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 +8 µC and +6 µC are 50 cm apart. At what distance x from the +8 µC charge, on the line between them, is the electric field zero?
q1 = +8 µCq2 = +6 µCd = 50 cm
x = cm
2.Coulomb’s law & electric fieldsHard
An electron enters the uniform field between two parallel plates moving parallel to them at 1.5 × 10⁷ m/s. The plates are 4 cm long and the field between them is 1200 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 = 1.5 × 10⁷ m/sL = 4 cmE = 1200 N/Ce = 1.602 × 10⁻¹⁹ Cme = 9.109 × 10⁻³¹ kg
y = mm
3.Series & parallel resistorsMedium
Find the equivalent resistance Req seen by the source V1.
V1 = 12 VR1 = 10 ΩR2 = 2 ΩR3 = 2 ΩR4 = 6 Ω
Req = Ω
4.Series & parallel resistorsHard
Find the current IR1 delivered by the source, which flows through R1.
V1 = 15 VR1 = 30 ΩR2 = 30 ΩR3 = 12 ΩR4 = 20 ΩR5 = 6 ΩR6 = 3 Ω
IR1 = 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 = 6 VR1 = 2.2 kΩR2 = 3.3 kΩC = 22 µF
τ = ms
6.RC & RL transientsHard
The switch has been closed for a long time, so the circuit is in steady state. At t = 0 the switch opens. Find the current i2 through R2 at t = 200 ms.
Vs = 20 VR1 = 2.2 kΩR2 = 10 kΩC = 68 µFt = 200 ms
i2 = mA
7.Magnetic fields & inductionMedium
A coil of 50 turns has an area of 40 cm². The coil sits in a uniform field of 0.3 T perpendicular to its plane, and the field is switched off in 20 ms. Find the average induced emf ε.
N = 50A = 40 cm²B = 0.3 TΔt = 20 ms
ε = V
8.Magnetic fields & inductionHard
A short coil of 50 turns is wound tightly round the middle of a long solenoid of 1000 turns, length 50 cm and diameter 2 cm. Find their mutual inductance M.
N1 = 1000N2 = 50l = 50 cm⌀ = 2 cmμ₀ = 4π × 10⁻⁷ T·m/A
M = µH
9.AC impedance & phasorsMedium
A series RLC circuit is connected to a 230 V (rms), 50 Hz supply. Find the real power P taken from the supply.
V = 230 V (rms)f = 50 HzR = 100 ΩL = 100 mHC = 47 µF
P = W
10.AC impedance & phasorsHard
A resistor R1 is connected in series with a parallel combination of R2 and a capacitor C, across a 50 V (rms), 500 Hz supply. Find the magnitude of the impedance |Z|.
V = 50 V (rms)f = 500 HzR1 = 15 ΩR2 = 30 ΩC = 22 µF
|Z| = Ω
3Solutions
1.Coulomb’s law & electric fields
Answer: x = 26.79 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
8/x² = 6/(50 − x)²
(50 − x)/x = √(6/8) = 0.866
Position
x = 50/(1 + 0.866) = 26.79 cm
Answer
x = 26.79 cm
The zero point is always closer to the smaller charge.
2.Coulomb’s law & electric fields
Answer: y = 0.7504 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 = 0.04/1.5 × 10⁷ = 2.667 × 10⁻⁹ s
Sideways acceleration
a = e·E/m = 1.602 × 10⁻¹⁹·1200/9.109 × 10⁻³¹ = 2.11 × 10¹⁴ m/s²
Deflection
y = ½·a·t² = ½·2.11 × 10¹⁴·(2.667 × 10⁻⁹)² = 0.0007504 m = 0.7504 mm
Answer
y = 0.7504 mm
The electron is pushed towards the positive plate, against the field direction.
3.Series & parallel resistors
Answer: Req = 11.6 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.
R3 + R4 = 2 + 6 = 8 Ω
R2 ∥ 8 = 2·8/(2+8) = 1.6 Ω
Req = R1 + 1.6 = 10 + 1.6 = 11.6 Ω
Answer
Req = 11.6 Ω
4.Series & parallel resistors
Answer: IR1 = 0.3629 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 = 6 + 3 = 9 Ω
R4 ∥ 9 = 6.207 Ω
R3 + 6.207 = 18.21 Ω
R2 ∥ 18.21 = 11.33 Ω
Req = R1 + 11.33 = 30 + 11.33 = 41.33 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 = 15/41.33 = 0.3629 A
Answer
IR1 = 0.3629 A
5.RC & RL transients
Answer: τ = 29.04 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 = 2.2·3.3/(2.2 + 3.3) = 1.32 kΩ
vC(∞) = 6·3.3/(2.2 + 3.3) = 3.6 V
Time constant
τ = Rth·C = 1.32·22 = 29.04 ms
Answer
τ = 29.04 ms
6.RC & RL transients
Answer: i2 = 1.222 mA
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 capacitor voltage
In steady state the capacitor is an open circuit, so R₁ and R₂ form a voltage divider. The capacitor voltage cannot jump when the switch opens.
vC(0) = 20·10/(2.2 + 10) = 16.39 V
Time constant after the switch opens
τ = R2·C = 10·68 = 680 ms
Discharge
vC(t) = 16.39·e−200/680 = 12.22 V
i2 = vC/R2 = 12.22/10 = 1.222 mA
Answer
i2 = 1.222 mA
7.Magnetic fields & induction
Answer: ε = 3 V
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 = 0.3·0.004 = 0.0012 Wb
Average emf
ε = NΔΦ/Δt = 50·0.0012/0.02 = 3 V
Answer
ε = 3 V
8.Magnetic fields & induction
Answer: M = 39.48 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).
Mutual inductance
M = μ₀N₁N₂A/l = 4π × 10⁻⁷·1000·50·0.0003142/0.5 = 39.48 µH
Answer
M = 39.48 µH
9.AC impedance & phasors
Answer: P = 467.4 W
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 = 31.42 Ω, XC = 67.73 Ω
Impedance
|Z| = √(100² + (−36.31)²) = 106.4 Ω
Real power
Inductors and capacitors only store and return energy; the resistor absorbs all of the real power.
I = 230/106.4 = 2.162 A
P = I²·R = 2.162²·100 = 467.4 W
Answer
P = 467.4 W
10.AC impedance & phasors
Answer: |Z| = 23.76 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·500·22·10⁻⁶) = 14.47 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 = 5.661 − j11.74 Ω
Total impedance
Z = 15 + 5.661 − j11.74 = 20.66 − j11.74 Ω
|Z| = √(20.66² + 11.74²) = 23.76 Ω
Answer
|Z| = 23.76 Ω
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