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
Charges of +2 µC and +4 µC are 40 cm apart. At what distance x from the +2 µC charge, on the line between them, is the electric field zero?
q1 = +2 µCq2 = +4 µCd = 40 cm
x = cm
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 current IR1 delivered by the source, which flows through R1.
V1 = 24 VR1 = 12 ΩR2 = 5 ΩR3 = 2 ΩR4 = 6 Ω
IR1 = A
4.Series & parallel resistorsHard
Find the current IR1 delivered by the source, which flows through R1.
V1 = 36 VR1 = 25 ΩR2 = 30 ΩR3 = 12 ΩR4 = 3 ΩR5 = 8 ΩR6 = 10 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 50% of its final value? Find t.
Vs = 5 VR = 15 ΩL = 68 mH
t = ms
6.RC & RL transientsHard
Just before the switch closes at t = 0, the capacitor holds 10 V with the polarity shown. Find the capacitor voltage vC at t = 300 ms.
Vs = 15 VR = 3.3 kΩC = 100 µFvC(0) = 10 Vt = 300 ms
vC = V
7.Magnetic fields & inductionMedium
A metal rod 60 cm long slides at 2 m/s along two rails in a uniform field of 0.5 T into the page. The rails are joined by a resistor of 1 Ω; the rod and rails have negligible resistance. Find the current I.
B = 0.5 TL = 60 cmv = 2 m/sR = 1 Ω
I = mA
8.Magnetic fields & inductionHard
A long air-core solenoid has 200 turns, a length of 20 cm and a diameter of 3 cm. Find its self-inductance L.
N = 200l = 20 cm⌀ = 3 cmμ₀ = 4π × 10⁻⁷ T·m/A
L = µH
9.AC impedance & phasorsMedium
A resistor and a capacitor are connected in parallel across a 120 V (rms), 200 Hz supply. Find the rms current I.
V = 120 V (rms)f = 200 HzR = 80 ΩC = 4.7 µF
I = A
10.AC impedance & phasorsHard
A series RLC circuit has R = 10 Ω, L = 5 mH and C = 10 µF. Find its quality factor Q at resonance.
R = 10 ΩL = 5 mHC = 10 µF
Q =
3Solutions
1.Coulomb’s law & electric fields
Answer: x = 16.57 cm
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.
Equal fields
2/x² = 4/(40 − x)²
(40 − x)/x = √(4/2) = 1.414
Position
x = 40/(1 + 1.414) = 16.57 cm
Answer
x = 16.57 cm
The zero point is always closer to the smaller charge.
2.Coulomb’s law & electric fields
Answer: E = 3.774 × 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 + 5²) = 7.81 cm
E1 = k·2 × 10⁻⁶/0.0781² = 2.948 × 10⁶ N/C
Components
The components along the line joining the charges cancel.
E = 2·E1·(h/r) = 2·2.948 × 10⁶·0.6402 = 3.774 × 10⁶ N/C
Answer
E = 3.774 × 10⁶ N/C
The field points along the bisector, away from the charges.
3.Series & parallel resistors
Answer: IR1 = 1.592 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 = 2 + 6 = 8 Ω
R2 ∥ 8 = 5·8/(5+8) = 3.077 Ω
Req = R1 + 3.077 = 12 + 3.077 = 15.08 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 = 24/15.08 = 1.592 A
Answer
IR1 = 1.592 A
4.Series & parallel resistors
Answer: IR1 = 1.034 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 = 8 + 10 = 18 Ω
R4 ∥ 18 = 2.571 Ω
R3 + 2.571 = 14.57 Ω
R2 ∥ 14.57 = 9.808 Ω
Req = R1 + 9.808 = 25 + 9.808 = 34.81 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 = 36/34.81 = 1.034 A
Answer
IR1 = 1.034 A
5.RC & RL transients
Answer: t = 3.142 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 = 5/15 = 0.3333 A
Solve for the time
0.5 = 1 − e−t/τ → t = −τ·ln(1 − 0.5)
t = −4.533·ln(0.5) = 3.142 ms
Answer
t = 3.142 ms
6.RC & RL transients
Answer: vC = 12.99 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) = 10 V, vC(∞) = Vs = 15 V
τ = R·C = 3.3 kΩ · 100 µF = 330 ms
vC(t) = 15 + (10 − 15)·e−t/τ
Voltage at the given time
vC(300) = 15 + (−5)·e−300/330 = 12.99 V
Answer
vC = 12.99 V
7.Magnetic fields & induction
Answer: I = 600 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 = 0.5·0.6·2 = 0.6 V
Current
I = ε/R = 0.6/1 = 0.6 A
Answer
I = 600 mA
8.Magnetic fields & induction
Answer: L = 177.7 µH
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).
Cross-section
A = π·(0.015)² = 0.0007069 m²
Self-inductance
L = μ₀N²A/l = 4π × 10⁻⁷·200²·0.0007069/0.2 = 0.0001777 H = 177.7 µH
Answer
L = 177.7 µH
9.AC impedance & phasors
Answer: I = 1.659 A
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|.
Reactance
XC = 1/(ωC) = 1/(2π·200·4.7·10⁻⁶) = 169.3 Ω
Branch currents
Both branches are connected straight across the supply, so each sees the full voltage V.
IR = V/R = 120/80 = 1.5 A
IX = V/XC = 120/169.3 = 0.7087 A
Total current
The resistor current is in phase with V and the reactive current is 90° out of phase, so they add like the sides of a right triangle.
I = √(IR² + IX²) = √(1.5² + 0.7087²) = 1.659 A
Answer
I = 1.659 A
10.AC impedance & phasors
Answer: Q = 2.236
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|.
Resonant frequency
f0 = 1/(2π·√(0.005·10·10⁻⁶)) = 711.8 Hz
Quality factor
Q compares the reactance at resonance with the resistance: Q = ω₀L/R.
Q = (1/R)·√(L/C) = (1/10)·√(0.005/(10·10⁻⁶)) = 2.236
Answer
Q = 2.236
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