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.

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1Formulas and methods

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

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

  1. Start at the end of the network furthest from the terminals.
  2. Combine resistors in series by adding them, and resistors in parallel with 1/R = 1/R₁ + 1/R₂ + …
  3. Redraw the circuit after each step.
  4. 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

  1. Find the initial value just after switching: a capacitor voltage and an inductor current cannot jump.
  2. Find the final value long after switching: a capacitor acts as an open circuit, an inductor as a short circuit.
  3. Find the time constant: τ = RC or τ = L/R, using the resistance seen by the capacitor or inductor.
  4. 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

  1. Convert every quantity to SI units: cm to m, cm² to m², mT and µT to T.
  2. 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.
  3. 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.
  4. 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

  1. Convert the frequency: ω = 2πf.
  2. Find each reactance: XL = ωL and XC = 1/(ωC).
  3. Add impedances like resistors, with Z = R + jX.
  4. 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
++8 µC−−6 µC++3 µC30 cm + 10 cm

Three point charges lie on a straight line: +8 µC on the left, +3 µC on the right 40 cm away, and −6 µC between them, 30 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 = +8 µCqR = +3 µCqM = −6 µCrL = 30 cmrR = 10 cm
F = N
2.Coulomb’s law & electric fieldsHard
+−P2 × 6 cm6 cm+2 µC−2 µC

Charges of +2 µC and −2 µC are 12 cm apart. Point P lies 6 cm from the midpoint between them, on the perpendicular bisector. Find the magnitude of the electric field E at P. Use k = 8.99 × 10⁹ N·m²/C².

q1 = +2 µCq2 = −2 µCd = 12 cmh = 6 cm
E = N/C
3.Series & parallel resistorsMedium
R1​ = 2 ΩR3​ = 12 Ω+−V1​24 VR2​12 ΩAR4​20 ΩB

Find the equivalent resistance Req seen by the source V1.

V1 = 24 VR1 = 2 ΩR2 = 12 ΩR3 = 12 ΩR4 = 20 Ω
Req = Ω
4.Series & parallel resistorsHard
R1​ = 3 ΩR3​ = 2 ΩR5​ = 5 Ω+−V1​10 VR2​2 ΩAR4​20 ΩBR6​8 Ω

Find the equivalent resistance Req seen by the source V1.

V1 = 10 VR1 = 3 ΩR2 = 2 ΩR3 = 2 ΩR4 = 20 ΩR5 = 5 ΩR6 = 8 Ω
Req = Ω
5.RC & RL transientsMedium
+−Vs​15 Vt = 0R = 10 ΩL47 mHi(t)

The inductor carries no current before the switch closes at t = 0. Find the energy W stored in the inductor at t = 5 ms, in mJ.

Vs = 15 VR = 10 ΩL = 47 mHt = 5 ms
W = mJ
6.RC & RL transientsHard
+−Vs​10 Vt = 0R = 3.3 kΩC33 µF+−vC​vC​(0) = 24 V

Just before the switch closes at t = 0, the capacitor holds 24 V with the polarity shown. Find the capacitor voltage vC at t = 150 ms.

Vs = 10 VR = 3.3 kΩC = 33 µFvC(0) = 24 Vt = 150 ms
vC = V
7.Magnetic fields & inductionMedium
R = 8 Ωv = 5 m/sL = 30 cmB = 0.8 T

A metal rod 30 cm long slides at 5 m/s along two rails in a uniform field of 0.8 T into the page. The rails are joined by a resistor of 8 Ω; the rod and rails have negligible resistance. Find the current I.

B = 0.8 TL = 30 cmv = 5 m/sR = 8 Ω
I = mA
8.Magnetic fields & inductionHard
I = 2 Al = 30 cmN = 1500, ⌀ 4 cm

A long air-core solenoid has 1500 turns, a length of 30 cm and a diameter of 4 cm. Find the energy W stored in its magnetic field when it carries 2 A.

N = 1500l = 30 cm⌀ = 4 cmI = 2 Aμ₀ = 4π × 10⁻⁷ T·m/A
W = mJ
9.AC impedance & phasorsMedium
Vf0​ = 60 HzRL = 2 mHC = ?

A series RLC circuit with L = 2 mH must resonate at 60 Hz. Find the capacitance C needed.

f0 = 60 HzL = 2 mH
C = µF
10.AC impedance & phasorsHard
24 V500 HzR = 40 ΩL = 50 mHC = 10 µF

A series RLC circuit is connected to a 24 V (rms), 500 Hz supply. Find the resonant frequency f0 of the circuit.

V = 24 V (rms)f = 500 HzR = 40 ΩL = 50 mHC = 10 µF
f0 = Hz

3Solutions

1.Coulomb’s law & electric fields

Answer: F = 11.39 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·8·6 × 10⁻¹²/0.3² = −4.795 N
FR = +k·3·6 × 10⁻¹²/0.1² = 16.18 N

Net force

F = −4.795 + (16.18) = 11.39 N → |F| = 11.39 N

Answer

F = 11.39 N

The net force points to the right.

2.Coulomb’s law & electric fields

Answer: E = 3.532 × 10⁶ 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.

Distance and field of one charge

r = √(6² + 6²) = 8.485 cm
E1 = k·2 × 10⁻⁶/0.08485² = 2.497 × 10⁶ N/C

Components

The components along the bisector cancel.

E = 2·E1·(a/r) = 2·2.497 × 10⁶·0.7071 = 3.532 × 10⁶ N/C

Answer

E = 3.532 × 10⁶ N/C

The field points parallel to the line from the positive towards the negative charge.

3.Series & parallel resistors

Answer: Req = 10.73 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.

R1​ = 2 ΩR3​ = 12 Ω+−V1​24 VR2​12 ΩAR4​20 ΩB

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 + 20 = 32 Ω
R2 ∥ 32 = 12·32/(12+32) = 8.727 Ω
Req = R1 + 8.727 = 2 + 8.727 = 10.73 Ω

Answer

Req = 10.73 Ω
4.Series & parallel resistors

Answer: Req = 4.663 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.

R1​ = 3 ΩR3​ = 2 ΩR5​ = 5 Ω+−V1​10 VR2​2 ΩAR4​20 ΩBR6​8 Ω

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 = 5 + 8 = 13 Ω
R4 ∥ 13 = 7.879 Ω
R3 + 7.879 = 9.879 Ω
R2 ∥ 9.879 = 1.663 Ω
Req = R1 + 1.663 = 3 + 1.663 = 4.663 Ω

Answer

Req = 4.663 Ω
5.RC & RL transients

Answer: W = 22.68 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 = 47 mH / 10 Ω = 4.7 ms
i(∞) = Vs/R = 15/10 = 1.5 A

Current at the given time

i(t) = 1.5·(1 − e−5/4.7) = 0.9823 A

Stored energy

W = ½·L·i² = ½·47·0.9823² = 22.68 mJ

Answer

W = 22.68 mJ
6.RC & RL transients

Answer: vC = 13.53 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) = 24 V, vC(∞) = Vs = 10 V
τ = R·C = 3.3 kΩ · 33 µF = 108.9 ms
vC(t) = 10 + (24 − 10)·e−t/τ

Voltage at the given time

vC(150) = 10 + (14)·e−150/108.9 = 13.53 V

Answer

vC = 13.53 V
7.Magnetic fields & induction

Answer: I = 150 mA

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 = 0.8·0.3·5 = 1.2 V

Current

I = ε/R = 1.2/8 = 0.15 A

Answer

I = 150 mA
8.Magnetic fields & induction

Answer: W = 23.69 mJ

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).

Self-inductance

L = μ₀N²A/l = 11.84 mH

Stored energy

W = ½LI² = ½·0.01184·2² = 0.02369 J = 23.69 mJ

Answer

W = 23.69 mJ
9.AC impedance & phasors

Answer: C = 3518 µF

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 condition

At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC).

ω₀ = 2π·60 = 377 rad/s
C = 1/(ω₀²·L) = 1/(377²·0.002) = 3518 µF

Answer

C = 3518 µF
10.AC impedance & phasors

Answer: f0 = 225.1 Hz

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 inductive and capacitive reactances cancel: ωL = 1/(ωC). The supply frequency does not matter.

f0 = 1/(2π√(LC)) = 1/(2π·√(0.05·10·10⁻⁶)) = 225.1 Hz

Answer

f0 = 225.1 Hz

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