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
A charge of +1 µC and a charge of +8 µC are 50 cm apart. Point P lies on the line between them, 20 cm from the first charge. Find the magnitude of the electric field E at P.
q1 = +1 µCq2 = +8 µCr1 = 20 cmr2 = 30 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 = 1 µC is 10 cm from it along one side and q₂ = 6 µC is 6 cm from it along the other side. Find the magnitude of the net electric force F on q₃.
q1 = 1 µCq2 = 6 µCq3 = 5 µCr13 = 10 cmr23 = 6 cm
F = N
3.Series & parallel resistorsMedium
Find the equivalent resistance Req seen by the source V1.
V1 = 12 VR1 = 4 ΩR2 = 4 ΩR3 = 6 ΩR4 = 6 Ω
Req = Ω
4.Series & parallel resistorsHard
Find the equivalent resistance Req seen by the source V1.
V1 = 36 VR1 = 20 ΩR2 = 10 ΩR3 = 5 ΩR4 = 5 ΩR5 = 3 ΩR6 = 12 Ω
Req = Ω
5.RC & RL transientsMedium
The source Vₛ charges the capacitor through R₁, and R₂ is connected in parallel with C. The capacitor is uncharged when the switch closes at t = 0. Find the capacitor voltage vC at t = 80 ms.
Vs = 30 VR1 = 2.2 kΩR2 = 3.3 kΩC = 100 µFt = 80 ms
vC = V
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 vC at t = 100 ms.
Vs = 12 VR1 = 1 kΩR2 = 2.2 kΩC = 100 µFt = 100 ms
vC = V
7.Magnetic fields & inductionMedium
A wire carries a current to the left through a uniform magnetic field pointing into the page. Click the direction of the magnetic force on the wire.
I: to the leftB: into the page
- 1up the page
- 2to the right
- 3down the page
- 4to the left
8.Magnetic fields & inductionHard
An alpha particle passes undeflected through a velocity selector with crossed fields E = 5 × 10⁴ V/m and B₁ = 0.2 T. It then enters a second region with only a magnetic field B₂ = 0.8 T, at right angles. Find the radius r of its path there.
E = 5 × 10⁴ V/mB1 = 0.2 TB2 = 0.8 Tm = 6.645 × 10⁻²⁷ kg|q| = 3.204 × 10⁻¹⁹ C
r = cm
9.AC impedance & phasorsMedium
A series RLC circuit is connected to a 24 V (rms), 1000 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).
V = 24 V (rms)f = 1000 HzR = 50 ΩL = 50 mHC = 4.7 µF
φ = °
10.AC impedance & phasorsHard
A series RLC circuit is connected to a 120 V (rms), 200 Hz supply. Find the rms voltage VC across the capacitor.
V = 120 V (rms)f = 200 HzR = 15 ΩL = 100 mHC = 47 µF
VC = V
3Solutions
1.Coulomb’s law & electric fields
Answer: E = 574361 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·1 × 10⁻⁶/0.2² = 224750 N/C
E2 = k·8 × 10⁻⁶/0.3² = 799111 N/C
Directions and sum
Both charges are positive: at P their fields point away from each charge, so in opposite directions.
E = |224750 − 799111| = 574361 N/C
Answer
E = 574361 N/C
2.Coulomb’s law & electric fields
Answer: F = 75.05 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.
Each force on q₃
F1 = k·1·5 × 10⁻¹²/0.1² = 4.495 N
F2 = k·6·5 × 10⁻¹²/0.06² = 74.92 N
Vector sum
F = √(4.495² + 74.92²) = 75.05 N
Answer
F = 75.05 N
3.Series & parallel resistors
Answer: Req = 7 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 = 6 + 6 = 12 Ω
R2 ∥ 12 = 4·12/(4+12) = 3 Ω
Req = R1 + 3 = 4 + 3 = 7 Ω
Answer
Req = 7 Ω
4.Series & parallel resistors
Answer: Req = 24.67 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 = 3 + 12 = 15 Ω
R4 ∥ 15 = 3.75 Ω
R3 + 3.75 = 8.75 Ω
R2 ∥ 8.75 = 4.667 Ω
Req = R1 + 4.667 = 20 + 4.667 = 24.67 Ω
Answer
Req = 24.67 Ω
5.RC & RL transients
Answer: vC = 8.181 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.
Thévenin equivalent seen by the capacitor
Short the source to find Rth; open the capacitor to find the voltage it will settle at.
Rth = R1 ∥ R2 = 2.2·3.3/(2.2 + 3.3) = 1.32 kΩ
vC(∞) = 30·3.3/(2.2 + 3.3) = 18 V
Time constant
τ = Rth·C = 1.32·100 = 132 ms
Charging curve
vC(t) = 18·(1 − e−80/132) = 8.181 V
Answer
vC = 8.181 V
6.RC & RL transients
Answer: vC = 5.237 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 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) = 12·2.2/(1 + 2.2) = 8.25 V
Time constant after the switch opens
τ = R2·C = 2.2·100 = 220 ms
Discharge
vC(t) = 8.25·e−100/220 = 5.237 V
Answer
vC = 5.237 V
7.Magnetic fields & induction
Answer: down 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.
I: to the left, B: into the page
Direction of the force
down the page
Answer
F: down the page
8.Magnetic fields & induction
Answer: r = 0.6481 cm
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).
Selected speed
Undeflected: the electric force qE balances the magnetic force qvB.
v = E/B1 = 5 × 10⁴/0.2 = 250000 m/s
Radius in the second field
r = mv/(|q|B2) = 0.006481 m = 0.6481 cm
Answer
r = 0.6481 cm
9.AC impedance & phasors
Answer: φ = 79.89 °
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·1000 = 6283 rad/s
XL = ωL = 6283·0.05 = 314.2 Ω
XC = 1/(ωC) = 1/(6283·4.7·10⁻⁶) = 33.86 Ω
Impedance
In series the impedances add as complex numbers: Z = R + j(XL − XC).
|Z| = √(R² + (XL − XC)²) = √(50² + (280.3)²) = 284.7 Ω
Phase angle
φ = arctan((XL − XC)/R) = arctan(280.3/50) = 79.89°
Answer
φ = 79.89°
10.AC impedance & phasors
Answer: VC = 18.51 V
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
XL = ωL = 1257·0.1 = 125.7 Ω
XC = 1/(ωC) = 1/(1257·47·10⁻⁶) = 16.93 Ω
Impedance
In series the impedances add as complex numbers: Z = R + j(XL − XC).
|Z| = √(R² + (XL − XC)²) = √(15² + (108.7)²) = 109.8 Ω
Current
The same current flows through every element of a series circuit.
I = V/|Z| = 120/109.8 = 1.093 A
Voltage across the capacitor
VC = I·XC = 1.093·16.93 = 18.51 V
Answer
VC = 18.51 V
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