Coulomb’s law & electric fields
- Coulomb’s law
- F = k·|q1q2|/r²
- Field of a point charge
- E = k·|q|/r²
- Constant
- k = 8.99 × 10⁹ N·m²/C²
- Force on a charge
- F = q·E
Physics 2: Electricity and Magnetism: cram pack
Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.
1Formulas and methods
Series & parallel resistors
- Resistors in series
- Req = R1 + R2 + …
- Resistors in parallel
- 1/Req = 1/R1 + 1/R2 + …
- Two in parallel
- Req = R1R2 / (R1 + R2)
- Ohm's law
- V = I·R
- Power
- P = V·I = I²R = V²/R
Method
- Start at the end of the network furthest from the terminals.
- Combine resistors in series by adding them, and resistors in parallel with 1/R = 1/R₁ + 1/R₂ + …
- Redraw the circuit after each step.
- Repeat until one resistor is left between the terminals.
RC & RL transients
- Time constant
- τ = RC | τ = L/R
- Response
- x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ
- Capacitor at DC
- open circuit; vC cannot jump
- Inductor at DC
- short circuit; iL cannot jump
- Settled
- t ≈ 5τ
Method
- Find the initial value just after switching: a capacitor voltage and an inductor current cannot jump.
- Find the final value long after switching: a capacitor acts as an open circuit, an inductor as a short circuit.
- Find the time constant: τ = RC or τ = L/R, using the resistance seen by the capacitor or inductor.
- Use x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ.
Magnetic fields & induction
- Long straight wire
- B = μ0I/(2πr)
- Centre of a flat coil
- B = μ0NI/(2R)
- Long solenoid
- B = μ0nI, n = N/L
- Constant
- μ0 = 4π × 10⁻⁷ T·m/A
- Force on a wire
- F = BIL·sin θ
- Force on a charge
- F = |q|vB·sin θ
- Circular motion
- r = mv/(|q|B), T = 2πm/(|q|B)
- Parallel wires
- F/L = μ0I1I2/(2πd)
- Flux
- Φ = BA·cos θ
- Faraday’s law
- ε = −N·ΔΦ/Δt
- Moving rod
- ε = BLv
- Generator
- E0 = NBAω, Erms = E0/√2
- Inductance
- ε = L·ΔI/Δt, L = μ0N²A/l, W = ½LI²
- Mutual inductance
- ε2 = M·di1/dt, M = k√(L1L2)
- Velocity selector and spectrometer
- v = E/B, r = √(2mV/|q|)/B
Method
- Convert every quantity to SI units: cm to m, cm² to m², mT and µT to T.
- Pick the right formula: B = μ₀I/(2πr) for a wire, μ₀NI/(2R) for the centre of a coil, μ₀nI inside a solenoid; F = BIL·sin θ or |q|vB·sin θ for forces.
- For induction, find the change of flux ΔΦ = Δ(BA·cos θ) and use ε = N·ΔΦ/Δt, or ε = BLv for a moving rod and NBAω for a rotating coil.
- Find directions separately with the right-hand rules, and Lenz’s law for induced currents.
AC impedance & phasors
- Impedances
- ZR = R, ZL = jωL, ZC = 1/(jωC)
- Angular frequency
- ω = 2πf
- Magnitude and angle
- |Z| = √(R² + X²), φ = atan(X/R)
- Phasor Ohm’s law
- V = I·Z
Method
- Convert the frequency: ω = 2πf.
- Find each reactance: XL = ωL and XC = 1/(ωC).
- Add impedances like resistors, with Z = R + jX.
- Find the magnitude |Z| = √(R² + X²) and the angle φ = atan(X/R), then use Ohm’s law with phasors.
2Practice problems
1.Coulomb’s law & electric fieldsMedium
A charge of +1 µC and a charge of +1 µC are 50 cm apart. Point P lies on the line between them, 37.5 cm from the first charge. Find the magnitude of the electric field E at P.
q1 = +1 µCq2 = +1 µCr1 = 37.5 cmr2 = 12.5 cm
E = N/C
2.Coulomb’s law & electric fieldsHard
Two identical small balls, each of mass 5 g, hang from the same point on insulating threads 100 cm long. They are given equal charges and each thread settles at 8° from the vertical. Use g = 9.81 m/s². Find the charge q on each ball in nC. Use k = 8.99 × 10⁹ N·m²/C².
m = 5 gL = 100 cmθ = 8°g = 9.81 m/s²
q = nC
3.Series & parallel resistorsMedium
Find the current IR2 through R2.
V1 = 6 VR1 = 20 ΩR2 = 5 ΩR3 = 15 ΩR4 = 5 Ω
IR2 = A
4.Series & parallel resistorsHard
Find the equivalent resistance Req seen by the source V1.
V1 = 36 VR1 = 3 ΩR2 = 8 ΩR3 = 8 ΩR4 = 20 ΩR5 = 4 ΩR6 = 5 Ω
Req = Ω
5.RC & RL transientsMedium
The inductor carries a steady current of 0.5 A. At t = 0 the switch closes the loop through R and the source is removed, so the inductor current decays through R. Find the voltage vR across R at t = 1 ms.
i(0) = 0.5 AR = 100 ΩL = 22 mHt = 1 ms
vR = V
6.RC & RL transientsHard
Just before the switch closes at t = 0, the capacitor holds 2 V with the polarity shown. Find the capacitor voltage vC at t = 150 ms.
Vs = 6 VR = 6.8 kΩC = 68 µFvC(0) = 2 Vt = 150 ms
vC = V
7.Magnetic fields & inductionMedium
Two long parallel wires 12 cm apart carry currents of 10 A and 20 A in opposite directions. Find the force per metre of length F/L between them.
I1 = 10 AI2 = 20 Ad = 12 cmμ₀ = 4π × 10⁻⁷ T·m/A
F/L = µN/m
8.Magnetic fields & inductionHard
An alpha particle starts from rest, is accelerated through a potential difference of 10000 V and then enters a uniform magnetic field of 0.2 T at right angles. Find the radius r of its path in the field.
m = 6.645 × 10⁻²⁷ kg|q| = 3.204 × 10⁻¹⁹ CV = 10000 VB = 0.2 T
r = cm
9.AC impedance & phasorsMedium
A resistor and an inductor are connected in parallel across a 230 V (rms), 200 Hz supply. Find the magnitude of the impedance |Z|.
V = 230 V (rms)f = 200 HzR = 100 ΩL = 100 mH
|Z| = Ω
10.AC impedance & phasorsHard
A resistor R1 is connected in series with a parallel combination of R2 and a capacitor C, across a 12 V (rms), 500 Hz supply. Find the rms current I.
V = 12 V (rms)f = 500 HzR1 = 40 ΩR2 = 80 ΩC = 10 µF
I = A
3Solutions
1.Coulomb’s law & electric fields
Answer: E = 511431 N/C
Coulomb’s law: two point charges attract or repel with F = k·|q₁·q₂|/r², with k = 8.99 × 10⁹ N·m²/C². The field of a point charge is E = k·|q|/r², pointing away from a positive charge and towards a negative one. Forces and fields from several charges add as vectors.
Field of each charge
E1 = k·1 × 10⁻⁶/0.375² = 63929 N/C
E2 = k·1 × 10⁻⁶/0.125² = 575360 N/C
Directions and sum
Both charges are positive: at P their fields point away from each charge, so in opposite directions.
E = |63929 − 575360| = 511431 N/C
Answer
E = 511431 N/C
2.Coulomb’s law & electric fields
Answer: q = 243.7 nC
Coulomb’s law: two point charges attract or repel with F = k·|q₁·q₂|/r², with k = 8.99 × 10⁹ N·m²/C². The field of a point charge is E = k·|q|/r², pointing away from a positive charge and towards a negative one. Forces and fields from several charges add as vectors.
Electric force from the equilibrium
The tension's horizontal part balances F, its vertical part balances m·g.
F = m·g·tan θ = 0.005·9.81·tan 8° = 0.006894 N
Separation
r = 2·L·sin θ = 2·1·sin 8° = 0.2783 m
Charge
F = k·q²/r² → q = r·√(F/k) = 0.2783·√(0.006894/8.99 × 10⁹) = 2.437 × 10⁻⁷ C = 243.7 nC
Answer
q = 243.7 nC
3.Series & parallel resistors
Answer: IR2 = 0.2 A
Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.
Combine resistors from the far end
Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.
R3 + R4 = 15 + 5 = 20 Ω
R2 ∥ 20 = 5·20/(5+20) = 4 Ω
Req = R1 + 4 = 20 + 4 = 24 Ω
Current from the source
The source sees only Req, so Ohm’s law gives the current it delivers. That current flows through R₁.
IR1 = V1/Req = 6/24 = 0.25 A
Current through R₂
Subtract the drop across R₁ to get the voltage at node A, which is the voltage across R₂.
VA = V1 − IR1·R1 = 6 − 0.25·20 = 1 V
IR2 = VA/R2 = 1/5 = 0.2 A
Answer
IR2 = 0.2 A
4.Series & parallel resistors
Answer: Req = 8.118 Ω
Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.
Combine resistors from the far end
Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.
R5 + R6 = 4 + 5 = 9 Ω
R4 ∥ 9 = 6.207 Ω
R3 + 6.207 = 14.21 Ω
R2 ∥ 14.21 = 5.118 Ω
Req = R1 + 5.118 = 3 + 5.118 = 8.118 Ω
Answer
Req = 8.118 Ω
5.RC & RL transients
Answer: vR = 0.5308 V
First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.
Time constant
τ = L/R = 22 mH / 100 Ω = 0.22 ms
Decay of the current
i(t) = 0.5·e−1/0.22 = 0.005308 A
vR = R·i = 100·0.005308 = 0.5308 V
Answer
vR = 0.5308 V
6.RC & RL transients
Answer: vC = 3.108 V
First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.
Initial value, final value, time constant
The capacitor voltage cannot jump, so it starts at its stored value and heads for the source voltage.
vC(0) = 2 V, vC(∞) = Vs = 6 V
τ = R·C = 6.8 kΩ · 68 µF = 462.4 ms
vC(t) = 6 + (2 − 6)·e−t/τ
Voltage at the given time
vC(150) = 6 + (−4)·e−150/462.4 = 3.108 V
Answer
vC = 3.108 V
7.Magnetic fields & induction
Answer: F/L = 333.3 µN/m
Magnetic forces: a straight wire carrying I in a field B feels F = BIL·sin θ, and a moving charge F = |q|vB·sin θ, where θ is the angle between the current (or velocity) and B. The force is perpendicular to both: point the fingers of the right hand along I or v, curl them towards B, and the thumb gives F for a positive charge; reverse it for a negative charge. Because F is perpendicular to v, a charge moving across a uniform field moves on a circle: |q|vB = mv²/r gives r = mv/(|q|B) and the period T = 2πm/(|q|B), which does not depend on the speed. Two parallel wires attract for currents in the same direction: F/L = μ₀I₁I₂/(2πd).
Force per metre
F/L = μ₀I₁I₂/(2πd) = 2 × 10⁻⁷·10·20/0.12 = 333.3 µN/m
Answer
F/L = 333.3 µN/m
Currents in opposite directions repel.
8.Magnetic fields & induction
Answer: r = 10.18 cm
Magnetic forces: a straight wire carrying I in a field B feels F = BIL·sin θ, and a moving charge F = |q|vB·sin θ, where θ is the angle between the current (or velocity) and B. The force is perpendicular to both: point the fingers of the right hand along I or v, curl them towards B, and the thumb gives F for a positive charge; reverse it for a negative charge. Because F is perpendicular to v, a charge moving across a uniform field moves on a circle: |q|vB = mv²/r gives r = mv/(|q|B) and the period T = 2πm/(|q|B), which does not depend on the speed. Two parallel wires attract for currents in the same direction: F/L = μ₀I₁I₂/(2πd).
Speed after the acceleration
v = √(2|q|V/m) = √(2·3.204 × 10⁻¹⁹·10000/(6.645 × 10⁻²⁷)) = 982005 m/s
Radius in the field
r = mv/(|q|B) = 0.1018 m = 10.18 cm
Answer
r = 10.18 cm
Combined: r = √(2mV/|q|)/B, so heavier ions of the same charge move on larger circles; this separates isotopes.
9.AC impedance & phasors
Answer: |Z| = 78.25 Ω
AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.
Reactance
XL = ωL = 2π·200·0.1 = 125.7 Ω
Branch currents
Both branches are connected straight across the supply, so each sees the full voltage V.
IR = V/R = 230/100 = 2.3 A
IX = V/XL = 230/125.7 = 1.83 A
Total current
The resistor current is in phase with V and the reactive current is 90° out of phase, so they add like the sides of a right triangle.
I = √(IR² + IX²) = √(2.3² + 1.83²) = 2.939 A
Impedance
|Z| = V/I = 230/2.939 = 78.25 Ω
Answer
|Z| = 78.25 Ω
10.AC impedance & phasors
Answer: I = 0.2073 A
AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.
Reactance
XC = 1/(2π·500·10·10⁻⁶) = 31.83 Ω
Parallel combination
Multiply top and bottom by the complex conjugate to split Zp into real and imaginary parts.
Zp = R₂·(−jXC)/(R₂ − jXC) = (R₂·XC² − jR₂²·XC)/(R₂² + XC²)
Zp = 10.93 − j27.48 Ω
Total impedance
Z = 40 + 10.93 − j27.48 = 50.93 − j27.48 Ω
|Z| = √(50.93² + 27.48²) = 57.87 Ω
Current
I = V/|Z| = 12/57.87 = 0.2073 A
Answer
I = 0.2073 A
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