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: R_eq = R₁ + R₂ + …
Resistors in parallel: 1/R_eq = 1/R₁ + 1/R₂ + …
Two in parallel: R_eq = 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; v_C cannot jump
Inductor at DC: short circuit; i_L 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, E_rms = 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: Z_R = R, Z_L = jωL, Z_C = 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: X_L = ωL and X_C = 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 +5 µC and a charge of +2 µC are 30 cm apart. Point P lies on the line between them, 7.5 cm from the first charge. Find the magnitude of the electric field E at P.

q₁ = +5 µCq₂ = +2 µCr₁ = 7.5 cmr₂ = 22.5 cm

E = N/C

2.Coulomb’s law & electric fieldsHard

Charges of +10 µC and +10 µC are 8 cm apart. Point P lies 8 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 = +10 µCq₂ = +10 µCd = 8 cmh = 8 cm

E = N/C

3.Series & parallel resistorsMedium

Find the current I_R1 delivered by the source, which flows through R₁.

V₁ = 9 VR₁ = 6 ΩR₂ = 10 ΩR₃ = 6 ΩR₄ = 3 Ω

I_R1 = A

4.Series & parallel resistorsHard

Find the current I_R1 delivered by the source, which flows through R₁.

V₁ = 12 VR₁ = 8 ΩR₂ = 6 ΩR₃ = 3 ΩR₄ = 30 ΩR₅ = 25 ΩR₆ = 8 Ω

I_R1 = A

5.RC & RL transientsMedium

The inductor carries no current before the switch closes at t = 0. Find the current i at t = 1 ms.

V_s = 12 VR = 47 ΩL = 10 mHt = 1 ms

i = A

6.RC & RL transientsHard

Just before the switch closes at t = 0, the capacitor holds 20 V with the polarity shown. How long does it take for v_C to reach 11 V? Find t.

V_s = 5 VR = 3.3 kΩC = 68 µFv_C(0) = 20 V

t = ms

7.Magnetic fields & inductionMedium

A circular loop lies in a uniform magnetic field pointing out of the page. The field strength is increasing. Click the direction of the induced current in the loop as seen in the drawing.

B: out of the pagedB/dt > 0

1clockwise
2counterclockwise

8.Magnetic fields & inductionHard

A long air-core solenoid has 2000 turns, a length of 30 cm and a diameter of 5 cm. Find its self-inductance L.

N = 2000l = 30 cm⌀ = 5 cmμ₀ = 4π × 10⁻⁷ T·m/A

L = mH

9.AC impedance & phasorsMedium

A series RLC circuit is connected to a 12 V (rms), 500 Hz supply. Find the real power P taken from the supply.

V = 12 V (rms)f = 500 HzR = 50 ΩL = 100 mHC = 10 µF

P = W

10.AC impedance & phasorsHard

A series RLC circuit is connected to a 50 V (rms), 60 Hz supply. Find the rms current I.

V = 50 V (rms)f = 60 HzR = 25 ΩL = 50 mHC = 100 µF

I = A

3Solutions

1.Coulomb’s law & electric fields

Answer: E = 7.636 × 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.

Field of each charge

E₁ = k·5 × 10⁻⁶/0.075² = 7.991 × 10⁶ N/C

E₂ = k·2 × 10⁻⁶/0.225² = 355160 N/C

Directions and sum

Both charges are positive: at P their fields point away from each charge, so in opposite directions.

E = |7.991 × 10⁶ − 355160| = 7.636 × 10⁶ N/C

Answer

E = 7.636 × 10⁶ N/C

2.Coulomb’s law & electric fields

Answer: E = 2.01 × 10⁷ N/C

Distance and field of one charge

r = √(4² + 8²) = 8.944 cm

E₁ = k·10 × 10⁻⁶/0.08944² = 1.124 × 10⁷ N/C

Components

The components along the line joining the charges cancel.

E = 2·E₁·(h/r) = 2·1.124 × 10⁷·0.8944 = 2.01 × 10⁷ N/C

Answer

E = 2.01 × 10⁷ N/C

The field points along the bisector, away from the charges.

3.Series & parallel resistors

Answer: I_R1 = 0.8382 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 = 6 + 3 = 9 Ω

R₂ ∥ 9 = 10·9/(10+9) = 4.737 Ω

R_eq = R₁ + 4.737 = 6 + 4.737 = 10.74 Ω

Current from the source

The source sees only R_eq, so Ohm’s law gives the current it delivers. That current flows through R₁.

I_R1 = V₁/R_eq = 9/10.74 = 0.8382 A

Answer

I_R1 = 0.8382 A

4.Series & parallel resistors

Answer: I_R1 = 0.9567 A

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 = 25 + 8 = 33 Ω

R₄ ∥ 33 = 15.71 Ω

R₃ + 15.71 = 18.71 Ω

R₂ ∥ 18.71 = 4.543 Ω

R_eq = R₁ + 4.543 = 8 + 4.543 = 12.54 Ω

Current from the source

The source sees only R_eq, so Ohm’s law gives the current it delivers. That current flows through R₁.

I_R1 = V₁/R_eq = 12/12.54 = 0.9567 A

Answer

I_R1 = 0.9567 A

5.RC & RL transients

Answer: i = 0.253 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 = 10 mH / 47 Ω = 0.2128 ms

i(∞) = V_s/R = 12/47 = 0.2553 A

Current at the given time

i(t) = i(∞)·(1 − e^−t/τ) = 0.2553·(1 − e^−1/0.2128) = 0.253 A

Answer

i = 0.253 A

6.RC & RL transients

Answer: t = 205.6 ms

Initial value, final value, time constant

The capacitor voltage cannot jump, so it starts at its stored value and heads for the source voltage.

v_C(0) = 20 V, v_C(∞) = V_s = 5 V

τ = R·C = 3.3 kΩ · 68 µF = 224.4 ms

v_C(t) = 5 + (20 − 5)·e^−t/τ

Solve for the time

11 = 5 + (15)·e^−t/τ → e^−t/τ = (11 − 5)/(20 − 5) = 0.4

t = τ·ln((20 − 5)/(11 − 5)) = 224.4·ln(2.5) = 205.6 ms

Answer

t = 205.6 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: L = 32.9 mH

Cross-section

A = π·(0.025)² = 0.001963 m²

Self-inductance

L = μ₀N²A/l = 4π × 10⁻⁷·2000²·0.001963/0.3 = 0.0329 H = 32.9 mH

Answer

L = 32.9 mH

9.AC impedance & phasors

Answer: P = 0.08758 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

X_L = 314.2 Ω, X_C = 31.83 Ω

Impedance

|Z| = √(50² + (282.3)²) = 286.7 Ω

Real power

Inductors and capacitors only store and return energy; the resistor absorbs all of the real power.

I = 12/286.7 = 0.04185 A

P = I²·R = 0.04185²·50 = 0.08758 W

Answer

P = 0.08758 W

10.AC impedance & phasors

Answer: I = 1.912 A

Reactances

ω = 2πf = 2π·60 = 377 rad/s

X_L = ωL = 377·0.05 = 18.85 Ω

X_C = 1/(ωC) = 1/(377·100·10⁻⁶) = 26.53 Ω

Impedance

In series the impedances add as complex numbers: Z = R + j(X_L − X_C).

|Z| = √(R² + (X_L − X_C)²) = √(25² + (−7.676)²) = 26.15 Ω

Current

The same current flows through every element of a series circuit.

I = V/|Z| = 50/26.15 = 1.912 A

Answer

I = 1.912 A

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