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
Charges of +4 µC and +2 µC are 60 cm apart. At what distance x from the +4 µC charge, on the line between them, is the electric field zero?
q1 = +4 µCq2 = +2 µCd = 60 cm
x = cm
2.Coulomb’s law & electric fieldsHard
An electron enters the uniform field between two parallel plates moving parallel to them at 5 × 10⁷ m/s. The plates are 10 cm long and the field between them is 4000 N/C. Electron charge 1.602 × 10⁻¹⁹ C, mass 9.109 × 10⁻³¹ kg; ignore gravity. How far y is it deflected sideways when it leaves the plates? Give the answer in mm.
v₀ = 5 × 10⁷ m/sL = 10 cmE = 4000 N/Ce = 1.602 × 10⁻¹⁹ Cme = 9.109 × 10⁻³¹ kg
y = mm
3.Series & parallel resistorsMedium
Find the current IR1 delivered by the source, which flows through R1.
V1 = 18 VR1 = 2 ΩR2 = 6 ΩR3 = 2 ΩR4 = 8 Ω
IR1 = A
4.Series & parallel resistorsHard
Find the current IR2 through R2.
V1 = 9 VR1 = 5 ΩR2 = 3 ΩR3 = 3 ΩR4 = 5 ΩR5 = 5 ΩR6 = 15 Ω
IR2 = A
5.RC & RL transientsMedium
The source Vₛ charges the capacitor through R₁, and R₂ is connected in parallel with C. The capacitor is uncharged when the switch closes at t = 0. Find the time constant τ of the circuit after the switch closes.
Vs = 24 VR1 = 2.2 kΩR2 = 6.8 kΩC = 22 µF
τ = ms
6.RC & RL transientsHard
Just before the switch closes at t = 0, the capacitor holds 4 V with the polarity shown. Find the capacitor voltage vC at t = 200 ms.
Vs = 9 VR = 6.8 kΩC = 68 µFvC(0) = 4 Vt = 200 ms
vC = V
7.Magnetic fields & inductionMedium
A coil of 500 turns, area 50 cm² and resistance 25 Ω forms a closed circuit. A uniform field perpendicular to the coil rises steadily from 0 T to 0.5 T in 200 ms. Find the average induced current I.
N = 500A = 50 cm²B: 0 T → 0.5 TΔt = 200 msR = 25 Ω
I = mA
8.Magnetic fields & inductionHard
A long air-core solenoid has 100 turns, a length of 40 cm and a diameter of 1 cm. Find its self-inductance L.
N = 100l = 40 cm⌀ = 1 cmμ₀ = 4π × 10⁻⁷ T·m/A
L = µH
9.AC impedance & phasorsMedium
A series RLC circuit is connected to a 100 V (rms), 100 Hz supply. Find the power factor cos φ of the circuit.
V = 100 V (rms)f = 100 HzR = 30 ΩL = 100 mHC = 22 µF
cos φ =
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), 100 Hz supply. Find the magnitude of the impedance |Z|.
V = 12 V (rms)f = 100 HzR1 = 25 ΩR2 = 50 ΩC = 10 µF
|Z| = Ω
3Solutions
1.Coulomb’s law & electric fields
Answer: x = 35.15 cm
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.
Equal fields
4/x² = 2/(60 − x)²
(60 − x)/x = √(2/4) = 0.7071
Position
x = 60/(1 + 0.7071) = 35.15 cm
Answer
x = 35.15 cm
The zero point is always closer to the smaller charge.
2.Coulomb’s law & electric fields
Answer: y = 1.407 mm
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.
Time between the plates
t = L/v₀ = 0.1/5 × 10⁷ = 2 × 10⁻⁹ s
Sideways acceleration
a = e·E/m = 1.602 × 10⁻¹⁹·4000/9.109 × 10⁻³¹ = 7.035 × 10¹⁴ m/s²
Deflection
y = ½·a·t² = ½·7.035 × 10¹⁴·(2 × 10⁻⁹)² = 0.001407 m = 1.407 mm
Answer
y = 1.407 mm
The electron is pushed towards the positive plate, against the field direction.
3.Series & parallel resistors
Answer: IR1 = 3.13 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 = 2 + 8 = 10 Ω
R2 ∥ 10 = 6·10/(6+10) = 3.75 Ω
Req = R1 + 3.75 = 2 + 3.75 = 5.75 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 = 18/5.75 = 3.13 A
Answer
IR1 = 3.13 A
4.Series & parallel resistors
Answer: IR2 = 0.8873 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.
R5 + R6 = 5 + 15 = 20 Ω
R4 ∥ 20 = 4 Ω
R3 + 4 = 7 Ω
R2 ∥ 7 = 2.1 Ω
Req = R1 + 2.1 = 5 + 2.1 = 7.1 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 = 9/7.1 = 1.268 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 = 9 − 1.268·5 = 2.662 V
IR2 = VA/R2 = 2.662/3 = 0.8873 A
Answer
IR2 = 0.8873 A
5.RC & RL transients
Answer: τ = 36.57 ms
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.
Thévenin equivalent seen by the capacitor
Short the source to find Rth; open the capacitor to find the voltage it will settle at.
Rth = R1 ∥ R2 = 2.2·6.8/(2.2 + 6.8) = 1.662 kΩ
vC(∞) = 24·6.8/(2.2 + 6.8) = 18.13 V
Time constant
τ = Rth·C = 1.662·22 = 36.57 ms
Answer
τ = 36.57 ms
6.RC & RL transients
Answer: vC = 5.756 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) = 4 V, vC(∞) = Vs = 9 V
τ = R·C = 6.8 kΩ · 68 µF = 462.4 ms
vC(t) = 9 + (4 − 9)·e−t/τ
Voltage at the given time
vC(200) = 9 + (−5)·e−200/462.4 = 5.756 V
Answer
vC = 5.756 V
7.Magnetic fields & induction
Answer: I = 250 mA
Electromagnetic induction: the flux through a loop is Φ = BA·cos θ, with θ between B and the normal of the loop. Faraday's law: a change of flux induces emf = −N·dΦ/dt in a coil of N turns, and Lenz's law says the induced current flows so that its own field opposes the change. A rod moving across a field gives emf = BLv. A coil rotating at ω gives e = NBAω·sin ωt, peak NBAω. Self-inductance: emf = −L·di/dt, with L = μ₀N²A/l for a long solenoid, and the energy stored is W = ½LI². Two coupled coils: emf₂ = −M·di₁/dt with M = k√(L₁L₂).
Change of flux per turn
ΔΦ = (0.5 − 0)·0.005 = 0.0025 Wb
Average emf
ε = NΔΦ/Δt = 500·0.0025/0.2 = 6.25 V
Current
I = ε/R = 6.25/25 = 0.25 A
Answer
I = 250 mA
8.Magnetic fields & induction
Answer: L = 2.467 µH
Electromagnetic induction: the flux through a loop is Φ = BA·cos θ, with θ between B and the normal of the loop. Faraday's law: a change of flux induces emf = −N·dΦ/dt in a coil of N turns, and Lenz's law says the induced current flows so that its own field opposes the change. A rod moving across a field gives emf = BLv. A coil rotating at ω gives e = NBAω·sin ωt, peak NBAω. Self-inductance: emf = −L·di/dt, with L = μ₀N²A/l for a long solenoid, and the energy stored is W = ½LI². Two coupled coils: emf₂ = −M·di₁/dt with M = k√(L₁L₂).
Cross-section
A = π·(0.005)² = 7.854 × 10⁻⁵ m²
Self-inductance
L = μ₀N²A/l = 4π × 10⁻⁷·100²·7.854 × 10⁻⁵/0.4 = 2.467 × 10⁻⁶ H = 2.467 µH
Answer
L = 2.467 µH
9.AC impedance & phasors
Answer: cos φ = 0.9532
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|.
Reactances
XL = 62.83 Ω, XC = 72.34 Ω
Impedance
|Z| = √(30² + (−9.511)²) = 31.47 Ω
Power factor
cos φ = R/|Z| = 30/31.47 = 0.9532
The circuit is capacitive, so the power factor is leading.
Answer
cos φ = 0.9532
10.AC impedance & phasors
Answer: |Z| = 71.94 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
XC = 1/(2π·100·10·10⁻⁶) = 159.2 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 = 45.51 − j14.3 Ω
Total impedance
Z = 25 + 45.51 − j14.3 = 70.51 − j14.3 Ω
|Z| = √(70.51² + 14.3²) = 71.94 Ω
Answer
|Z| = 71.94 Ω
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