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 −6 µC and +5 µC. They are touched together and then put back 30 cm apart. Find the magnitude of the force F between them now. Use k = 8.99 × 10⁹ N·m²/C².
q1 = −6 µCq2 = +5 µCr = 30 cm
F = N
2.Coulomb’s law & electric fieldsHard
Charges of +4 µC and +4 µC are 10 cm apart. Point P lies 12 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 = +4 µCq2 = +4 µCd = 10 cmh = 12 cm
E = N/C
3.Series & parallel resistorsMedium
Find the equivalent resistance Req seen by the source V1.
V1 = 18 VR1 = 10 ΩR2 = 3 ΩR3 = 8 ΩR4 = 4 Ω
Req = Ω
4.Series & parallel resistorsHard
Find the equivalent resistance Req seen by the source V1.
V1 = 10 VR1 = 20 ΩR2 = 2 ΩR3 = 8 ΩR4 = 10 ΩR5 = 5 ΩR6 = 15 Ω
Req = Ω
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 0.5 A? Find t.
i(0) = 1 AR = 47 ΩL = 10 mH
t = ms
6.RC & RL transientsHard
Just before the switch closes at t = 0, the capacitor holds 6 V with the polarity shown. How long does it take for vC to reach 5.7 V? Find t.
Vs = 5 VR = 6.8 kΩC = 220 µFvC(0) = 6 V
t = ms
7.Magnetic fields & inductionMedium
An electron moves to the right through a uniform magnetic field pointing out of the page. Click the direction of the magnetic force on it.
v: to the rightB: out of the page
- 1up the page
- 2to the right
- 3down the page
- 4to the left
8.Magnetic fields & inductionHard
A short coil of 200 turns is wound tightly round the middle of a long solenoid of 2000 turns, length 40 cm and diameter 4 cm. Find their mutual inductance M.
N1 = 2000N2 = 200l = 40 cm⌀ = 4 cmμ₀ = 4π × 10⁻⁷ T·m/A
M = mH
9.AC impedance & phasorsMedium
A series RLC circuit is connected to a 50 V (rms), 100 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).
V = 50 V (rms)f = 100 HzR = 50 ΩL = 250 mHC = 4.7 µF
φ = °
10.AC impedance & phasorsHard
A series RLC circuit is connected to a 230 V (rms), 200 Hz supply. Find the resonant frequency f0 of the circuit.
V = 230 V (rms)f = 200 HzR = 10 ΩL = 100 mHC = 22 µF
f0 = Hz
3Solutions
1.Coulomb’s law & electric fields
Answer: F = 0.02497 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′ = (−6 + 5)/2 = −0.5 µC
New force
F = 8.99 × 10⁹·(0.5 × 10⁻⁶)²/0.3² = 0.02497 N
Answer
F = 0.02497 N
Both spheres now carry the same sign, so they repel.
2.Coulomb’s law & electric fields
Answer: E = 3.928 × 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 = √(5² + 12²) = 13 cm
E1 = k·4 × 10⁻⁶/0.13² = 2.128 × 10⁶ N/C
Components
The components along the line joining the charges cancel.
E = 2·E1·(h/r) = 2·2.128 × 10⁶·0.9231 = 3.928 × 10⁶ N/C
Answer
E = 3.928 × 10⁶ N/C
The field points along the bisector, away from the charges.
3.Series & parallel resistors
Answer: Req = 12.4 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 = 8 + 4 = 12 Ω
R2 ∥ 12 = 3·12/(3+12) = 2.4 Ω
Req = R1 + 2.4 = 10 + 2.4 = 12.4 Ω
Answer
Req = 12.4 Ω
4.Series & parallel resistors
Answer: Req = 21.76 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.
R5 + R6 = 5 + 15 = 20 Ω
R4 ∥ 20 = 6.667 Ω
R3 + 6.667 = 14.67 Ω
R2 ∥ 14.67 = 1.76 Ω
Req = R1 + 1.76 = 20 + 1.76 = 21.76 Ω
Answer
Req = 21.76 Ω
5.RC & RL transients
Answer: t = 0.1475 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 = 10 mH / 47 Ω = 0.2128 ms
Solve for the time
0.5 = 1·e−t/τ → t = τ·ln(1/0.5)
t = 0.2128·ln(2) = 0.1475 ms
Answer
t = 0.1475 ms
6.RC & RL transients
Answer: t = 533.6 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.
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) = 6 V, vC(∞) = Vs = 5 V
τ = R·C = 6.8 kΩ · 220 µF = 1496 ms
vC(t) = 5 + (6 − 5)·e−t/τ
Solve for the time
5.7 = 5 + (1)·e−t/τ → e−t/τ = (5.7 − 5)/(6 − 5) = 0.7
t = τ·ln((6 − 5)/(5.7 − 5)) = 1496·ln(1.429) = 533.6 ms
Answer
t = 533.6 ms
7.Magnetic fields & induction
Answer: up the page
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).
Right-hand rule
Point the fingers along the current or velocity and curl them towards B; the thumb gives the force on a positive charge.
v: to the right, B: out of the page
Negative charge
The force on an electron is opposite to that on a positive charge.
Direction of the force
up the page
Answer
F: up the page
8.Magnetic fields & induction
Answer: M = 1.579 mH
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).
Mutual inductance
M = μ₀N₁N₂A/l = 4π × 10⁻⁷·2000·200·0.001257/0.4 = 1.579 mH
Answer
M = 1.579 mH
9.AC impedance & phasors
Answer: φ = −74.6 °
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·100 = 628.3 rad/s
XL = ωL = 628.3·0.25 = 157.1 Ω
XC = 1/(ωC) = 1/(628.3·4.7·10⁻⁶) = 338.6 Ω
Impedance
In series the impedances add as complex numbers: Z = R + j(XL − XC).
|Z| = √(R² + (XL − XC)²) = √(50² + (−181.5)²) = 188.3 Ω
Phase angle
φ = arctan((XL − XC)/R) = arctan(−181.5/50) = −74.6°
Answer
φ = −74.6°
10.AC impedance & phasors
Answer: f0 = 107.3 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·√(0.1·22·10⁻⁶)) = 107.3 Hz
Answer
f0 = 107.3 Hz
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