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 +3 µC and +4 µC are 60 cm apart. At what distance x from the +3 µC charge, on the line between them, is the electric field zero?
q1 = +3 µCq2 = +4 µCd = 60 cm
x = cm
2.Coulomb’s law & electric fieldsHard
Three positive point charges sit at the corners of a right triangle: q₃ = 10 µC is at the right-angle corner, q₁ = 1 µC is 6 cm from it along one side and q₂ = 10 µC is 8 cm from it along the other side. Find the magnitude of the net electric force F on q₃.
q1 = 1 µCq2 = 10 µCq3 = 10 µCr13 = 6 cmr23 = 8 cm
F = N
3.Series & parallel resistorsMedium
Find the current IR2 through R2.
V1 = 15 VR1 = 5 ΩR2 = 3 ΩR3 = 12 ΩR4 = 12 Ω
IR2 = A
4.Series & parallel resistorsHard
Find the equivalent resistance Req seen by the source V1.
V1 = 18 VR1 = 4 ΩR2 = 3 ΩR3 = 6 ΩR4 = 4 ΩR5 = 3 ΩR6 = 10 Ω
Req = Ω
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 capacitor voltage vC at t = 15 ms.
Vs = 10 VR1 = 1 kΩR2 = 10 kΩC = 22 µFt = 15 ms
vC = V
6.RC & RL transientsHard
Just before the switch closes at t = 0, the capacitor holds 10 V with the polarity shown. How long does it take for vC to reach 13 V? Find t.
Vs = 20 VR = 10 kΩC = 220 µFvC(0) = 10 V
t = ms
7.Magnetic fields & inductionMedium
A coil of 50 turns has an area of 40 cm². A uniform field perpendicular to the coil rises steadily from 0.1 T to 0.6 T in 10 ms. Find the average induced emf ε.
N = 50A = 40 cm²B: 0.1 T → 0.6 TΔt = 10 ms
ε = V
8.Magnetic fields & inductionHard
A coil of 100 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 71.09 V?
N = 100A = 8 cm × 10 cmB = 0.4 TErms = 71.09 V
n = rpm
9.AC impedance & phasorsMedium
A series RLC circuit is connected to a 50 V (rms), 60 Hz supply. Find the power factor cos φ of the circuit.
V = 50 V (rms)f = 60 HzR = 80 ΩL = 250 mHC = 10 µF
cos φ =
10.AC impedance & phasorsHard
A series RLC circuit is connected to a 24 V (rms), 50 Hz supply. Find the rms current I.
V = 24 V (rms)f = 50 HzR = 60 ΩL = 20 mHC = 22 µF
I = A
3Solutions
1.Coulomb’s law & electric fields
Answer: x = 27.85 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
3/x² = 4/(60 − x)²
(60 − x)/x = √(4/3) = 1.155
Position
x = 60/(1 + 1.155) = 27.85 cm
Answer
x = 27.85 cm
The zero point is always closer to the smaller charge.
2.Coulomb’s law & electric fields
Answer: F = 142.7 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·1·10 × 10⁻¹²/0.06² = 24.97 N
F2 = k·10·10 × 10⁻¹²/0.08² = 140.5 N
Vector sum
F = √(24.97² + 140.5²) = 142.7 N
Answer
F = 142.7 N
3.Series & parallel resistors
Answer: IR2 = 1.739 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 = 3·24/(3+24) = 2.667 Ω
Req = R1 + 2.667 = 5 + 2.667 = 7.667 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 = 15/7.667 = 1.957 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 = 15 − 1.957·5 = 5.217 V
IR2 = VA/R2 = 5.217/3 = 1.739 A
Answer
IR2 = 1.739 A
4.Series & parallel resistors
Answer: Req = 6.254 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 = 3 + 10 = 13 Ω
R4 ∥ 13 = 3.059 Ω
R3 + 3.059 = 9.059 Ω
R2 ∥ 9.059 = 2.254 Ω
Req = R1 + 2.254 = 4 + 2.254 = 6.254 Ω
Answer
Req = 6.254 Ω
5.RC & RL transients
Answer: vC = 4.797 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.
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 = 1·10/(1 + 10) = 0.9091 kΩ
vC(∞) = 10·10/(1 + 10) = 9.091 V
Time constant
τ = Rth·C = 0.9091·22 = 20 ms
Charging curve
vC(t) = 9.091·(1 − e−15/20) = 4.797 V
Answer
vC = 4.797 V
6.RC & RL transients
Answer: t = 784.7 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) = 10 V, vC(∞) = Vs = 20 V
τ = R·C = 10 kΩ · 220 µF = 2200 ms
vC(t) = 20 + (10 − 20)·e−t/τ
Solve for the time
13 = 20 + (−10)·e−t/τ → e−t/τ = (13 − 20)/(10 − 20) = 0.7
t = τ·ln((10 − 20)/(13 − 20)) = 2200·ln(1.429) = 784.7 ms
Answer
t = 784.7 ms
7.Magnetic fields & induction
Answer: ε = 10 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 = (0.6 − 0.1)·0.004 = 0.002 Wb
Average emf
ε = NΔΦ/Δt = 50·0.002/0.01 = 10 V
Answer
ε = 10 V
8.Magnetic fields & induction
Answer: n = 3000 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 = 100.5 V
Angular speed
ω = E₀/(NBA) = 100.5/(100·0.4·0.008) = 314.2 rad/s
Revolutions per minute
n = 60·314.2/(2π) = 3000 rpm
Answer
n = 3000 rpm
9.AC impedance & phasors
Answer: cos φ = 0.4237
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 = 94.25 Ω, XC = 265.3 Ω
Impedance
|Z| = √(80² + (−171)²) = 188.8 Ω
Power factor
cos φ = R/|Z| = 80/188.8 = 0.4237
The circuit is capacitive, so the power factor is leading.
Answer
cos φ = 0.4237
10.AC impedance & phasors
Answer: I = 0.1591 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·50 = 314.2 rad/s
XL = ωL = 314.2·0.02 = 6.283 Ω
XC = 1/(ωC) = 1/(314.2·22·10⁻⁶) = 144.7 Ω
Impedance
In series the impedances add as complex numbers: Z = R + j(XL − XC).
|Z| = √(R² + (XL − XC)²) = √(60² + (−138.4)²) = 150.8 Ω
Current
The same current flows through every element of a series circuit.
I = V/|Z| = 24/150.8 = 0.1591 A
Answer
I = 0.1591 A
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