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
Three point charges lie on a straight line: +1 µC on the left, +4 µC on the right 40 cm away, and −4 µC between them, 10 cm from the left charge. Find the magnitude of the net electric force F on the middle charge. Use k = 8.99 × 10⁹ N·m²/C².
qL = +1 µCqR = +4 µCqM = −4 µCrL = 10 cmrR = 30 cm
F = N
2.Coulomb’s law & electric fieldsHard
Three positive point charges sit at the corners of a right triangle: q₃ = 6 µC is at the right-angle corner, q₁ = 1 µC is 6 cm from it along one side and q₂ = 5 µC is 4 cm from it along the other side. Find the magnitude of the net electric force F on q₃.
q1 = 1 µCq2 = 5 µCq3 = 6 µCr13 = 6 cmr23 = 4 cm
F = N
3.Series & parallel resistorsMedium
Find the current IR2 through R2.
V1 = 30 VR1 = 15 ΩR2 = 10 ΩR3 = 4 ΩR4 = 2 Ω
IR2 = A
4.Series & parallel resistorsHard
Find the equivalent resistance Req seen by the source V1.
V1 = 6 VR1 = 3 ΩR2 = 2 ΩR3 = 3 ΩR4 = 30 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 = 1 ms.
Vs = 20 VR = 33 ΩL = 33 mHt = 1 ms
i = A
6.RC & RL transientsHard
Just before the switch closes at t = 0, the capacitor holds 15 V with the polarity shown. Find the capacitor voltage vC at t = 100 ms.
Vs = 24 VR = 4.7 kΩC = 68 µFvC(0) = 15 Vt = 100 ms
vC = V
7.Magnetic fields & inductionMedium
A metal rod 25 cm long slides at 4 m/s along two rails in a uniform field of 0.2 T into the page. Find the induced emf ε between its ends.
B = 0.2 TL = 25 cmv = 4 m/s
ε = mV
8.Magnetic fields & inductionHard
A short coil of 200 turns is wound tightly round the middle of a long solenoid of 500 turns, length 20 cm and diameter 4 cm. Find their mutual inductance M.
N1 = 500N2 = 200l = 20 cm⌀ = 4 cmμ₀ = 4π × 10⁻⁷ T·m/A
M = µH
9.AC impedance & phasorsMedium
A series RLC circuit is connected to a 120 V (rms), 50 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).
V = 120 V (rms)f = 50 HzR = 80 ΩL = 250 mHC = 100 µF
φ = °
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), 60 Hz supply. Find the rms current I.
V = 50 V (rms)f = 60 HzR1 = 50 ΩR2 = 150 ΩC = 4.7 µF
I = A
3Solutions
1.Coulomb’s law & electric fields
Answer: F = 1.998 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.
Force from each charge
Right is positive.
FL = −k·1·4 × 10⁻¹²/0.1² = −3.596 N
FR = +k·4·4 × 10⁻¹²/0.3² = 1.598 N
Net force
F = −3.596 + (1.598) = −1.998 N → |F| = 1.998 N
Answer
F = 1.998 N
The net force points to the left.
2.Coulomb’s law & electric fields
Answer: F = 169.2 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·1·6 × 10⁻¹²/0.06² = 14.98 N
F2 = k·5·6 × 10⁻¹²/0.04² = 168.6 N
Vector sum
F = √(14.98² + 168.6²) = 169.2 N
Answer
F = 169.2 N
3.Series & parallel resistors
Answer: IR2 = 0.6 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 = 4 + 2 = 6 Ω
R2 ∥ 6 = 10·6/(10+6) = 3.75 Ω
Req = R1 + 3.75 = 15 + 3.75 = 18.75 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 = 30/18.75 = 1.6 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 = 30 − 1.6·15 = 6 V
IR2 = VA/R2 = 6/10 = 0.6 A
Answer
IR2 = 0.6 A
4.Series & parallel resistors
Answer: Req = 4.647 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 = 6.316 Ω
R3 + 6.316 = 9.316 Ω
R2 ∥ 9.316 = 1.647 Ω
Req = R1 + 1.647 = 3 + 1.647 = 4.647 Ω
Answer
Req = 4.647 Ω
5.RC & RL transients
Answer: i = 0.3831 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 = 33 mH / 33 Ω = 1 ms
i(∞) = Vs/R = 20/33 = 0.6061 A
Current at the given time
i(t) = i(∞)·(1 − e−t/τ) = 0.6061·(1 − e−1/1) = 0.3831 A
Answer
i = 0.3831 A
6.RC & RL transients
Answer: vC = 17.42 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) = 15 V, vC(∞) = Vs = 24 V
τ = R·C = 4.7 kΩ · 68 µF = 319.6 ms
vC(t) = 24 + (15 − 24)·e−t/τ
Voltage at the given time
vC(100) = 24 + (−9)·e−100/319.6 = 17.42 V
Answer
vC = 17.42 V
7.Magnetic fields & induction
Answer: ε = 200 mV
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).
Motional emf
ε = BLv = 0.2·0.25·4 = 0.2 V
Answer
ε = 200 mV
8.Magnetic fields & induction
Answer: M = 789.6 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 × 10⁻⁷·500·200·0.001257/0.2 = 789.6 µH
Answer
M = 789.6 µH
9.AC impedance & phasors
Answer: φ = 30.28 °
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 = 314.2·0.25 = 78.54 Ω
XC = 1/(ωC) = 1/(314.2·100·10⁻⁶) = 31.83 Ω
Impedance
In series the impedances add as complex numbers: Z = R + j(XL − XC).
|Z| = √(R² + (XL − XC)²) = √(80² + (46.71)²) = 92.64 Ω
Phase angle
φ = arctan((XL − XC)/R) = arctan(46.71/80) = 30.28°
Answer
φ = 30.28°
10.AC impedance & phasors
Answer: I = 0.2581 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|.
Reactance
XC = 1/(2π·60·4.7·10⁻⁶) = 564.4 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 = 140.1 − j37.24 Ω
Total impedance
Z = 50 + 140.1 − j37.24 = 190.1 − j37.24 Ω
|Z| = √(190.1² + 37.24²) = 193.7 Ω
Current
I = V/|Z| = 50/193.7 = 0.2581 A
Answer
I = 0.2581 A
ExamProblems · examproblems.com/courses/physics-2-electricity-and-magnetism