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 +4 µC and +10 µC. They are touched together and then put back 40 cm apart. Find the magnitude of the force F between them now. Use k = 8.99 × 10⁹ N·m²/C².
q1 = +4 µCq2 = +10 µCr = 40 cm
F = N
2.Coulomb’s law & electric fieldsHard
An electron enters the uniform field between two parallel plates moving parallel to them at 4.5 × 10⁷ m/s. The plates are 5 cm long and the field between them is 4800 N/C. Electron charge 1.602 × 10⁻¹⁹ C, mass 9.109 × 10⁻³¹ kg; ignore gravity. How far y is it deflected sideways when it leaves the plates? Give the answer in mm.
v₀ = 4.5 × 10⁷ m/sL = 5 cmE = 4800 N/Ce = 1.602 × 10⁻¹⁹ Cme = 9.109 × 10⁻³¹ kg
y = mm
3.Series & parallel resistorsMedium
Find the current IR1 delivered by the source, which flows through R1.
V1 = 15 VR1 = 5 ΩR2 = 5 ΩR3 = 6 ΩR4 = 4 Ω
IR1 = A
4.Series & parallel resistorsHard
Find the current IR2 through R2.
V1 = 6 VR1 = 5 ΩR2 = 25 ΩR3 = 4 ΩR4 = 3 ΩR5 = 10 ΩR6 = 25 Ω
IR2 = A
5.RC & RL transientsMedium
The inductor carries no current before the switch closes at t = 0. Find the voltage vL across the inductor at t = 10 ms.
Vs = 15 VR = 50 ΩL = 220 mHt = 10 ms
vL = 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 = 75 ms.
Vs = 10 VR1 = 3.3 kΩR2 = 2.2 kΩC = 68 µFt = 75 ms
vC = V
7.Magnetic fields & inductionMedium
An alpha particle moves at 6 × 10⁵ m/s perpendicular to a uniform magnetic field of 1 T. Find the radius r of its circular path.
m = 6.645 × 10⁻²⁷ kg|q| = 3.204 × 10⁻¹⁹ Cv = 6 × 10⁵ m/sB = 1 T
r = cm
8.Magnetic fields & inductionHard
An alpha particle passes undeflected through a velocity selector with crossed fields E = 5 × 10⁴ V/m and B₁ = 0.05 T. It then enters a second region with only a magnetic field B₂ = 1 T, at right angles. Find the radius r of its path there.
E = 5 × 10⁴ V/mB1 = 0.05 TB2 = 1 Tm = 6.645 × 10⁻²⁷ kg|q| = 3.204 × 10⁻¹⁹ C
r = cm
9.AC impedance & phasorsMedium
A resistor and an inductor are connected in parallel across a 230 V (rms), 60 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).
V = 230 V (rms)f = 60 HzR = 60 ΩL = 200 mH
φ = °
10.AC impedance & phasorsHard
A series RLC circuit is connected to a 120 V (rms), 50 Hz supply. Find the resonant frequency f0 of the circuit.
V = 120 V (rms)f = 50 HzR = 50 ΩL = 200 mHC = 4.7 µF
f0 = Hz
3Solutions
1.Coulomb’s law & electric fields
Answer: F = 2.753 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′ = (4 + 10)/2 = 7 µC
New force
F = 8.99 × 10⁹·(7 × 10⁻⁶)²/0.4² = 2.753 N
Answer
F = 2.753 N
Both spheres now carry the same sign, so they repel.
2.Coulomb’s law & electric fields
Answer: y = 0.5211 mm
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.
Time between the plates
t = L/v₀ = 0.05/4.5 × 10⁷ = 1.111 × 10⁻⁹ s
Sideways acceleration
a = e·E/m = 1.602 × 10⁻¹⁹·4800/9.109 × 10⁻³¹ = 8.442 × 10¹⁴ m/s²
Deflection
y = ½·a·t² = ½·8.442 × 10¹⁴·(1.111 × 10⁻⁹)² = 0.0005211 m = 0.5211 mm
Answer
y = 0.5211 mm
The electron is pushed towards the positive plate, against the field direction.
3.Series & parallel resistors
Answer: IR1 = 1.8 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 = 6 + 4 = 10 Ω
R2 ∥ 10 = 5·10/(5+10) = 3.333 Ω
Req = R1 + 3.333 = 5 + 3.333 = 8.333 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 = 15/8.333 = 1.8 A
Answer
IR1 = 1.8 A
4.Series & parallel resistors
Answer: IR2 = 0.1238 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 = 10 + 25 = 35 Ω
R4 ∥ 35 = 2.763 Ω
R3 + 2.763 = 6.763 Ω
R2 ∥ 6.763 = 5.323 Ω
Req = R1 + 5.323 = 5 + 5.323 = 10.32 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 = 6/10.32 = 0.5812 A
Current through R₂
Subtract the drop across R₁ to get the voltage at node A, which is the voltage across R₂.
VA = V1 − IR1·R1 = 6 − 0.5812·5 = 3.094 V
IR2 = VA/R2 = 3.094/25 = 0.1238 A
Answer
IR2 = 0.1238 A
5.RC & RL transients
Answer: vL = 1.545 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.
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 = 220 mH / 50 Ω = 4.4 ms
i(∞) = Vs/R = 15/50 = 0.3 A
Inductor voltage
At t = 0 the whole supply voltage appears across the inductor; it then decays to zero.
vL(t) = Vs·e−t/τ = 15·e−10/4.4 = 1.545 V
Answer
vL = 1.545 V
6.RC & RL transients
Answer: vC = 2.423 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) = 10·2.2/(3.3 + 2.2) = 4 V
Time constant after the switch opens
τ = R2·C = 2.2·68 = 149.6 ms
Discharge
vC(t) = 4·e−75/149.6 = 2.423 V
Answer
vC = 2.423 V
7.Magnetic fields & induction
Answer: r = 1.244 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).
Force balance
The magnetic force is always perpendicular to the velocity, so it bends the path into a circle.
|q|vB = mv²/r → r = mv/(|q|B)
Radius
r = 6.645 × 10⁻²⁷·6 × 10⁵/(3.204 × 10⁻¹⁹·1) = 0.01244 m = 1.244 cm
Answer
r = 1.244 cm
8.Magnetic fields & induction
Answer: r = 2.074 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 × 10⁴/0.05 = 1 × 10⁶ m/s
Radius in the second field
r = mv/(|q|B2) = 0.02074 m = 2.074 cm
Answer
r = 2.074 cm
9.AC impedance & phasors
Answer: φ = 38.51 °
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
XL = ωL = 2π·60·0.2 = 75.4 Ω
Branch currents
Both branches are connected straight across the supply, so each sees the full voltage V.
IR = V/R = 230/60 = 3.833 A
IX = V/XL = 230/75.4 = 3.05 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) = √(3.833² + 3.05²) = 4.899 A
Phase angle
The inductor current lags V, so the total current lags and the angle is positive.
φ = arctan(IX/IR) = arctan(3.05/3.833) = 38.51°
Answer
φ = 38.51°
10.AC impedance & phasors
Answer: f0 = 164.2 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·√(0.2·4.7·10⁻⁶)) = 164.2 Hz
Answer
f0 = 164.2 Hz
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