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

Three point charges lie on a straight line: −6 µC on the left, −1 µC on the right 10 cm away, and −8 µC between them, 2.5 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.

q_L = −6 µCq_R = −1 µCq_M = −8 µCr_L = 2.5 cmr_R = 7.5 cm

F = N

2.Coulomb’s law & electric fieldsHard

Charges of +2 µC and −2 µC are 12 cm apart. Point P lies 5 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 = 5 cm

E = N/C

3.Series & parallel resistorsMedium

Find the equivalent resistance R_eq seen by the source V₁.

V₁ = 18 VR₁ = 4 ΩR₂ = 10 ΩR₃ = 20 ΩR₄ = 15 Ω

R_eq = Ω

4.Series & parallel resistorsHard

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

V₁ = 9 VR₁ = 30 ΩR₂ = 5 ΩR₃ = 2 ΩR₄ = 3 ΩR₅ = 3 ΩR₆ = 4 Ω

I_R1 = A

5.RC & RL transientsMedium

The inductor carries no current before the switch closes at t = 0. How long after the switch closes does the current reach 99% of its final value? Find t.

V_s = 24 VR = 33 ΩL = 47 mH

t = ms

6.RC & RL transientsHard

Just before the switch closes at t = 0, the capacitor holds 20 V with the polarity shown. Find the capacitor voltage v_C at t = 750 ms.

V_s = 12 VR = 6.8 kΩC = 100 µFv_C(0) = 20 Vt = 750 ms

v_C = V

7.Magnetic fields & inductionMedium

A metal rod 60 cm long slides at 1 m/s along two rails in a uniform field of 0.1 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 = 0.1 TL = 60 cmv = 1 m/sR = 8 Ω

I = mA

8.Magnetic fields & inductionHard

A long air-core solenoid has 200 turns, a length of 40 cm and a diameter of 2 cm. Find the energy W stored in its magnetic field when it carries 1.5 A.

N = 200l = 40 cm⌀ = 2 cmI = 1.5 Aμ₀ = 4π × 10⁻⁷ T·m/A

W = µJ

9.AC impedance & phasorsMedium

A series RLC circuit is connected to a 10 V (rms), 200 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 10 V (rms)f = 200 HzR = 25 ΩL = 20 mHC = 4.7 µF

φ = °

10.AC impedance & phasorsHard

A series RLC circuit is driven by a 24 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage V_L across the inductor.

V = 24 V (rms)R = 2 ΩL = 5 mHC = 10 µF

V_L = V

3Solutions

1.Coulomb’s law & electric fields

Answer: F = 677.6 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.

F_L = +k·6·8 × 10⁻¹²/0.025² = 690.4 N

F_R = −k·1·8 × 10⁻¹²/0.075² = −12.79 N

Net force

F = 690.4 + (−12.79) = 677.6 N → |F| = 677.6 N

Answer

F = 677.6 N

The net force points to the right.

2.Coulomb’s law & electric fields

Answer: E = 4.529 × 10⁶ N/C

Distance and field of one charge

r = √(6² + 5²) = 7.81 cm

E₁ = k·2 × 10⁻⁶/0.0781² = 2.948 × 10⁶ N/C

Components

The components along the bisector cancel.

E = 2·E₁·(a/r) = 2·2.948 × 10⁶·0.7682 = 4.529 × 10⁶ N/C

Answer

E = 4.529 × 10⁶ N/C

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

3.Series & parallel resistors

Answer: R_eq = 11.78 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.

R₃ + R₄ = 20 + 15 = 35 Ω

R₂ ∥ 35 = 10·35/(10+35) = 7.778 Ω

R_eq = R₁ + 7.778 = 4 + 7.778 = 11.78 Ω

Answer

R_eq = 11.78 Ω

4.Series & parallel resistors

Answer: I_R1 = 0.279 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 = 3 + 4 = 7 Ω

R₄ ∥ 7 = 2.1 Ω

R₃ + 2.1 = 4.1 Ω

R₂ ∥ 4.1 = 2.253 Ω

R_eq = R₁ + 2.253 = 30 + 2.253 = 32.25 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 = 9/32.25 = 0.279 A

Answer

I_R1 = 0.279 A

5.RC & RL transients

Answer: t = 6.559 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.

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 / 33 Ω = 1.424 ms

i(∞) = V_s/R = 24/33 = 0.7273 A

Solve for the time

0.99 = 1 − e^−t/τ → t = −τ·ln(1 − 0.99)

t = −1.424·ln(0.01) = 6.559 ms

Answer

t = 6.559 ms

6.RC & RL transients

Answer: v_C = 14.66 V

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 = 12 V

τ = R·C = 6.8 kΩ · 100 µF = 680 ms

v_C(t) = 12 + (20 − 12)·e^−t/τ

Voltage at the given time

v_C(750) = 12 + (8)·e^−750/680 = 14.66 V

Answer

v_C = 14.66 V

7.Magnetic fields & induction

Answer: I = 7.5 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 = 0.1·0.6·1 = 0.06 V

Current

I = ε/R = 0.06/8 = 0.0075 A

Answer

I = 7.5 mA

8.Magnetic fields & induction

Answer: W = 44.41 µJ

Self-inductance

L = μ₀N²A/l = 39.48 µH

Stored energy

W = ½LI² = ½·3.948 × 10⁻⁵·1.5² = 4.441 × 10⁻⁵ J = 44.41 µJ

Answer

W = 44.41 µJ

9.AC impedance & phasors

Answer: φ = −80.16 °

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·200 = 1257 rad/s

X_L = ωL = 1257·0.02 = 25.13 Ω

X_C = 1/(ωC) = 1/(1257·4.7·10⁻⁶) = 169.3 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 + (−144.2)²) = 146.3 Ω

Phase angle

φ = arctan((X_L − X_C)/R) = arctan(−144.2/25) = −80.16°

Answer

φ = −80.16°

10.AC impedance & phasors

Answer: V_L = 268.3 V

Resonance

At resonance the reactances cancel, so the impedance is just R.

ω₀ = 1/√(LC) = 1/√(0.005·10·10⁻⁶) = 4472 rad/s

I = V/R = 24/2 = 12 A

Voltage across the inductor

X_L = ω₀L = 4472·0.005 = 22.36 Ω

V_L = I·X_L = 12·22.36 = 268.3 V

Answer

V_L = 268.3 V

This is Q = 11.18 times the supply voltage.

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