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 +8 µC and a charge of −6 µC are 40 cm apart. Point P lies on the line between them, 30 cm from the first charge. Find the magnitude of the electric field E at P.
q1 = +8 µCq2 = −6 µCr1 = 30 cmr2 = 10 cm
E = N/C
2.Coulomb’s law & electric fieldsHard
Three positive point charges sit at the corners of a right triangle: q₃ = 6 µC is at the right-angle corner, q₁ = 8 µC is 5 cm from it along one side and q₂ = 6 µC is 10 cm from it along the other side. Find the magnitude of the net electric force F on q₃.
q1 = 8 µCq2 = 6 µCq3 = 6 µCr13 = 5 cmr23 = 10 cm
F = N
3.Series & parallel resistorsMedium
Find the current IR2 through R2.
V1 = 24 VR1 = 6 ΩR2 = 8 ΩR3 = 12 ΩR4 = 2 Ω
IR2 = A
4.Series & parallel resistorsHard
Find the current IR2 through R2.
V1 = 9 VR1 = 6 ΩR2 = 20 ΩR3 = 30 ΩR4 = 4 ΩR5 = 5 ΩR6 = 10 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 = 1 ms.
Vs = 10 VR = 20 ΩL = 22 mHt = 1 ms
vL = V
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 = 250 ms.
Vs = 9 VR = 2.2 kΩC = 100 µFvC(0) = 10 Vt = 250 ms
vC = V
7.Magnetic fields & inductionMedium
A proton moving at 3 × 10⁶ m/s enters a uniform magnetic field at right angles and moves on a circle of radius 6.266 cm. Find the magnetic flux density B.
m = 1.673 × 10⁻²⁷ kg|q| = 1.602 × 10⁻¹⁹ Cv = 3 × 10⁶ m/sr = 6.266 cm
B = T
8.Magnetic fields & inductionHard
An alpha particle starts from rest, is accelerated through a potential difference of 2500 V and then enters a uniform magnetic field of 0.8 T at right angles. Find the radius r of its path in the field.
m = 6.645 × 10⁻²⁷ kg|q| = 3.204 × 10⁻¹⁹ CV = 2500 VB = 0.8 T
r = cm
9.AC impedance & phasorsMedium
A series RLC circuit is connected to a 120 V (rms), 200 Hz supply. Find the power factor cos φ of the circuit.
V = 120 V (rms)f = 200 HzR = 40 ΩL = 50 mHC = 2.2 µF
cos φ =
10.AC impedance & phasorsHard
A series RLC circuit is connected to a 100 V (rms), 200 Hz supply. Find the resonant frequency f0 of the circuit.
V = 100 V (rms)f = 200 HzR = 80 ΩL = 50 mHC = 1 µF
f0 = Hz
3Solutions
1.Coulomb’s law & electric fields
Answer: E = 6.193 × 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.
Field of each charge
E1 = k·8 × 10⁻⁶/0.3² = 799111 N/C
E2 = k·6 × 10⁻⁶/0.1² = 5.394 × 10⁶ N/C
Directions and sum
The first field points away from the positive charge and the second towards the negative charge: both point the same way.
E = |799111 + 5.394 × 10⁶| = 6.193 × 10⁶ N/C
Answer
E = 6.193 × 10⁶ N/C
2.Coulomb’s law & electric fields
Answer: F = 175.6 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·8·6 × 10⁻¹²/0.05² = 172.6 N
F2 = k·6·6 × 10⁻¹²/0.1² = 32.36 N
Vector sum
F = √(172.6² + 32.36²) = 175.6 N
Answer
F = 175.6 N
3.Series & parallel resistors
Answer: IR2 = 1.377 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 = 12 + 2 = 14 Ω
R2 ∥ 14 = 8·14/(8+14) = 5.091 Ω
Req = R1 + 5.091 = 6 + 5.091 = 11.09 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 = 24/11.09 = 2.164 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 = 24 − 2.164·6 = 11.02 V
IR2 = VA/R2 = 11.02/8 = 1.377 A
Answer
IR2 = 1.377 A
4.Series & parallel resistors
Answer: IR2 = 0.3039 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 = 5 + 10 = 15 Ω
R4 ∥ 15 = 3.158 Ω
R3 + 3.158 = 33.16 Ω
R2 ∥ 33.16 = 12.48 Ω
Req = R1 + 12.48 = 6 + 12.48 = 18.48 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 = 9/18.48 = 0.4871 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 = 9 − 0.4871·6 = 6.077 V
IR2 = VA/R2 = 6.077/20 = 0.3039 A
Answer
IR2 = 0.3039 A
5.RC & RL transients
Answer: vL = 4.029 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 = 22 mH / 20 Ω = 1.1 ms
i(∞) = Vs/R = 10/20 = 0.5 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 = 10·e−1/1.1 = 4.029 V
Answer
vL = 4.029 V
6.RC & RL transients
Answer: vC = 9.321 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 = 9 V
τ = R·C = 2.2 kΩ · 100 µF = 220 ms
vC(t) = 9 + (10 − 9)·e−t/τ
Voltage at the given time
vC(250) = 9 + (1)·e−250/220 = 9.321 V
Answer
vC = 9.321 V
7.Magnetic fields & induction
Answer: B = 0.5 T
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).
Field from the radius
B = mv/(|q|r) = 1.673 × 10⁻²⁷·3 × 10⁶/(1.602 × 10⁻¹⁹·0.06266) = 0.5 T
Answer
B = 0.5 T
8.Magnetic fields & induction
Answer: r = 1.273 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).
Speed after the acceleration
v = √(2|q|V/m) = √(2·3.204 × 10⁻¹⁹·2500/(6.645 × 10⁻²⁷)) = 491003 m/s
Radius in the field
r = mv/(|q|B) = 0.01273 m = 1.273 cm
Answer
r = 1.273 cm
Combined: r = √(2mV/|q|)/B, so heavier ions of the same charge move on larger circles; this separates isotopes.
9.AC impedance & phasors
Answer: cos φ = 0.1326
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
XL = 62.83 Ω, XC = 361.7 Ω
Impedance
|Z| = √(40² + (−298.9)²) = 301.5 Ω
Power factor
cos φ = R/|Z| = 40/301.5 = 0.1326
The circuit is capacitive, so the power factor is leading.
Answer
cos φ = 0.1326
10.AC impedance & phasors
Answer: f0 = 711.8 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·√(0.05·1·10⁻⁶)) = 711.8 Hz
Answer
f0 = 711.8 Hz
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