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
Two identical small metal spheres carry charges of +3 µC and −4 µC. They are touched together and then put back 5 cm apart. Find the magnitude of the force F between them now. Use k = 8.99 × 10⁹ N·m²/C².
q1 = +3 µCq2 = −4 µCr = 5 cm
F = N
2.Coulomb’s law & electric fieldsHard
Charges of +6 µC and +6 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 = +6 µCq2 = +6 µCd = 12 cmh = 5 cm
E = N/C
3.Series & parallel resistorsMedium
Find the current IR1 delivered by the source, which flows through R1.
V1 = 15 VR1 = 2 ΩR2 = 8 ΩR3 = 10 ΩR4 = 8 Ω
IR1 = A
4.Series & parallel resistorsHard
Find the current IR1 delivered by the source, which flows through R1.
V1 = 6 VR1 = 30 ΩR2 = 10 ΩR3 = 30 ΩR4 = 4 ΩR5 = 15 ΩR6 = 15 Ω
IR1 = 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 80% of its final value? Find t.
Vs = 20 VR = 15 ΩL = 68 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 current i2 through R2 at t = 250 ms.
Vs = 10 VR1 = 3.3 kΩR2 = 4.7 kΩC = 100 µFt = 250 ms
i2 = mA
7.Magnetic fields & inductionMedium
An electron moves on a circle in a uniform magnetic field of 2 mT. Find the time T it takes for one revolution.
m = 9.109 × 10⁻³¹ kg|q| = 1.602 × 10⁻¹⁹ CB = 2 mT
T = ns
8.Magnetic fields & inductionHard
The current in a 10 mH inductor changes steadily from 0 A to 4 A in 10 ms. Find the magnitude of the self-induced emf ε.
L = 10 mH0 A → 4 AΔt = 10 ms
ε = V
9.AC impedance & phasorsMedium
A series RLC circuit with L = 2 mH must resonate at 100 Hz. Find the capacitance C needed.
f0 = 100 HzL = 2 mH
C = µF
10.AC impedance & phasorsHard
A series RLC circuit is connected to a 50 V (rms), 500 Hz supply. Find the resonant frequency f0 of the circuit.
V = 50 V (rms)f = 500 HzR = 100 ΩL = 20 mHC = 22 µF
f0 = Hz
3Solutions
1.Coulomb’s law & electric fields
Answer: F = 0.899 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.
Charge after touching
q′ = (3 + −4)/2 = −0.5 µC
New force
F = 8.99 × 10⁹·(0.5 × 10⁻⁶)²/0.05² = 0.899 N
Answer
F = 0.899 N
Both spheres now carry the same sign, so they repel.
2.Coulomb’s law & electric fields
Answer: E = 1.132 × 10⁷ 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 + 5²) = 7.81 cm
E1 = k·6 × 10⁻⁶/0.0781² = 8.843 × 10⁶ N/C
Components
The components along the line joining the charges cancel.
E = 2·E1·(h/r) = 2·8.843 × 10⁶·0.6402 = 1.132 × 10⁷ N/C
Answer
E = 1.132 × 10⁷ N/C
The field points along the bisector, away from the charges.
3.Series & parallel resistors
Answer: IR1 = 1.99 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 + 8 = 18 Ω
R2 ∥ 18 = 8·18/(8+18) = 5.538 Ω
Req = R1 + 5.538 = 2 + 5.538 = 7.538 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 = 15/7.538 = 1.99 A
Answer
IR1 = 1.99 A
4.Series & parallel resistors
Answer: IR1 = 0.1591 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.
R5 + R6 = 15 + 15 = 30 Ω
R4 ∥ 30 = 3.529 Ω
R3 + 3.529 = 33.53 Ω
R2 ∥ 33.53 = 7.703 Ω
Req = R1 + 7.703 = 30 + 7.703 = 37.7 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 = 6/37.7 = 0.1591 A
Answer
IR1 = 0.1591 A
5.RC & RL transients
Answer: t = 7.296 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 = 68 mH / 15 Ω = 4.533 ms
i(∞) = Vs/R = 20/15 = 1.333 A
Solve for the time
0.8 = 1 − e−t/τ → t = −τ·ln(1 − 0.8)
t = −4.533·ln(0.2) = 7.296 ms
Answer
t = 7.296 ms
6.RC & RL transients
Answer: i2 = 0.7343 mA
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 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.
vC(0) = 10·4.7/(3.3 + 4.7) = 5.875 V
Time constant after the switch opens
τ = R2·C = 4.7·100 = 470 ms
Discharge
vC(t) = 5.875·e−250/470 = 3.451 V
i2 = vC/R2 = 3.451/4.7 = 0.7343 mA
Answer
i2 = 0.7343 mA
7.Magnetic fields & induction
Answer: T = 17.86 ns
Magnetic forces: a straight wire carrying I in a field B feels F = BIL·sin θ, and a moving charge F = |q|vB·sin θ, where θ is the angle between the current (or velocity) and B. The force is perpendicular to both: point the fingers of the right hand along I or v, curl them towards B, and the thumb gives F for a positive charge; reverse it for a negative charge. Because F is perpendicular to v, a charge moving across a uniform field moves on a circle: |q|vB = mv²/r gives r = mv/(|q|B) and the period T = 2πm/(|q|B), which does not depend on the speed. Two parallel wires attract for currents in the same direction: F/L = μ₀I₁I₂/(2πd).
Period of the circular motion
A faster particle moves on a larger circle, so the time per turn stays the same.
T = 2πm/(|q|B) = 2π·9.109 × 10⁻³¹/(1.602 × 10⁻¹⁹·0.002) = 17.86 ns
Answer
T = 17.86 ns
The speed is not needed: this is why a cyclotron can use one fixed frequency.
8.Magnetic fields & induction
Answer: ε = 4 V
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-induced emf
The emf opposes the change of current (Lenz), so its sign is opposite to ΔI; the size is L·|ΔI|/Δt.
ε = L·ΔI/Δt = 0.01·4/0.01 = 4 V
Answer
ε = 4 V
9.AC impedance & phasors
Answer: C = 1267 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·100 = 628.3 rad/s
C = 1/(ω₀²·L) = 1/(628.3²·0.002) = 1267 µF
Answer
C = 1267 µF
10.AC impedance & phasors
Answer: f0 = 239.9 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·√(0.02·22·10⁻⁶)) = 239.9 Hz
Answer
f0 = 239.9 Hz
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