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 +3 µC are 40 cm apart. Point P lies on the line between them, 10 cm from the first charge. Find the magnitude of the electric field E at P.
q1 = +1 µCq2 = +3 µCr1 = 10 cmr2 = 30 cm
E = N/C
2.Coulomb’s law & electric fieldsHard
Three positive point charges sit at the corners of a right triangle: q₃ = 5 µC is at the right-angle corner, q₁ = 1 µC is 8 cm from it along one side and q₂ = 2 µC is 4 cm from it along the other side. Find the magnitude of the net electric force F on q₃.
q1 = 1 µCq2 = 2 µCq3 = 5 µCr13 = 8 cmr23 = 4 cm
F = N
3.Series & parallel resistorsMedium
Find the equivalent resistance Req seen by the source V1.
V1 = 30 VR1 = 4 ΩR2 = 10 ΩR3 = 2 ΩR4 = 8 Ω
Req = Ω
4.Series & parallel resistorsHard
Find the equivalent resistance Req seen by the source V1.
V1 = 15 VR1 = 10 ΩR2 = 2 ΩR3 = 20 ΩR4 = 8 ΩR5 = 2 ΩR6 = 6 Ω
Req = Ω
5.RC & RL transientsMedium
The inductor carries no current before the switch closes at t = 0. Find the current i at t = 5 ms.
Vs = 9 VR = 22 ΩL = 47 mHt = 5 ms
i = A
6.RC & RL transientsHard
Just before the switch closes at t = 0, the capacitor holds 30 V with the polarity shown. How long does it take for vC to reach 27 V? Find t.
Vs = 20 VR = 2.2 kΩC = 220 µFvC(0) = 30 V
t = ms
7.Magnetic fields & inductionMedium
A square loop moves to the right into a region with a uniform magnetic field pointing out of the page. Click the direction of the induced current in the loop as seen in the drawing, while the loop is entering the field.
B: out of the pageentering the field
- 1clockwise
- 2counterclockwise
8.Magnetic fields & inductionHard
A coil of 20 turns, 15 cm × 10 cm, rotates at 300 rpm in a uniform field of 0.4 T about an axis perpendicular to the field. Find the peak emf E0.
N = 20A = 15 cm × 10 cmB = 0.4 Tn = 300 rpm
E0 = V
9.AC impedance & phasorsMedium
A series RLC circuit is connected to a 24 V (rms), 500 Hz supply. Find the real power P taken from the supply.
V = 24 V (rms)f = 500 HzR = 100 ΩL = 250 mHC = 47 µF
P = W
10.AC impedance & phasorsHard
A series RLC circuit is driven by a 5 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.
V = 5 V (rms)R = 2 ΩL = 20 mHC = 2.2 µF
VL = V
3Solutions
1.Coulomb’s law & electric fields
Answer: E = 599333 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 × 10⁻⁶/0.1² = 899000 N/C
E2 = k·3 × 10⁻⁶/0.3² = 299667 N/C
Directions and sum
Both charges are positive: at P their fields point away from each charge, so in opposite directions.
E = |899000 − 299667| = 599333 N/C
Answer
E = 599333 N/C
2.Coulomb’s law & electric fields
Answer: F = 56.62 N
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.
Each force on q₃
F1 = k·1·5 × 10⁻¹²/0.08² = 7.023 N
F2 = k·2·5 × 10⁻¹²/0.04² = 56.19 N
Vector sum
F = √(7.023² + 56.19²) = 56.62 N
Answer
F = 56.62 N
3.Series & parallel resistors
Answer: Req = 9 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 + 8 = 10 Ω
R2 ∥ 10 = 10·10/(10+10) = 5 Ω
Req = R1 + 5 = 4 + 5 = 9 Ω
Answer
Req = 9 Ω
4.Series & parallel resistors
Answer: Req = 11.85 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 = 2 + 6 = 8 Ω
R4 ∥ 8 = 4 Ω
R3 + 4 = 24 Ω
R2 ∥ 24 = 1.846 Ω
Req = R1 + 1.846 = 10 + 1.846 = 11.85 Ω
Answer
Req = 11.85 Ω
5.RC & RL transients
Answer: i = 0.3697 A
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 and final current
Long after the switch closes the inductor behaves like a short circuit, so the final current is set by R alone.
τ = L/R = 47 mH / 22 Ω = 2.136 ms
i(∞) = Vs/R = 9/22 = 0.4091 A
Current at the given time
i(t) = i(∞)·(1 − e−t/τ) = 0.4091·(1 − e−5/2.136) = 0.3697 A
Answer
i = 0.3697 A
6.RC & RL transients
Answer: t = 172.6 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.
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) = 30 V, vC(∞) = Vs = 20 V
τ = R·C = 2.2 kΩ · 220 µF = 484 ms
vC(t) = 20 + (30 − 20)·e−t/τ
Solve for the time
27 = 20 + (10)·e−t/τ → e−t/τ = (27 − 20)/(30 − 20) = 0.7
t = τ·ln((30 − 20)/(27 − 20)) = 484·ln(1.429) = 172.6 ms
Answer
t = 172.6 ms
7.Magnetic fields & induction
Answer: clockwise
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
The flux out of the page increases, or the flux into the page decreases.
Field of the induced current
The induced current opposes the change.
Induced field inside the loop: into the page.
Direction of the current
Right-hand grip rule: a counterclockwise current gives a field out of the page inside the loop.
clockwise
Answer
clockwise
8.Magnetic fields & induction
Answer: E0 = 3.77 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).
Angular speed and area
ω = 2π·300/60 = 31.42 rad/s
A = 0.015 m²
Peak emf
E₀ = NBAω = 20·0.4·0.015·31.42 = 3.77 V
Answer
E0 = 3.77 V
9.AC impedance & phasors
Answer: P = 0.09347 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 = 785.4 Ω, XC = 6.773 Ω
Impedance
|Z| = √(100² + (778.6)²) = 785 Ω
Real power
Inductors and capacitors only store and return energy; the resistor absorbs all of the real power.
I = 24/785 = 0.03057 A
P = I²·R = 0.03057²·100 = 0.09347 W
Answer
P = 0.09347 W
10.AC impedance & phasors
Answer: VL = 238.4 V
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|.
Resonance
At resonance the reactances cancel, so the impedance is just R.
ω₀ = 1/√(LC) = 1/√(0.02·2.2·10⁻⁶) = 4767 rad/s
I = V/R = 5/2 = 2.5 A
Voltage across the inductor
XL = ω₀L = 4767·0.02 = 95.35 Ω
VL = I·XL = 2.5·95.35 = 238.4 V
Answer
VL = 238.4 V
This is Q = 47.67 times the supply voltage.
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