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 +3 µC and a charge of +2 µC are 50 cm apart. Point P lies on the line between them, 25 cm from the first charge. Find the magnitude of the electric field E at P.
q1 = +3 µCq2 = +2 µCr1 = 25 cmr2 = 25 cm
E = N/C
2.Coulomb’s law & electric fieldsHard
An electron enters the uniform field between two parallel plates moving parallel to them at 1 × 10⁷ m/s. The plates are 2 cm long and the field between them is 2300 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 × 10⁷ m/sL = 2 cmE = 2300 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 = 30 VR1 = 20 ΩR2 = 20 ΩR3 = 12 ΩR4 = 12 Ω
IR1 = A
4.Series & parallel resistorsHard
Find the equivalent resistance Req seen by the source V1.
V1 = 30 VR1 = 5 ΩR2 = 2 ΩR3 = 4 ΩR4 = 3 ΩR5 = 15 ΩR6 = 10 Ω
Req = Ω
5.RC & RL transientsMedium
The inductor carries no current before the switch closes at t = 0. Find the energy W stored in the inductor at t = 1 ms, in mJ.
Vs = 24 VR = 15 ΩL = 22 mHt = 1 ms
W = mJ
6.RC & RL transientsHard
Just before the switch closes at t = 0, the capacitor holds 2 V with the polarity shown. How long does it take for vC to reach 6.8 V? Find t.
Vs = 10 VR = 10 kΩC = 22 µFvC(0) = 2 V
t = ms
7.Magnetic fields & inductionMedium
A coil of 500 turns has an area of 40 cm². The coil sits in a uniform field of 0.5 T perpendicular to its plane, and the field is switched off in 50 ms. Find the average induced emf ε.
N = 500A = 40 cm²B = 0.5 TΔt = 50 ms
ε = V
8.Magnetic fields & inductionHard
A coil of 150 turns, 8 cm × 10 cm, rotates in a uniform field of 0.4 T about an axis perpendicular to the field. At what speed n (in revolutions per minute) must it turn to give an rms emf of 10.66 V?
N = 150A = 8 cm × 10 cmB = 0.4 TErms = 10.66 V
n = rpm
9.AC impedance & phasorsMedium
A series RLC circuit is connected to a 50 V (rms), 100 Hz supply. Find the rms current I.
V = 50 V (rms)f = 100 HzR = 10 ΩL = 50 mHC = 10 µF
I = A
10.AC impedance & phasorsHard
A series RLC circuit is connected to a 10 V (rms), 60 Hz supply. Find the rms current I.
V = 10 V (rms)f = 60 HzR = 30 ΩL = 50 mHC = 47 µF
I = A
3Solutions
1.Coulomb’s law & electric fields
Answer: E = 143840 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·3 × 10⁻⁶/0.25² = 431520 N/C
E2 = k·2 × 10⁻⁶/0.25² = 287680 N/C
Directions and sum
Both charges are positive: at P their fields point away from each charge, so in opposite directions.
E = |431520 − 287680| = 143840 N/C
Answer
E = 143840 N/C
2.Coulomb’s law & electric fields
Answer: y = 0.809 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 = 0.02/1 × 10⁷ = 2 × 10⁻⁹ s
Sideways acceleration
a = e·E/m = 1.602 × 10⁻¹⁹·2300/9.109 × 10⁻³¹ = 4.045 × 10¹⁴ m/s²
Deflection
y = ½·a·t² = ½·4.045 × 10¹⁴·(2 × 10⁻⁹)² = 0.000809 m = 0.809 mm
Answer
y = 0.809 mm
The electron is pushed towards the positive plate, against the field direction.
3.Series & parallel resistors
Answer: IR1 = 0.9706 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 = 12 + 12 = 24 Ω
R2 ∥ 24 = 20·24/(20+24) = 10.91 Ω
Req = R1 + 10.91 = 20 + 10.91 = 30.91 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 = 30/30.91 = 0.9706 A
Answer
IR1 = 0.9706 A
4.Series & parallel resistors
Answer: Req = 6.539 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 = 15 + 10 = 25 Ω
R4 ∥ 25 = 2.679 Ω
R3 + 2.679 = 6.679 Ω
R2 ∥ 6.679 = 1.539 Ω
Req = R1 + 1.539 = 5 + 1.539 = 6.539 Ω
Answer
Req = 6.539 Ω
5.RC & RL transients
Answer: W = 6.88 mJ
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 = 22 mH / 15 Ω = 1.467 ms
i(∞) = Vs/R = 24/15 = 1.6 A
Current at the given time
i(t) = 1.6·(1 − e−1/1.467) = 0.7909 A
Stored energy
W = ½·L·i² = ½·22·0.7909² = 6.88 mJ
Answer
W = 6.88 mJ
6.RC & RL transients
Answer: t = 201.6 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) = 2 V, vC(∞) = Vs = 10 V
τ = R·C = 10 kΩ · 22 µF = 220 ms
vC(t) = 10 + (2 − 10)·e−t/τ
Solve for the time
6.8 = 10 + (−8)·e−t/τ → e−t/τ = (6.8 − 10)/(2 − 10) = 0.4
t = τ·ln((2 − 10)/(6.8 − 10)) = 220·ln(2.5) = 201.6 ms
Answer
t = 201.6 ms
7.Magnetic fields & induction
Answer: ε = 20 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 = 0.5·0.004 = 0.002 Wb
Average emf
ε = NΔΦ/Δt = 500·0.002/0.05 = 20 V
Answer
ε = 20 V
8.Magnetic fields & induction
Answer: n = 299.9 rpm
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).
Peak emf
E₀ = √2·Erms = 15.08 V
Angular speed
ω = E₀/(NBA) = 15.08/(150·0.4·0.008) = 31.41 rad/s
Revolutions per minute
n = 60·31.41/(2π) = 299.9 rpm
Answer
n = 299.9 rpm
9.AC impedance & phasors
Answer: I = 0.3902 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|.
Reactances
ω = 2πf = 2π·100 = 628.3 rad/s
XL = ωL = 628.3·0.05 = 31.42 Ω
XC = 1/(ωC) = 1/(628.3·10·10⁻⁶) = 159.2 Ω
Impedance
In series the impedances add as complex numbers: Z = R + j(XL − XC).
|Z| = √(R² + (XL − XC)²) = √(10² + (−127.7)²) = 128.1 Ω
Current
The same current flows through every element of a series circuit.
I = V/|Z| = 50/128.1 = 0.3902 A
Answer
I = 0.3902 A
10.AC impedance & phasors
Answer: I = 0.2079 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|.
Reactances
ω = 2πf = 2π·60 = 377 rad/s
XL = ωL = 377·0.05 = 18.85 Ω
XC = 1/(ωC) = 1/(377·47·10⁻⁶) = 56.44 Ω
Impedance
In series the impedances add as complex numbers: Z = R + j(XL − XC).
|Z| = √(R² + (XL − XC)²) = √(30² + (−37.59)²) = 48.09 Ω
Current
The same current flows through every element of a series circuit.
I = V/|Z| = 10/48.09 = 0.2079 A
Answer
I = 0.2079 A
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