1True or false
Electric field lines may cross at a point where the fields of two charges meet.
Show answer
Answer: False
At any point the resultant field has only one direction, so only one line can pass through it.
!Common mistakeLearners sometimes draw each charge's lines separately and let them cross; the diagram must show the resultant field.
2True or false · ★ Challenge
Sliding a sheet of paper between the plates of an isolated charged capacitor lowers the p.d. between the plates.
Show answer
Answer: True
The dielectric raises C while Q cannot change (isolated), so V = Q/C falls.
!Common mistakeLearners often think V is fixed; it is fixed only if a battery stays connected.
3True or false · ★ Challenge
At a point where the resultant electric field is zero, the electric potential must also be zero.
Show answer
Answer: False
Potential is a scalar that adds with signs; midway between two equal positive charges E = 0 but V = 2kq/r.
!Common mistakeTreating field and potential as the same quantity is a common mix-up; one is a vector, the other a scalar.
4True or false
The electric field at a point near a charge exists even if no other charge is placed there to feel it.
Show answer
Answer: True
The field is a property of the space around the source charge; a test charge only reveals it.
!Common mistakeSome learners think the field appears only when a test charge is present; the field is there anyway.
5True or false · ★ Challenge
An oil drop could carry a charge of 2.5 × 10⁻¹⁹ C.
Show answer
Answer: False
Charge comes in whole multiples of e = 1.6 × 10⁻¹⁹ C; 2.5 × 10⁻¹⁹ ÷ 1.6 × 10⁻¹⁹ ≈ 1.6 is not a whole number.
!Common mistakeLearners often treat charge as able to take any value; Millikan's oil-drop results showed it is quantised.
6True or false
The electronvolt is a unit of potential difference.
Show answer
Answer: False
The electronvolt is a unit of energy: 1 eV = 1.6 × 10⁻¹⁹ J, the energy one electron gains through 1 V.
!Common mistakeThe word "volt" in the name misleads many learners; the unit measures energy, not p.d.
7Multiple choice · ★ Challenge
An electron enters the space between two horizontal charged plates moving horizontally. Ignoring gravity, what path does it follow between the plates?
- AA parabola curving towards the positive plate
- BA straight line along the field lines
- CA circle about the centre of the plates
- DA parabola curving towards the negative plate
Show answer
Answer: A. A parabola curving towards the positive plate
The force is constant and perpendicular to its initial velocity, like gravity on a projectile, so the path is a parabola, bending towards the positive plate.
!Common mistakeChoosing the negative plate forgets that the electron is attracted to the positive plate.
8Multiple choice · ★ Challenge
In a field-line diagram for two point charges, the lines from both charges bend away from each other, and there is a point between them with no lines through it. What can you conclude?
- AThe two charges have the same sign
- BThe two charges have opposite signs
- COnly one of the objects is charged
- DBoth objects are neutral metal spheres
Show answer
Answer: A. The two charges have the same sign
Like charges repel, so their lines push apart and there is a neutral point between them where the fields cancel.
!Common mistakeChoosing opposite signs is wrong: lines from unlike charges join from one charge to the other.
9True or false
In a photocopier, black toner powder is attracted only to the charged parts of the drum, which match the dark parts of the original page.
Show answer
Answer: True
Light from the white parts of the page discharges the drum; the dark parts stay charged and attract the oppositely charged toner.
!Common mistakeSome learners think the toner is attracted to the light parts; light removes the charge, so those parts stay clean.
10Multiple choice · ★ Challenge
A negatively charged rod is held near an uncharged metal sphere on an insulating stand. The sphere is touched with a finger, the finger is removed, and then the rod is taken away. What charge does the sphere end with?
- ANegative, the same sign as the rod
- BNo charge, as the rod never touched it
- CPositive, spread evenly over its surface
- DPositive on one side, negative on the other
Show answer
Answer: C. Positive, spread evenly over its surface
The rod repels electrons from the sphere through the finger to earth; with the finger removed they cannot return, so the sphere is left positive.
!Common mistakeChoosing "no charge" assumes charging needs contact with the rod; in induction the earth connection supplies or removes the electrons.
11True or false
When a polythene rod is rubbed with a woollen cloth and becomes negative, it has gained electrons from the wool, and the wool becomes positive.
Show answer
Answer: True
Electrons transfer from the wool to the polythene; the wool is left with a positive charge of the same size.
!Common mistakeSome learners think only the rod gets charged; the cloth always gets an equal and opposite charge.
12Multiple choice · ★ Challenge
During a thunderstorm, why is it safer to sit inside a car with a metal body than to shelter under a lone tree?
- AThe rubber tyres insulate the car completely from the ground below
- BLightning cannot strike moving or parked metal objects at all
- CThe metal body attracts lightning away from the area around it
- DCharge stays on the outside of the metal shell and flows round it to earth
Show answer
Answer: D. Charge stays on the outside of the metal shell and flows round it to earth
The field inside a closed conductor is zero; a strike flows over the metal skin to the ground. A tall tree attracts strikes and current can jump from it to a person.
!Common mistakeChoosing the tyres is a common myth: lightning has already crossed kilometres of air, so a few centimetres of rubber cannot stop it.
13Fill in the blank · ★ Challenge
A 15 μF and a 30 μF capacitor are connected in series. Their combined capacitance is ______ μF.
Show answer
Answer: 10
1/C = 1/15 + 1/30 = 3/30, so C = 10 μF.
!Common mistakeWriting 45 adds them as for parallel; in series the combination is smaller than either capacitor.
14Multiple choice
Which change increases the capacitance of a parallel-plate capacitor?
- AMoving the plates farther apart
- BRaising the p.d. across the plates
- CUsing plates of smaller area
- DMoving the plates closer together
Show answer
Answer: D. Moving the plates closer together
C = ε₀εᵣA/d: a smaller gap d gives a larger C.
!Common mistakeChoosing "raising the p.d." confuses capacitance with charge; C depends only on the shape, size and the material between the plates.
15Fill in the blank · ★ Challenge
A rubbed balloon has a charge of −6.4 nC (e = 1.6 × 10⁻¹⁹ C). It has gained ______ electrons.
Show answer
Answer: 4.0 × 10¹⁰
n = q/e = 6.4 × 10⁻⁹ ÷ 1.6 × 10⁻¹⁹ = 4.0 × 10¹⁰.
!Common mistakeForgetting that nC means 10⁻⁹ C gives an answer a thousand million times too big.
16Multiple choice
A plastic comb rubbed on dry hair attracts small uncharged bits of paper. Why?
- AThe comb induces opposite charge on the near side of each bit of paper
- BThe paper is already charged with the opposite sign to the comb
- CThe comb gives the paper some of its charge through the air
- DRubbing makes the comb magnetic, so it pulls on the paper
Show answer
Answer: A. The comb induces opposite charge on the near side of each bit of paper
The charged comb pulls opposite charge to the near side of the paper and pushes like charge to the far side; the nearer charge is attracted more strongly.
!Common mistakeChoosing "the paper is already charged" is tempting because opposite charges attract, but neutral objects are also attracted by induction.
17Multiple choice · ★ Challenge
At 0.20 m from a point charge the electric field is 900 N/C. What is the field 0.60 m from the charge?
- A300 N/C
- B2700 N/C
- C100 N/C
- D8100 N/C
Show answer
Answer: C. 100 N/C
E ∝ 1/r²; the distance is 3 times larger, so E = 900 ÷ 3² = 100 N/C.
!Common mistakeChoosing 300 N/C uses E ∝ 1/r, which is how potential varies, not field.
18Fill in the blank · ★ Challenge
An electron (e = 1.6 × 10⁻¹⁹ C) is between plates where the field is 5.0 × 10³ N/C. The electric force on it is ______ N.
Show answer
Answer: 8.0 × 10⁻¹⁶
F = eE = 1.6 × 10⁻¹⁹ × 5.0 × 10³ = 8.0 × 10⁻¹⁶ N.
!Common mistakeDividing by the charge instead of multiplying gives 3 × 10²² N, which is absurd for one electron.
19Multiple choice
Why must the hose nozzle be earthed when a fuel tanker fills the underground tanks at a petrol station?
- ASo charge made by flowing fuel goes to earth before a spark forms
- BSo the fuel flows faster by being pulled down to the earth
- CSo that electricity from the pump cannot flow into the fuel
- DSo the fuel becomes charged and sticks to the tank walls
Show answer
Answer: A. So charge made by flowing fuel goes to earth before a spark forms
Fuel rubbing through the hose separates charge; earthing lets it flow away so no spark can ignite the fuel vapour.
!Common mistakeChoosing "fuel flows faster" confuses earthing with gravity; earthing only provides a path for charge.
20Short answer · ★ Challenge
Explain how electrostatic paint spraying gives an even coat on a bicycle frame and wastes little paint.
Show answer
Model answer: The paint droplets are given the same charge, so they repel each other and form a fine, even spray. The frame is earthed (or given the opposite charge), so the droplets are attracted to it and follow the field lines, even round to the back of the tubes. Little paint misses the frame, and thin areas attract more droplets.
!Common mistakeSaying the droplets are attracted "because they are wet" misses the electric force; uncharged droplets would mostly miss the thin tubes.
21Multiple choice · ★ Challenge
An electron is accelerated from rest through a potential difference of 500 V (e = 1.6 × 10⁻¹⁹ C). What kinetic energy does it gain?
- A3.2 × 10⁻²² J, which is 500 eV
- B500 J, which is 3.1 × 10²¹ eV
- C8.0 × 10⁻¹⁷ J, which is 500 eV
- D8.0 × 10⁻¹⁷ J, which is 8.0 × 10⁻¹⁷ eV
Show answer
Answer: C. 8.0 × 10⁻¹⁷ J, which is 500 eV
W = qV = 1.6 × 10⁻¹⁹ × 500 = 8.0 × 10⁻¹⁷ J; one electron through 500 V gains 500 eV.
!Common mistakeChoosing 3.2 × 10⁻²² J divides by 500 instead of multiplying; the energy is charge × p.d.
22Multiple choice
Which describes the electric field lines between two large parallel plates, one positive and one negative, away from the edges?
- ACurved lines bulging outwards from the − plate to the + plate
- BLines spreading out radially from the centre of the + plate
- CParallel, evenly spaced lines from the + plate to the − plate
- DClosed circles running round the two plates
Show answer
Answer: C. Parallel, evenly spaced lines from the + plate to the − plate
Between the plates the field is uniform: same size and direction everywhere, shown by parallel equally spaced lines from + to −.
!Common mistakeChoosing radial lines copies the pattern of a point charge; large plates give a uniform field.
23Short answer · ★ Challenge
Explain why a metal rod held in the bare hand cannot be charged by rubbing, but the same rod can be charged if it is held by a plastic handle.
Show answer
Model answer: Metal is a conductor and the human body also conducts, so any charge produced flows through the hand and body to earth. With an insulating handle the charge has no path to earth and stays on the rod.
!Common mistakeSaying "metals cannot be charged" is wrong; metals can hold charge as long as they are insulated from earth.
24True or false
To find the total force on a charge from two other charges, you may simply add the sizes of the two forces, whatever their directions.
Show answer
Answer: False
Forces are vectors; they add by size only if they act in the same direction, otherwise use vector addition.
!Common mistakeAdding sizes is the most common error with several charges; always draw the force arrows first.
25Short answer · ★ Challenge
Explain why there is a point of zero electric field between two equal positive charges, but no such point between a positive and a negative charge.
Show answer
Model answer: Between two positive charges the fields point away from each charge, so they are in opposite directions and cancel at the midpoint. Between + and −, the field of the positive charge points towards the negative charge and so does the field of the negative charge, so the fields add and can never be zero there.
!Common mistakeSaying "the fields always cancel halfway between two charges" ignores direction; it depends on whether the charges are alike.
26Short answer
Looking at a diagram of electric field lines, how can you tell (a) where the field is strongest, and (b) which way the force on an electron points?
Show answer
Model answer: (a) The field is strongest where the lines are closest together. (b) The force on an electron is opposite to the arrows on the field lines, because the arrows show the force on a positive charge.
!Common mistakeSaying the electron moves along the arrows forgets that the electron is negative.
27Multiple choice · ★ Challenge
Two 22 μF capacitors are connected in series to a 15 V supply. What is the charge on each and the p.d. across each?
- A330 μC on each; 15 V across each
- B165 μC on each; 15 V across each
- C165 μC on each; 7.5 V across each
- D660 μC on each; 7.5 V across each
Show answer
Answer: C. 165 μC on each; 7.5 V across each
Series: C = 22 × 22 ÷ 44 = 11 μF; Q = CV = 11 × 15 = 165 μC, the same on each; V = Q/C = 165 ÷ 22 = 7.5 V each.
!Common mistakeChoosing 330 μC uses 22 μF × 15 V for each capacitor, as if each had the full 15 V; in series the p.d. is shared.
28Multiple choice · ★ Challenge
Charge +2.0 μC is at A and charge −2.0 μC is at B, 0.40 m apart. Point P is on the line AB, 0.10 m beyond B (k = 9 × 10⁹ N m²/C²). What is the resultant field at P?
- A1.9 × 10⁶ N/C towards B
- B1.7 × 10⁶ N/C away from B
- C1.8 × 10⁶ N/C towards B
- D1.7 × 10⁶ N/C towards B
Show answer
Answer: D. 1.7 × 10⁶ N/C towards B
From A (0.50 m): 9 × 10⁹ × 2.0 × 10⁻⁶ ÷ 0.25 = 7.2 × 10⁴ N/C away from A. From B (0.10 m): 1.8 × 10⁶ N/C towards B. These are opposite, so E = 1.8 × 10⁶ − 0.072 × 10⁶ ≈ 1.7 × 10⁶ N/C towards B.
!Common mistakeChoosing 1.9 × 10⁶ N/C adds the fields; outside the pair they point in opposite directions and must be subtracted.
29Multiple choice
Capacitors of 4.7 μF and 2.2 μF are connected in parallel. What is their combined capacitance?
- A6.9 μF
- B1.5 μF
- C2.5 μF
- D10.3 μF
Show answer
Answer: A. 6.9 μF
In parallel, capacitances add: C = 4.7 + 2.2 = 6.9 μF.
!Common mistakeChoosing 1.5 μF uses the product-over-sum rule, which is for capacitors in SERIES (the opposite of resistors).
30Multiple choice · ★ Challenge
In a hydrogen atom the electron is about 5.3 × 10⁻¹¹ m from the proton (e = 1.6 × 10⁻¹⁹ C, k = 9 × 10⁹ N m²/C²). What is the size of the electric force between them?
- A4.3 × 10⁻¹⁸ N
- B5.1 × 10¹¹ N
- C9.1 × 10⁻¹⁸ N
- D8.2 × 10⁻⁸ N
Show answer
Answer: D. 8.2 × 10⁻⁸ N
F = ke²/r² = 9 × 10⁹ × (1.6 × 10⁻¹⁹)² ÷ (5.3 × 10⁻¹¹)² ≈ 8.2 × 10⁻⁸ N.
!Common mistakeChoosing 4.3 × 10⁻¹⁸ N comes from forgetting to square r; Coulomb's law uses r².
31Fill in the blank
A 470 μF capacitor in a radio is charged to 6.0 V. It stores a charge of ______ mC (2 s.f.).
Show answer
Answer: 2.8
Q = CV = 470 × 10⁻⁶ × 6.0 = 2.82 × 10⁻³ C ≈ 2.8 mC.
!Common mistakeDividing C by V (78 μC) is a common slip; charge grows with both C and V, so Q = CV.
32Fill in the blank · ★ Challenge
A +1.0 μC charge lies on a straight line midway between a +5.0 μC charge 0.10 m to its left and a −5.0 μC charge 0.10 m to its right (k = 9 × 10⁹ N m²/C²). The resultant force on it is ______ N, towards the negative charge.
Show answer
Answer: 9.0
Each force = 9 × 10⁹ × 5.0 × 10⁻⁶ × 1.0 × 10⁻⁶ ÷ 0.10² = 4.5 N; both push/pull to the right, so F = 9.0 N.
!Common mistakeWriting 0 assumes the forces cancel as they would for two like charges; here one repels and one attracts, so both point the same way.
33Multiple choice · ★ Challenge
A parallel-plate capacitor has plates of area 0.020 m² separated by 0.50 mm of air (ε₀ = 8.85 × 10⁻¹² F/m). What is its capacitance?
- A0.35 pF
- B350 pF
- C3500 pF
- D35 pF
Show answer
Answer: B. 350 pF
C = ε₀A/d = 8.85 × 10⁻¹² × 0.020 ÷ 0.50 × 10⁻³ = 3.5 × 10⁻¹⁰ F = 350 pF.
!Common mistakeChoosing 0.35 pF divides by 0.50 instead of 0.50 × 10⁻³ m; convert mm to m.
34Short answer
A glass rod rubbed with silk gains a charge of +5.0 nC. What charge does the silk gain? Name the principle you used.
Show answer
Model answer: The silk gains −5.0 nC. Conservation of charge: charge is only moved from one object to the other, so the total stays zero.
!Common mistakeAnswering "+5.0 nC as well" breaks conservation of charge; rubbing separates charge, it does not make new charge.
35Multiple choice · ★ Challenge
A +1.0 μC charge sits at point P. A +5.0 μC charge is 0.25 m west of P and a −4.0 μC charge is 0.40 m north of P (k = 9 × 10⁹ N m²/C²). What is the size of the resultant force on the +1.0 μC charge?
- A0.95 N
- B0.50 N
- C0.72 N
- D0.75 N
Show answer
Answer: D. 0.75 N
From +5.0 μC: 9 × 10⁹ × 5.0 × 10⁻¹² ÷ 0.0625 = 0.72 N east. From −4.0 μC: 9 × 10⁹ × 4.0 × 10⁻¹² ÷ 0.16 = 0.225 N north. F = √(0.72² + 0.225²) ≈ 0.75 N.
!Common mistakeChoosing 0.95 N adds the two forces as numbers; they act at right angles, so use Pythagoras.
36Short answer · ★ Challenge
You have three 10 μF capacitors. Show how to connect all three to obtain 15 μF, and give the other two values you could make using all three.
Show answer
Model answer: Connect two in series (5 μF) and put the third in parallel with that pair: 5 + 10 = 15 μF. Other combinations: all three in parallel = 30 μF; all three in series = 3.3 μF (also two in parallel in series with one = 6.7 μF).
!Common mistakeSwapping the rules for resistors and capacitors is the common mistake: capacitors ADD in parallel.
37Multiple choice
A small object carries a charge of −4.8 × 10⁻¹⁹ C. What does this tell you?
- AIt has three fewer electrons than protons
- BIt has three more electrons than protons
- CIt has lost 4.8 × 10⁻¹⁹ electrons
- DIt has gained thirty electrons
Show answer
Answer: B. It has three more electrons than protons
n = q/e = 4.8 × 10⁻¹⁹ ÷ 1.6 × 10⁻¹⁹ = 3; the negative sign means extra electrons.
!Common mistakeChoosing "three fewer electrons" ignores the sign: a negative object has a surplus of electrons.
38Short answer · ★ Challenge
A parallel-plate capacitor is charged by a battery and then disconnected. The plates are then pulled apart to twice their separation. State what happens to C, Q, V and the stored energy, and explain where any extra energy comes from.
Show answer
Model answer: C halves (C ∝ 1/d). Q stays the same, since the capacitor is isolated. V = Q/C doubles. Energy = Q²/2C doubles. The extra energy comes from the work done pulling the plates apart against their attraction.
!Common mistakeKeeping V constant is the common error; that is only true while the battery is still connected.
39Multiple choice
Two parallel plates 5.0 cm apart are connected to a 2000 V supply. What is the electric field strength between them?
- A400 V/m
- B4.0 × 10⁴ V/m
- C100 V/m
- D1.0 × 10⁵ V/m
Show answer
Answer: B. 4.0 × 10⁴ V/m
E = V/d = 2000 ÷ 0.050 = 4.0 × 10⁴ V/m.
!Common mistakeChoosing 400 V/m divides by 5.0 without changing centimetres into metres.
40Multiple choice · ★ Challenge
A 1000 μF capacitor discharges through a 4.7 kΩ resistor. A capacitor is usually taken as fully discharged after about 5 time constants. Roughly how long is that?
- A4.7 s
- B0.0047 s
- C24 s
- D4700 s
Show answer
Answer: C. 24 s
τ = RC = 4.7 × 10³ × 1000 × 10⁻⁶ = 4.7 s; 5τ ≈ 23.5 s ≈ 24 s.
!Common mistakeChoosing 4.7 s gives one time constant only; after one τ the capacitor still holds 37% of its p.d.
41Short answer · ★ Challenge
Three equal positive charges q sit at the corners of an equilateral triangle of side a. Describe the direction of the resultant force on one of them and show that its size is √3 kq²/a².
Show answer
Model answer: Each of the other two charges repels it with F = kq²/a², along the two sides, 60° apart. The components across the bisector cancel; along the bisector (pointing away from the centre of the triangle) each gives F cos 30°. Resultant = 2F cos 30° = √3 kq²/a².
!Common mistakeAdding the two forces to get 2kq²/a² ignores the 60° angle between them; only their components along the bisector add.
42Multiple choice
A charge of −2.0 μC placed at a point feels a force of 0.010 N towards the east. What is the electric field at that point?
- A5000 N/C towards the west
- B5000 N/C towards the east
- C2.0 × 10⁻⁸ N/C towards the east
- D200 N/C towards the west
Show answer
Answer: A. 5000 N/C towards the west
E = F/q = 0.010 ÷ 2.0 × 10⁻⁶ = 5000 N/C; the field points opposite to the force on a negative charge, so west.
!Common mistakeChoosing east forgets that field direction is defined by the force on a POSITIVE charge.
43Short answer · ★ Challenge
Describe how you could use a small charged polystyrene ball hanging on a nylon thread to compare the electric field strength at different distances from the charged dome of a Van de Graaff generator. Why must the ball's charge be small?
Show answer
Model answer: Hang the ball at measured distances from the dome and measure the angle the thread is pushed from the vertical; tan θ = F/mg, so a larger angle means a larger force F and a stronger field E = F/q. The charge must be small so that it does not change (disturb) the charges on the dome and therefore the field being measured.
!Common mistakeUsing a large charge on the test ball is a common design error; it would move the charges on the dome and change the field.
44Short answer
Describe how the current in the circuit and the p.d. across the capacitor change while a capacitor charges from a battery through a resistor.
Show answer
Model answer: At the start the capacitor is empty, so the current is largest (V/R). As charge builds up, the capacitor p.d. rises, opposing the battery, so the current falls, quickly at first and then more slowly, towards zero. The capacitor p.d. rises from zero towards the battery voltage, quickly at first and then more slowly.
!Common mistakeSaying the current stays constant until the capacitor is full ignores that the growing capacitor p.d. reduces the p.d. across the resistor.
45Multiple choice · ★ Challenge
The charge–voltage graph of a capacitor is a straight line from the origin to (12 V, 600 μC). What are its capacitance and the energy it stores at 12 V?
- A50 μF and 7.2 mJ
- B0.020 μF and 3.6 mJ
- C7200 μF and 3.6 mJ
- D50 μF and 3.6 mJ
Show answer
Answer: D. 50 μF and 3.6 mJ
C = Q/V = 600 μC ÷ 12 V = 50 μF. Energy = area under graph = ½QV = ½ × 600 × 10⁻⁶ × 12 = 3.6 × 10⁻³ J.
!Common mistakeChoosing 7.2 mJ uses QV and forgets the ½; the area under a straight line from the origin is a triangle.
46Multiple choice
How much work is done in moving a charge along an equipotential surface?
- AWork equal to the charge times the field strength
- BNone, because the potential is the same all along it
- CWork equal to the potential times the distance moved
- DWork that depends on the speed at which it is moved
Show answer
Answer: B. None, because the potential is the same all along it
W = qΔV and ΔV = 0 along an equipotential, so no work is done.
!Common mistakeChoosing "charge times field" mixes up force (qE) with work; along an equipotential the force is perpendicular to the motion.
47Short answer · ★ Challenge
At 0.10 m from a positive point charge, the potential is 9000 V. Find the potential and the field strength at 0.30 m from the charge, and explain why they fall by different factors.
Show answer
Model answer: V ∝ 1/r: V = 9000 ÷ 3 = 3000 V. At 0.10 m, E = V/r = 9000 ÷ 0.10 = 9.0 × 10⁴ V/m; E ∝ 1/r², so at 0.30 m, E = 9.0 × 10⁴ ÷ 9 = 1.0 × 10⁴ V/m. Potential is kQ/r but field is kQ/r², so tripling r divides V by 3 and E by 9.
!Common mistakeDividing both by 9 treats potential like field; only the field has r².
48Multiple choice · ★ Challenge
For two fixed point charges, a graph of force F against 1/r² is a straight line through the origin with gradient 0.072 N m². One charge is 2.0 μC (k = 9 × 10⁹ N m²/C²). What is the other charge?
- A8.0 μC
- B4.0 μC
- C0.40 μC
- D8.0 pC
Show answer
Answer: B. 4.0 μC
Gradient = kq₁q₂, so q₂ = 0.072 ÷ (9 × 10⁹ × 2.0 × 10⁻⁶) = 4.0 × 10⁻⁶ C = 4.0 μC.
!Common mistakeChoosing 8.0 pC divides the gradient by k only and forgets that the gradient also contains the known charge q₁.
49Short answer
Two small spheres of equal mass hang from threads and push each other apart. Sphere A carries three times the charge of sphere B. Explain why both threads still make the same angle with the vertical.
Show answer
Model answer: By Newton's third law the electric force of A on B equals the force of B on A in size (F = kqAqB/r² is the same for both). With equal weights and equal sideways forces, both threads tilt by the same angle.
!Common mistakeThinking the more highly charged sphere feels the bigger force forgets that the force depends on the product qAqB, which is the same for both.
50Short answer · ★ Challenge
Before a storm, the potential difference between the base of a thundercloud and the ground 1.5 km below is 4.5 × 10⁸ V. Find the average electric field and compare it with the field needed for air to break down (about 3 × 10⁶ V/m). Why can lightning still start?
Show answer
Model answer: E = V/d = 4.5 × 10⁸ ÷ 1500 = 3.0 × 10⁵ V/m, about ten times smaller than the breakdown field. Lightning starts where the field is much stronger than the average: near pockets of charge in the cloud and near sharp points such as trees and masts, where the field is concentrated.
!Common mistakeUsing d = 1.5 instead of 1500 m gives a field a thousand times too large; convert km to m.