and charge does not vanish in a bulb. Every coulomb that enters a bulb leaves it, so in series the current is the same at every point. What the bulb takes is energy: each coulomb hands over in it some of the energy it gained in the battery.
- Wrong It grows after each bulb
  Feedback
  Nothing in a bulb adds charge. As much as goes in each second comes out.
Once the prediction is locked in, the Scene changes
Measure the circuit
Currents
The Scene then shows: 0.6667 A · Current c/nodes/board/outputs/currentC
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Every tag shows the same current. A bulb does not use up current; it takes energy from the charge passing through it, and the charge carries on.

Worked solution
1. Current is the amount of charge passing a point each second.
2. Charge is not made or destroyed, and it does not pile up in a bulb. So every coulomb that enters a bulb leaves it.
3. In series there is one path, so the same current passes every point on it: $\dfrac{12}{18} \approx 0.67\ \mathrm{A}$ ⁠, before the bulbs, between them and after them.
What a bulb takes is energy, not charge: each coulomb hands over in it some of the energy it gained in the battery.
Step 3 of 6
Open this step
The voltage is shared
Practice · Calculate
Question
In series: A and B of 6 Ω, C of 12 Ω, and a 12 V battery. What is the voltage across bulb C?

Correct answer
6 V ± 2%
Setup, choices and feedback
The Scene at this step
Wiring
Seriesfixed
Switch for bulb C
onfixed
Resistance of bulb C
12 Ωfixed
Measure the circuit
Circuitfixed
Accepted within
± 2%
Computed from the Scene at this step’s setting
Voltage c/nodes/board/outputs/voltageC
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 4 V ± 2%
  Misconception: Thinks the battery's voltage is shared equally between bulbs in series, whatever their resistance.
  Feedback from the catalogue
  In series the current is the same in every bulb, and the voltage across each is $V = IR$ ⁠. So the voltage is shared in proportion to the resistances: a bulb with twice the resistance gets twice the voltage. Only when the resistances are equal are the voltages equal.
- 12 V ± 2%
  Misconception: Thinks each bulb in series gets the battery's full voltage.
  Feedback from the catalogue
  Each coulomb gains $12\ \mathrm{J}$ ⁠ in the battery and hands all of it over on the way round. In series it passes through every bulb, so they share the $12\ \mathrm{V}$ ⁠: $V_A + V_B + V_C = 12\ \mathrm{V}$ ⁠. Each bulb has the full voltage only in parallel.
Feedback for a right answer
$I = 0.5\ \mathrm{A}$ ⁠, so $V_C = 0.5 \cdot 12 = 6\ \mathrm{V}$ ⁠: C, with twice the resistance, gets twice the voltage of A or B. Move to Voltages and see its big step down.

Hint
First the current, from the total resistance. Then $V_C = I R_C$ ⁠.

Worked solution
1. The total resistance: $R = 6 + 6 + 12 = 24\ \Omega$ ⁠.
2. The current, the same in every bulb: $I = \dfrac{12}{24} = 0.5\ \mathrm{A}$ ⁠.
3. The voltage across C: $V_C = I R_C = 0.5 \cdot 12 = 6\ \mathrm{V}$ ⁠.
4. Check: $V_A = V_B = 0.5 \cdot 6 = 3\ \mathrm{V}$ ⁠, and $3 + 3 + 6 = 12\ \mathrm{V}$ ⁠, the battery’s voltage.
If you answered $4\ \mathrm{V}$ ⁠, you split the voltage equally; it is shared in proportion to the resistances. If you answered $12\ \mathrm{V}$ ⁠, you gave C the whole battery voltage, as in parallel.
Step 4 of 6
Open this step
In parallel
Practice · Predict, then look
Question
A and B are in parallel, and the battery gives 4 A. The switch is closed, adding C, of 6 Ω, in parallel too. What will happen to the current from the battery?

Correct answer
- Correct It will grow
Setup, choices and feedback
The Scene at this step
Wiring
Parallelfixed
Switch for bulb C
off
Resistance of bulb C
6 Ωfixed
Measure the circuit
Currents
Choices
- Correct It will grow
- Wrong It will shrink, because another bulb is more resistance
  Misconception: Thinks adding resistors in parallel raises the total resistance.
  Feedback from the catalogue
  Every branch in parallel is another path for the current, like another checkout opening at the supermarket. So the total resistance is smaller than any one of the resistances: $\dfrac{1}{R} = \dfrac{1}{R_A} + \dfrac{1}{R_B} + \dfrac{1}{R_C}$ ⁠. Adding the resistances themselves is only for series.
- Wrong It will not change, because the battery always gives the same current
  Misconception: Thinks a battery always supplies the same current, whatever the circuit.
  Feedback from the catalogue
  A battery gives a fixed voltage, a fixed energy per coulomb. How many coulombs pass each second depends on the circuit: $I = \dfrac{V}{R}$ ⁠. More resistance, less current; another branch in parallel, more current.
Once the prediction is locked in, the Scene changes
Switch for bulb C
on
The Scene then shows: 6 A · Current from the battery/nodes/board/outputs/totalCurrent
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
C has its own path, and it takes another 2 A. A and B have not changed, and the battery now gives 6 A.

Worked solution
1. In parallel, each bulb is connected straight to the battery, so each has 12 V across it.
2. In each branch: $I = \dfrac{12}{6} = 2\ \mathrm{A}$ ⁠. Adding C does not change the currents in A and B.
3. At the junction the currents add: before, $2 + 2 = 4\ \mathrm{A}$ ⁠; after, $2 + 2 + 2 = 6\ \mathrm{A}$ ⁠.
Another branch in parallel is another path, so the total resistance falls, from 3 Ω to 2 Ω, and the current from the battery rises.
Step 5 of 6
Open this step
Resistance in parallel
Practice · Calculate
Question
In parallel: A and B of 6 Ω, and C of 3 Ω. What total resistance does the battery see?

Correct answer
1.5 Ω ± 3%
Setup, choices and feedback
The Scene at this step
Wiring
Parallelfixed
Switch for bulb C
onfixed
Resistance of bulb C
3 Ωfixed
Measure the circuit
Circuitfixed
Accepted within
± 3%
Computed from the Scene at this step’s setting
Resistance the battery sees; 0 stands for no closed path/nodes/board/outputs/totalResistance
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 15 Ω ± 3%
  Misconception: Thinks adding resistors in parallel raises the total resistance.
  Feedback from the catalogue
  Every branch in parallel is another path for the current, like another checkout opening at the supermarket. So the total resistance is smaller than any one of the resistances: $\dfrac{1}{R} = \dfrac{1}{R_A} + \dfrac{1}{R_B} + \dfrac{1}{R_C}$ ⁠. Adding the resistances themselves is only for series.
- 0.6667 Ω ± 3%
  Misconception: Stops at 1/R in the parallel formula and forgets to invert it.
  Feedback from the catalogue
  The sum $\dfrac{1}{R_A} + \dfrac{1}{R_B} + \dfrac{1}{R_C}$ ⁠ is $\dfrac{1}{R}$ ⁠, not $R$ ⁠. After adding, turn the fraction over. A quick check: the total resistance in parallel is less than the smallest one, but not far less when there are only a few branches.
Feedback for a right answer
$\dfrac{1}{R} = \dfrac{4}{6}$ ⁠, so $R = 1.5\ \Omega$ ⁠, less than any one bulb: the battery gives 8 A.

Hint
$\dfrac{1}{R} = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{3}$ ⁠, and do not forget to turn it over at the end.

Worked solution
1. The formula: $\dfrac{1}{R} = \dfrac{1}{R_A} + \dfrac{1}{R_B} + \dfrac{1}{R_C}$ ⁠.
2. Substituting: $\dfrac{1}{R} = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{2}{6} = \dfrac{4}{6}$ ⁠.
3. Turn it over: $R = \dfrac{6}{4} = 1.5\ \Omega$ ⁠.
4. Check with the currents: $2 + 2 + 4 = 8\ \mathrm{A}$ ⁠, and $\dfrac{12}{8} = 1.5\ \Omega$ ⁠.
The result is less than 3 Ω, the smallest resistance, as it must be. If you answered $15$ ⁠, you added as in series. If you answered about $0.67$ ⁠, you stopped at $\dfrac{1}{R}$ ⁠.
Step 6 of 6
Open this step
Mixed
Practice · Predict, then look
Question
In the mixed circuit, A is in series with B and C, which are in parallel. The switch for C is opened. What will happen to bulb B?

Correct answer
- Correct It will shine brighter
Setup, choices and feedback
The Scene at this step
Wiring
Mixedfixed
Switch for bulb C
on
Resistance of bulb C
6 Ωfixed
Measure the circuit
Currents
Choices
- Correct It will shine brighter
- Wrong It will not change, because it has a branch of its own
  Misconception: Thinks a bulb is unaffected by changes elsewhere in a mixed circuit.
  Feedback from the catalogue
  Only bulbs in parallel straight across the battery are independent. In a mixed circuit, bulb A is in series with everything else: switching C off raises the total resistance, the current through A falls, less voltage drops across it, and more is left for B. A change in one branch reaches the whole circuit.
- Wrong It will dim
  Feedback
  The current from the battery does shrink, but now all of it passes through B, not just half.
- Wrong It will go out
  Feedback
  The switch is only in C’s branch. There is still a closed path through A and B.
Once the prediction is locked in, the Scene changes
Switch for bulb C
off
The Scene then shows: 6 W · Power b/nodes/board/outputs/powerB
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
B rose from about 2.7 W to 6 W: now all the current passes through it, and a larger share of the battery’s voltage falls across it. A dimmed.

Worked solution
1. Before: B and C in parallel are $\dfrac{6}{2} = 3\ \Omega$ ⁠, and with A in series $R = 6 + 3 = 9\ \Omega$ ⁠. The current $\dfrac{12}{9} \approx 1.33\ \mathrm{A}$ ⁠ splits, and half of it, about $0.67\ \mathrm{A}$ ⁠, passes through B. Its power is $P = I^2 R \approx 0.67^2 \cdot 6 \approx 2.7\ \mathrm{W}$ ⁠.
2. After: A and B are left in series, $R = 12\ \Omega$ ⁠, and the current is $\dfrac{12}{12} = 1\ \mathrm{A}$ ⁠, all of it through B. Its power is $1^2 \cdot 6 = 6\ \mathrm{W}$ ⁠.
3. B shines brighter, and A, whose current fell from $1.33$ ⁠ to $1\ \mathrm{A}$ ⁠, dims.
In a mixed circuit the bulbs depend on each other: a change in one branch changes the total current, and with it the voltage across all the rest.

Lesson 5 of 6 · About 25 minutes

Capacitors

Two metal plates on a battery: equal and opposite charge on the plates, capacitance and what sets it, what an insulator between the plates changes, and the energy a capacitor stores.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Two plates and a battery
Practice · Choose
Question
The capacitor is connected to the battery, and each plate carries a charge of size $Q$ ⁠. What is the total charge of the two plates together?

Correct answer
- Correct Zero: $+Q$ ⁠ on one plate and $-Q$ ⁠ on the other
Setup, choices and feedback
The Scene at this step
Plate area
400 cm²
Gap between the plates
1.5 cm
Between the plates
Air
Show charge and field
0.4 (Plates → Charge, field and energy)
Choices
Shown in a random order.
- Correct Zero: $+Q$ ⁠ on one plate and $-Q$ ⁠ on the other
- Wrong $+Q$ ⁠: the battery filled the capacitor with charge
  Misconception: Thinks a charged capacitor carries a net charge.
  Feedback from the catalogue
  The battery does not add charge to the capacitor: it moves electrons from one plate to the other. So one plate carries $+Q$ ⁠ and the other $-Q$ ⁠, and the total charge is zero. “The charge of the capacitor” is the charge on each plate.
- Wrong $2Q$ ⁠: each plate added $Q$ ⁠
  Misconception: Thinks a charged capacitor carries a net charge.
  Feedback from the catalogue
  The battery does not add charge to the capacitor: it moves electrons from one plate to the other. So one plate carries $+Q$ ⁠ and the other $-Q$ ⁠, and the total charge is zero. “The charge of the capacitor” is the charge on each plate.
Feedback for a right answer
Count the marks: each plate has as many, + on one and − on the other. Together their charge is zero, as before the battery was connected.

Worked solution
1. The battery does not make charge. It only moves electrons from plate to plate, through the circuit.
2. Every electron that leaves the positive plate arrives on the negative plate. So the positive plate loses exactly the charge the negative plate gains.
3. Altogether: $+Q + (-Q) = 0$ ⁠. The capacitor as a whole stays neutral.
When we say “the charge of the capacitor is $Q$ ⁠”, we mean the charge on each plate. What a capacitor stores is a separation of charges, and with it energy.
Step 2 of 5
Open this step
Capacitance
Practice · Calculate
Question
Two plates of 400 cm² each, 1.5 cm apart, with air between them ( $\kappa = 1$ ⁠). What is the capacitance? Use $\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{F/m}$ ⁠.

Correct answer
23.63 pF ± 3%
Setup, choices and feedback
The Scene at this step
Plate area
400 cm²fixed
Gap between the plates
1.5 cmfixed
Between the plates
Airfixed
Show charge and field
Plates
Other settings
Readings/nodes/cap/props/readings = false
Accepted within
± 3%
Computed from the Scene at this step’s setting
Κ·ε₀·A/d/nodes/cap/outputs/capacitance
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 2360 pF ± 3%
  Misconception: Puts centimetres (or square centimetres) into a formula that needs metres.
  Feedback from the catalogue
  The constants $k$ ⁠ and $\varepsilon_0$ ⁠ are given in SI units, so the distances must be in metres too: $5\ \mathrm{cm} = 0.05\ \mathrm{m}$ ⁠. An area converts with the square: $1\ \mathrm{cm^2} = 10^{-4}\ \mathrm{m^2}$ ⁠, so $400\ \mathrm{cm^2} = 0.04\ \mathrm{m^2}$ ⁠.
- 0.236 pF ± 3%
  Misconception: Puts centimetres (or square centimetres) into a formula that needs metres.
  Feedback from the catalogue
  The constants $k$ ⁠ and $\varepsilon_0$ ⁠ are given in SI units, so the distances must be in metres too: $5\ \mathrm{cm} = 0.05\ \mathrm{m}$ ⁠. An area converts with the square: $1\ \mathrm{cm^2} = 10^{-4}\ \mathrm{m^2}$ ⁠, so $400\ \mathrm{cm^2} = 0.04\ \mathrm{m^2}$ ⁠.
Feedback for a right answer
$C = 8.85 \times 10^{-12} \cdot \dfrac{0.04}{0.015} \approx 23.6\ \mathrm{pF}$ ⁠: plates the size of a page, and a tiny capacitance. In real capacitors the plates are thin, rolled up and very close together.

Hint
Convert to metres: $400\ \mathrm{cm^2} = 0.04\ \mathrm{m^2}$ ⁠ and $1.5\ \mathrm{cm} = 0.015\ \mathrm{m}$ ⁠.

Worked solution
1. The givens in SI units: $A = 400\ \mathrm{cm^2} = 400 \times 10^{-4}\ \mathrm{m^2} = 0.04\ \mathrm{m^2}$ ⁠, $d = 0.015\ \mathrm{m}$ ⁠, $\kappa = 1$ ⁠.
2. The formula: $C = \kappa \varepsilon_0 \dfrac{A}{d}$ ⁠.
3. Substituting: $C = 8.85 \times 10^{-12} \cdot \dfrac{0.04}{0.015}$ ⁠.
4. The result: $C \approx 2.36 \times 10^{-11}\ \mathrm{F} = 23.6\ \mathrm{pF}$ ⁠.
If you got about 2360 pF, you put in centimetres instead of metres. A square centimetre is $10^{-4}$ ⁠ of a square metre, not $10^{-2}$ ⁠. If you got about 0.24 pF, you converted the area but not the gap.
Step 3 of 5
Open this step
Storing a charge
Practice · Reach the target
Question
The battery is 9 V. Change Plate area and Gap between the plates until each plate carries a charge of 100 pC, that is $10^{-10}\ \mathrm{C}$ ⁠, and press Check.

Goal
- 1 × 10⁻¹⁰ C ± 5%
  At the start of the step: 2.126 × 10⁻¹⁰ C
Setup, choices and feedback
The Scene at this step
Plate area
400 cm²
Gap between the plates
1.5 cm
Between the plates
Airfixed
Show charge and field
Plates
Read from the Scene at
Charge on each plate, C·V/nodes/cap/outputs/charge
One setting that reaches it
Plate area
125 cm²
Gap between the plates
1 cm
Feedback for a right answer
The charge on the plates is about 100 pC. Shrinking the area and widening the gap work the same way: both make $\dfrac{A}{d}$ ⁠ smaller.

Feedback for a wrong answer
Not yet. If the charge is too large, shrink the area or widen the gap; if it is too small, do the opposite.

Hint
$Q = CV$ ⁠, so you need $C = \dfrac{Q}{V} \approx 11\ \mathrm{pF}$ ⁠: about half the capacitance there is now. Shrink the area, or move the plates apart.

Worked solution
1. The capacitance needed: $C = \dfrac{Q}{V} = \dfrac{10^{-10}}{9} \approx 1.11 \times 10^{-11}\ \mathrm{F}$ ⁠.
2. From $C = \varepsilon_0 \dfrac{A}{d}$ ⁠: $\dfrac{A}{d} = \dfrac{C}{\varepsilon_0} \approx 1.25\ \mathrm{m}$ ⁠, so the area in square centimetres divided by the gap in centimetres has to be about $125$ ⁠.
3. For example: 250 cm² with a gap of 2 cm, or 400 cm² with a gap of 3.25 cm. Both give about 100 pC.
The plates hold 213 pC now, so the capacitance has to be about halved: halve the area, or double the gap.
Step 4 of 5
Open this step
Between the plates
Practice · Predict, then look
Question
A glass slab ( $\kappa = 4.7$ ⁠) is slid between the plates, and the battery stays connected. What will happen to the charge on the plates?

Correct answer
- Correct It will grow 4.7 times
Setup, choices and feedback
The Scene at this step
Plate area
400 cm²fixed
Gap between the plates
1.5 cmfixed
Between the plates
Air
Show charge and field
0.4 (Plates → Charge, field and energy)
Other settings
Readings/nodes/cap/props/readings = false
Choices
- Correct It will grow 4.7 times
- Wrong It will not change: the glass only insulates, and no current passes through it anyway
  Misconception: Thinks an insulator between the plates changes nothing because no current crosses it.
  Feedback from the catalogue
  An insulator between the plates does not conduct, but the charges in its molecules shift in the field, and it is polarised. The charge that appears on its faces cancels part of the field, so the battery can push more charge onto the plates. The capacitance grows $\kappa$ ⁠ times: 4.7 times with glass.
- Wrong It will shrink, because the glass blocks the field
  Feedback
  The glass does weaken the field of the charge on the plates, and that is exactly why the battery can push more charge on, until the field is back to $\dfrac{V}{d}$ ⁠.
Once the prediction is locked in, the Scene changes
Between the plates
Glass
Other settings
Readings/nodes/cap/props/readings = true
The Scene then shows: 998.8 pC · Charge on each plate, C·V/nodes/cap/outputs/charge
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The capacitance grew 4.7 times, and the charge with it: from about 213 pC to about 999 pC. The field between the plates stays 600 V/m, because the voltage and the gap have not changed.

Worked solution
1. The battery is connected, so the voltage between the plates stays $9\ \mathrm{V}$ ⁠.
2. The glass is polarised: the charges in its molecules shift a little, and on its faces a charge appears opposite to that of the plate next to it. It cancels part of the field between the plates.
3. To bring the voltage back to $9\ \mathrm{V}$ ⁠, the battery pushes more charge onto the plates. In numbers: $C = \kappa \varepsilon_0 \dfrac{A}{d}$ ⁠ grows 4.7 times, and so does $Q = CV$ ⁠.
4. The result: from about 213 pC to about 999 pC.
The material between the plates does not conduct, but it takes part: its polarisation is what lets the plates hold more charge.
Step 5 of 5
Open this step
The energy in a capacitor
Practice · Calculate
Question
The capacitance is 23.6 pF and the battery 9 V. How much energy is stored in the capacitor?

Correct answer
0.9568 nJ ± 3%
Setup, choices and feedback
The Scene at this step
Plate area
400 cm²fixed
Gap between the plates
1.5 cmfixed
Between the plates
Airfixed
Show charge and field
Platesfixed
Other settings
Readings/nodes/cap/props/readings = false
Accepted within
± 3%
Computed from the Scene at this step’s setting
Stored energy, ½·C·V²/nodes/cap/outputs/energy
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 1.912 nJ ± 3%
  Misconception: Uses U = CV² (or QV) for the energy stored in a capacitor, without the half.
  Feedback from the catalogue
  The battery does not move all the charge at the full voltage. At first the plates are empty and the voltage between them is zero, and it rises in a straight line with the charge. On average the charge is moved across $\dfrac{V}{2}$ ⁠, so $U = \tfrac12 QV = \tfrac12 CV^2$ ⁠: the area of the triangle under the graph.
- 0.1062 nJ ± 3%
  Misconception: Uses U = ½CV, forgetting to square the voltage.
  Feedback from the catalogue
  The voltage appears twice in a capacitor’s energy: once in the charge, $Q = CV$ ⁠, and once in the voltage the charge is moved across. So $U = \tfrac12 QV = \tfrac12 CV^2$ ⁠. Check with the units: farads times volts squared are coulombs times volts, joules. Farads times volts are only coulombs.
Feedback for a right answer
$U = \tfrac12 \cdot 2.36 \times 10^{-11} \cdot 9^2 \approx 0.96\ \mathrm{nJ}$ ⁠: very little. A camera flash capacitor, a few hundred microfarads at a few hundred volts, stores billions of times more.

Hint
$U = \tfrac12 C V^2$ ⁠, with $C$ ⁠ in farads.

Worked solution
1. The givens: $C = 23.6\ \mathrm{pF} = 2.36 \times 10^{-11}\ \mathrm{F}$ ⁠ and $V = 9\ \mathrm{V}$ ⁠.
2. The formula: $U = \tfrac12 C V^2$ ⁠.
3. Substituting: $U = \tfrac12 \cdot 2.36 \times 10^{-11} \cdot 81$ ⁠.
4. The result: $U \approx 9.6 \times 10^{-10}\ \mathrm{J} \approx 0.96\ \mathrm{nJ}$ ⁠.
If you got about 1.9 nJ, you left out the half. It is there because the voltage rose gradually from $0$ ⁠ to $9\ \mathrm{V}$ ⁠ while the charge built up. If you got about 0.11 nJ, you did not square the voltage.

Lesson 6 of 6 · About 25 minutes

Conductors, insulators and semiconductors

Why copper conducts, diamond does not, and silicon sits in between: energy bands and the gap between them, heat lifting electrons across the gap and leaving holes, and why a semiconductor conducts better when it is warm.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Bands of levels
Practice · Choose
Question
In a Metal the bands overlap: the conduction band starts below the top of the valence band. Why does a metal conduct?

Correct answer
- Correct Right above the filled levels there are empty ones, so an electric field can speed electrons up
Setup, choices and feedback
The Scene at this step
Band gap
-1 eV
Temperature
300 K
Show the energy levels
On
Choices
Shown in a random order.
- Correct Right above the filled levels there are empty ones, so an electric field can speed electrons up
- Wrong It has more electrons than silicon or diamond
  Misconception: Thinks a material conducts because it has more electrons, rather than because its electrons have empty levels to move into.
  Feedback from the catalogue
  Every solid is full of electrons. What decides whether it conducts is whether its electrons have empty levels just above their own. In a full band no electron can take energy from an electric field, so a full band does not conduct, however many electrons it holds.
- Wrong Its atoms move freely and carry the charge
  Feedback
  The atoms in a solid metal only vibrate about their places. The current is carried by electrons, and the question is whether they have anywhere to go.
Feedback for a right answer
An electron that speeds up gains energy, so it needs an empty level a little above its own. A metal always has those, even without heat: here about one free electron per atom.

Worked solution
1. In a current, an electric field speeds electrons up, and a faster electron sits at a higher energy level.
2. A full band cannot conduct: every level in it is taken, and an electron has nowhere to move up to.
3. In a metal the bands overlap, so there are empty levels right next to the filled ones, and the electrons move easily.
Silicon and diamond have plenty of electrons, but all of them are in a full band. It is not the number of electrons that matters, but whether they have empty levels.
Step 2 of 6
Open this step
The thermal shaking
Practice · Calculate
Question
The atoms in the crystal shake, and the typical energy of the shaking is $kT$ ⁠, with k = 8.62 × 10⁻⁵ eV/K. What is $kT$ ⁠ at Room temperature, 300 K?

Correct answer
0.02585 eV ± 0.0005 eV
Setup, choices and feedback
The Scene at this step
Band gap
1.12 eV
Temperature
300 Kfixed
Show the energy levels
On
Accepted within
± 0.0005 eV
Computed from the Scene at this step’s setting
KT, the typical energy of the atoms’ thermal shaking/nodes/solid/outputs/thermalEnergy
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
$kT \approx 0.026$ ⁠ electronvolts, about $43$ ⁠ times less than silicon’s band gap, 1.12 eV. Only a very rare large shake lifts an electron into the conduction band.

Hint
Multiply $k$ ⁠ by $300$ ⁠. The answer is in electronvolts.

Worked solution
1. Given: k = 8.62 × 10⁻⁵ eV/K, T = 300 K.
2. $kT = 8.62 \times 10^{-5} \times 300 = 0.0259$ ⁠.
3. That is about 0.026 eV.
Compare it with the gap: $\dfrac{1.12}{0.026} \approx 43$ ⁠. Ordinary warmth falls far short of that energy, which is why silicon has so few free electrons.
Step 3 of 6
Open this step
Warm silicon
Practice · Predict, then look
Question
Silicon at 300 K has about $10^{10}$ ⁠ free electrons in every cubic centimetre. We warm it to 373 K, the temperature of boiling water. What will happen to the number of free electrons?

Correct answer
- Correct It will grow about a hundred times
Setup, choices and feedback
The Scene at this step
Band gap
1.12 eVfixed
Temperature
300 K
Show the energy levels
On
Choices
- Correct It will grow about a hundred times
- Wrong It will grow about $1.24$ ⁠ times, like the temperature
  Misconception: Thinks the number of free electrons in a semiconductor grows in proportion to the temperature.
  Feedback from the catalogue
  The share of electrons that heat lifts across the gap goes as $e^{-E_g/2kT}$ ⁠: an exponential. A little warming changes the exponent a lot, so the number of free electrons grows tens and hundreds of times, not in the same proportion as the temperature.
- Wrong It will fall: heating spoils conduction, as in a metal
  Misconception: Thinks a semiconductor conducts worse when it is warmer, as a metal does.
  Feedback from the catalogue
  In a metal the number of free electrons is fixed, and warming only adds shaking, so the resistance rises. In a semiconductor warming lifts more and more electrons across the gap, and that effect is far larger: it conducts better when warm.
Once the prediction is locked in, the Scene changes
Temperature
373 K
The Scene then shows: 12.05 · Log₁₀ of the free electrons per cm³ (as many holes in a semiconductor): 10 means 10¹⁰/nodes/solid/outputs/logFreeElectrons
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
At 373 K there are about $10^{12}$ ⁠ free electrons, about a hundred times more. A semiconductor conducts far better when it is warm, which is why electronics need cooling.

Worked solution
1. The share of electrons that heat lifts across the gap goes as $e^{-E_g/2kT}$ ⁠: an exponential, not a simple proportion.
2. At 300 K: $\dfrac{E_g}{2kT} = \dfrac{1.12}{2 \times 0.0259} \approx 21.7$ ⁠. At 373 K: $\dfrac{1.12}{2 \times 0.0321} \approx 17.4$ ⁠.
3. The exponent drops by about $4.3$ ⁠, and $e^{4.3} \approx 74$ ⁠. With the small change in $T^{3/2}$ ⁠ that is about a hundred times.
In a metal the number of free electrons does not depend on heat, so there warming only adds shaking and raises the resistance. In a semiconductor the effect on the number of electrons is far larger.
Step 4 of 6
Open this step
The size of the gap
Practice · Reach the target
Question
Temperature is fixed at 300 K. Change Band gap until the crystal has $10^{12}$ ⁠ free electrons per cubic centimetre, a hundred times more than silicon, and press Check.

Goal
- 12 ± 0.1
  At the start of the step: 10.07
Setup, choices and feedback
The Scene at this step
Band gap
1.12 eV
Temperature
300 Kfixed
Show the energy levels
On
Read from the Scene at
Log₁₀ of the free electrons per cm³ (as many holes in a semiconductor): 10 means 10¹⁰/nodes/solid/outputs/logFreeElectrons
One setting that reaches it
Band gap
0.88 eV
Feedback for a right answer
At a gap of about 0.88 eV. Lowering it by only a quarter of an electronvolt adds a hundred times the electrons, because the number depends on the gap through an exponential.

Feedback for a wrong answer
Not yet. Each drop of about $0.12$ ⁠ electronvolts in the gap multiplies the number about tenfold.

Hint
A smaller gap, more electrons. Read the number on the board.

Worked solution
1. The number goes as $e^{-E_g/2kT}$ ⁠, and $2kT \approx 0.052$ ⁠ electronvolts.
2. A hundredfold increase needs the exponent to fall by $\ln 100 \approx 4.6$ ⁠.
3. $\Delta E_g = 4.6 \times 0.052 \approx 0.24$ ⁠, so a gap of $1.12 - 0.24 \approx 0.88$ ⁠ electronvolts.
That is between silicon and germanium, 0.66 eV, which has almost $10^{14}$ ⁠.
Step 5 of 6
Open this step
Insulators
Practice · Predict, then look
Question
Diamond has a gap of 5.47 eV, and at room temperature it has almost no free electrons. We heat it to 1000 K, Red hot. What will happen?

Correct answer
- Correct A few free electrons will appear, but it will stay an insulator
Setup, choices and feedback
The Scene at this step
Band gap
5.47 eVfixed
Temperature
300 K
Show the energy levels
On
Choices
- Correct A few free electrons will appear, but it will stay an insulator
- Wrong It will become a conductor, like a metal
  Feedback
  Even at 1000 K, $kT$ ⁠ is only about 0.086 electronvolts, $60$ ⁠ times less than the gap. The diamond gains very few free electrons, nowhere near one per atom.
- Wrong Nothing: an insulator has no free electrons at all, at any temperature
  Misconception: Thinks an insulator has exactly no free electrons, at any temperature.
  Feedback from the catalogue
  An insulator is a material with a wide gap, not a different kind of thing. Heat lifts electrons across any gap, only very rarely when the gap is wide: red-hot diamond has millions of free electrons per cubic centimetre, far fewer than a metal but not zero.
Once the prediction is locked in, the Scene changes
Temperature
1000 K
The Scene then shows: 6.474 · Log₁₀ of the free electrons per cm³ (as many holes in a semiconductor): 10 means 10¹⁰/nodes/solid/outputs/logFreeElectrons
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
At 1000 K diamond has a few million free electrons per cubic centimetre, about $10^{6.5}$ ⁠. A metal has about $10^{22.7}$ ⁠: $10^{16}$ ⁠ times more. The difference between an insulator and a semiconductor is the size of the gap, not a different principle.

Worked solution
1. At 1000 K: $kT \approx 0.086$ ⁠ electronvolts, and $\dfrac{E_g}{2kT} \approx \dfrac{5.47}{0.172} \approx 32$ ⁠.
2. $e^{-32} \approx 10^{-14}$ ⁠: very small, but not zero.
3. So a few free electrons appear, millions per cubic centimetre, and diamond is still an insulator.
An insulator is simply a material with a wide gap. Heat lifts electrons across any gap, only at very different rates.
Step 6 of 6
Open this step
A cold metal
Practice · Choose
Question
We cool the Metal to Liquid nitrogen, 77 K. What happens to its number of free electrons?

Correct answer
- Correct It stays almost the same, about one per atom
Setup, choices and feedback
The Scene at this step
Band gap
-1 eV
Temperature
77 K
Show the energy levels
On
Choices
Shown in a random order.
- Correct It stays almost the same, about one per atom
- Wrong It plunges, as in silicon
  Misconception: Thinks a metal loses its free electrons when cooled, as a semiconductor does.
  Feedback from the catalogue
  In a metal the bands overlap, so it has free electrons with no help from heat, about one per atom. Cooling does not reduce their number. It only calms the atoms’ shaking, which is why a cold metal’s resistance is lower.
- Wrong It grows, so the resistance falls
  Feedback
  A metal’s resistance does fall as it cools, but not because there are more electrons: the atoms shake less, and the electrons bump into them less.
Feedback for a right answer
In a metal the free electrons need no heat: the bands overlap, and they have empty levels at any temperature. What cooling changes is the shaking, so a metal’s resistance falls as it cools.

Worked solution
1. In a metal the bands overlap, so electrons spill into the empty band with no help from heat.
2. The number on the board stays about $5 \times 10^{22}$ ⁠ per cubic centimetre, even at 77 K.
3. In silicon at 77 K, by contrast, almost all the free electrons vanish.
So a metal conducts better when it is cold, and a semiconductor conducts better when it is warm.

Across the unit · 34 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Electric charge

Rubbing moves electrons

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Counting electrons

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Neutral paper

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Polarisation

Question

Correct answer

The Scene at this step

Choices

Worked solution

Coulomb's law, field and potential

Coulomb's law

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The square of the distance

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The electric field

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Fields add as vectors

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Potential

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Zero potential

Question

Correct answer

The Scene at this step

Choices

Worked solution

Current, voltage and resistance

What flows

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Ohm's law

Question