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, 18 cm from the first charge. Find the magnitude of the electric field E at P.

q₁ = +5 µCq₂ = −2 µCr₁ = 18 cmr₂ = 12 cm

E = N/C

2.Coulomb’s law & electric fieldsHard

Three positive point charges sit at the corners of a right triangle: q₃ = 5 µC is at the right-angle corner, q₁ = 4 µC is 4 cm from it along one side and q₂ = 4 µC is 4 cm from it along the other side. Find the magnitude of the net electric force F on q₃.

q₁ = 4 µCq₂ = 4 µCq₃ = 5 µCr₁₃ = 4 cmr₂₃ = 4 cm

F = N

3.Series & parallel resistorsMedium

Find the current I_R2 through R₂.

V₁ = 12 VR₁ = 10 ΩR₂ = 5 ΩR₃ = 10 ΩR₄ = 2 Ω

I_R2 = A

4.Series & parallel resistorsHard

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

V₁ = 10 VR₁ = 4 ΩR₂ = 20 ΩR₃ = 10 ΩR₄ = 12 ΩR₅ = 3 ΩR₆ = 20 Ω

I_R1 = A

5.RC & RL transientsMedium

The inductor carries a steady current of 1 A. At t = 0 the switch closes the loop through R and the source is removed, so the inductor current decays through R. How long does it take for the current to fall to 0.1 A? Find t.

i(0) = 1 AR = 10 ΩL = 22 mH

t = ms

6.RC & RL transientsHard

The switch has been closed for a long time, so the circuit is in steady state. At t = 0 the switch opens. Find the capacitor voltage v_C at t = 100 ms.

V_s = 24 VR₁ = 6.8 kΩR₂ = 10 kΩC = 33 µFt = 100 ms

v_C = V

7.Magnetic fields & inductionMedium

A metal rod 80 cm long slides at 5 m/s along two rails in a uniform field of 0.1 T into the page. The rails are joined by a resistor of 10 Ω; the rod and rails have negligible resistance. Find the current I.

B = 0.1 TL = 80 cmv = 5 m/sR = 10 Ω

I = mA

8.Magnetic fields & inductionHard

A coil of 200 turns, 15 cm × 20 cm, rotates at 1500 rpm in a uniform field of 0.25 T about an axis perpendicular to the field. Find the peak emf E₀.

N = 200A = 15 cm × 20 cmB = 0.25 Tn = 1500 rpm

E₀ = V

9.AC impedance & phasorsMedium

A series RLC circuit with L = 100 mH must resonate at 500 Hz. Find the capacitance C needed.

f₀ = 500 HzL = 100 mH

C = µF

10.AC impedance & phasorsHard

A series RLC circuit is connected to a 10 V (rms), 200 Hz supply. Find the resonant frequency f₀ of the circuit.

V = 10 V (rms)f = 200 HzR = 100 ΩL = 50 mHC = 100 µF

f₀ = Hz

3Solutions

1.Coulomb’s law & electric fields

Answer: E = 2.636 × 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 × 10⁻⁶/0.18² = 1.387 × 10⁶ N/C

E₂ = k·2 × 10⁻⁶/0.12² = 1.249 × 10⁶ N/C

Directions and sum

The first field points away from the positive charge and the second towards the negative charge: both point the same way.

E = |1.387 × 10⁶ + 1.249 × 10⁶| = 2.636 × 10⁶ N/C

Answer

E = 2.636 × 10⁶ N/C

2.Coulomb’s law & electric fields

Answer: F = 158.9 N

Each force on q₃

F₁ = k·4·5 × 10⁻¹²/0.04² = 112.4 N

F₂ = k·4·5 × 10⁻¹²/0.04² = 112.4 N

Vector sum

F = √(112.4² + 112.4²) = 158.9 N

Answer

F = 158.9 N

3.Series & parallel resistors

Answer: I_R2 = 0.6261 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 = 10 + 2 = 12 Ω

R₂ ∥ 12 = 5·12/(5+12) = 3.529 Ω

R_eq = R₁ + 3.529 = 10 + 3.529 = 13.53 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 = 12/13.53 = 0.887 A

Current through R₂

Subtract the drop across R₁ to get the voltage at node A, which is the voltage across R₂.

V_A = V₁ − I_R1·R₁ = 12 − 0.887·10 = 3.13 V

I_R2 = V_A/R₂ = 3.13/5 = 0.6261 A

Answer

I_R2 = 0.6261 A

4.Series & parallel resistors

Answer: I_R1 = 0.7439 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 + 20 = 23 Ω

R₄ ∥ 23 = 7.886 Ω

R₃ + 7.886 = 17.89 Ω

R₂ ∥ 17.89 = 9.442 Ω

R_eq = R₁ + 9.442 = 4 + 9.442 = 13.44 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 = 10/13.44 = 0.7439 A

Answer

I_R1 = 0.7439 A

5.RC & RL transients

Answer: t = 5.066 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

τ = L/R = 22 mH / 10 Ω = 2.2 ms

Solve for the time

0.1 = 1·e^−t/τ → t = τ·ln(1/0.1)

t = 2.2·ln(10) = 5.066 ms

Answer

t = 5.066 ms

6.RC & RL transients

Answer: v_C = 10.55 V

Initial capacitor voltage

In steady state the capacitor is an open circuit, so R₁ and R₂ form a voltage divider. The capacitor voltage cannot jump when the switch opens.

v_C(0) = 24·10/(6.8 + 10) = 14.29 V

Time constant after the switch opens

τ = R₂·C = 10·33 = 330 ms

Discharge

v_C(t) = 14.29·e^−100/330 = 10.55 V

Answer

v_C = 10.55 V

7.Magnetic fields & induction

Answer: I = 40 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 = 0.1·0.8·5 = 0.4 V

Current

I = ε/R = 0.4/10 = 0.04 A

Answer

I = 40 mA

8.Magnetic fields & induction

Answer: E₀ = 235.6 V

Angular speed and area

ω = 2π·1500/60 = 157.1 rad/s

A = 0.03 m²

Peak emf

E₀ = NBAω = 200·0.25·0.03·157.1 = 235.6 V

Answer

E₀ = 235.6 V

9.AC impedance & phasors

Answer: C = 1.013 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·500 = 3142 rad/s

C = 1/(ω₀²·L) = 1/(3142²·0.1) = 1.013 µF

Answer

C = 1.013 µF

10.AC impedance & phasors

Answer: f₀ = 71.18 Hz

Resonance

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

f₀ = 1/(2π√(LC)) = 1/(2π·√(0.05·100·10⁻⁶)) = 71.18 Hz

Answer

f₀ = 71.18 Hz

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