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 +3 µC and +2 µC. They are touched together and then put back 10 cm apart. Find the magnitude of the force F between them now. Use k = 8.99 × 10⁹ N·m²/C².
q1 = +3 µCq2 = +2 µCr = 10 cm
F = N
2.Coulomb’s law & electric fieldsHard
Charges of +10 µC and +10 µC are 12 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 = +10 µCq2 = +10 µCd = 12 cmh = 12 cm
E = N/C
3.Series & parallel resistorsMedium
Find the equivalent resistance Req seen by the source V1.
V1 = 10 VR1 = 15 ΩR2 = 12 ΩR3 = 4 ΩR4 = 10 Ω
Req = Ω
4.Series & parallel resistorsHard
Find the current IR2 through R2.
V1 = 12 VR1 = 6 ΩR2 = 6 ΩR3 = 30 ΩR4 = 4 ΩR5 = 6 ΩR6 = 8 Ω
IR2 = 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 99% of its final value? Find t.
Vs = 9 VR = 22 ΩL = 220 mH
t = ms
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 = 60 ms.
Vs = 9 VR1 = 6.8 kΩR2 = 10 kΩC = 22 µFt = 60 ms
vC = V
7.Magnetic fields & inductionMedium
Two long parallel wires 15 cm apart carry currents of 8 A and 12 A in opposite directions. Find the magnitude of the magnetic flux density B at point P, halfway between them.
I1 = 8 AI2 = 12 Ad = 15 cmμ₀ = 4π × 10⁻⁷ T·m/A
B = µT
8.Magnetic fields & inductionHard
A long air-core solenoid has 100 turns, a length of 50 cm and a diameter of 2.5 cm. Find the energy W stored in its magnetic field when it carries 1.5 A.
N = 100l = 50 cm⌀ = 2.5 cmI = 1.5 Aμ₀ = 4π × 10⁻⁷ T·m/A
W = µJ
9.AC impedance & phasorsMedium
A series RLC circuit is connected to a 10 V (rms), 200 Hz supply. Find the rms current I.
V = 10 V (rms)f = 200 HzR = 10 ΩL = 150 mHC = 10 µF
I = A
10.AC impedance & phasorsHard
A resistor R1 is connected in series with a parallel combination of R2 and a capacitor C, across a 24 V (rms), 100 Hz supply. Find the magnitude of the impedance |Z|.
V = 24 V (rms)f = 100 HzR1 = 40 ΩR2 = 20 ΩC = 47 µF
|Z| = Ω
3Solutions
1.Coulomb’s law & electric fields
Answer: F = 5.619 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′ = (3 + 2)/2 = 2.5 µC
New force
F = 8.99 × 10⁹·(2.5 × 10⁻⁶)²/0.1² = 5.619 N
Answer
F = 5.619 N
Both spheres now carry the same sign, so they repel.
2.Coulomb’s law & electric fields
Answer: E = 8.934 × 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 + 12²) = 13.42 cm
E1 = k·10 × 10⁻⁶/0.1342² = 4.994 × 10⁶ N/C
Components
The components along the line joining the charges cancel.
E = 2·E1·(h/r) = 2·4.994 × 10⁶·0.8944 = 8.934 × 10⁶ N/C
Answer
E = 8.934 × 10⁶ N/C
The field points along the bisector, away from the charges.
3.Series & parallel resistors
Answer: Req = 21.46 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 = 4 + 10 = 14 Ω
R2 ∥ 14 = 12·14/(12+14) = 6.462 Ω
Req = R1 + 6.462 = 15 + 6.462 = 21.46 Ω
Answer
Req = 21.46 Ω
4.Series & parallel resistors
Answer: IR2 = 0.9169 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 = 6 + 8 = 14 Ω
R4 ∥ 14 = 3.111 Ω
R3 + 3.111 = 33.11 Ω
R2 ∥ 33.11 = 5.08 Ω
Req = R1 + 5.08 = 6 + 5.08 = 11.08 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 = 12/11.08 = 1.083 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 = 12 − 1.083·6 = 5.502 V
IR2 = VA/R2 = 5.502/6 = 0.9169 A
Answer
IR2 = 0.9169 A
5.RC & RL transients
Answer: t = 46.05 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 = 220 mH / 22 Ω = 10 ms
i(∞) = Vs/R = 9/22 = 0.4091 A
Solve for the time
0.99 = 1 − e−t/τ → t = −τ·ln(1 − 0.99)
t = −10·ln(0.01) = 46.05 ms
Answer
t = 46.05 ms
6.RC & RL transients
Answer: vC = 4.078 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) = 9·10/(6.8 + 10) = 5.357 V
Time constant after the switch opens
τ = R2·C = 10·22 = 220 ms
Discharge
vC(t) = 5.357·e−60/220 = 4.078 V
Answer
vC = 4.078 V
7.Magnetic fields & induction
Answer: B = 53.33 µT
Magnetic field of currents: around a long straight wire the field lines are circles (right-hand grip rule) with B = μ₀I/(2πr). At the centre of a flat circular coil of N turns, B = μ₀NI/(2R). Inside a long solenoid the field is uniform, B = μ₀nI with n = N/L turns per metre, whatever the radius. μ₀ = 4π × 10⁻⁷ T·m/A. Work in SI units (metres and amperes), and add the fields of several sources as vectors.
Field of each wire at P
B1 = 2 × 10⁻⁷·8/0.075 = 21.33 µT
B2 = 2 × 10⁻⁷·12/0.075 = 32 µT
Combine
B = B₁ + B₂ = 53.33 µT
Answer
B = 53.33 µT
8.Magnetic fields & induction
Answer: W = 13.88 µJ
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).
Self-inductance
L = μ₀N²A/l = 12.34 µH
Stored energy
W = ½LI² = ½·1.234 × 10⁻⁵·1.5² = 1.388 × 10⁻⁵ J = 13.88 µJ
Answer
W = 13.88 µJ
9.AC impedance & phasors
Answer: I = 0.09143 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|.
Reactances
ω = 2πf = 2π·200 = 1257 rad/s
XL = ωL = 1257·0.15 = 188.5 Ω
XC = 1/(ωC) = 1/(1257·10·10⁻⁶) = 79.58 Ω
Impedance
In series the impedances add as complex numbers: Z = R + j(XL − XC).
|Z| = √(R² + (XL − XC)²) = √(10² + (108.9)²) = 109.4 Ω
Current
The same current flows through every element of a series circuit.
I = V/|Z| = 10/109.4 = 0.09143 A
Answer
I = 0.09143 A
10.AC impedance & phasors
Answer: |Z| = 55.52 Ω
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/(2π·100·47·10⁻⁶) = 33.86 Ω
Parallel combination
Multiply top and bottom by the complex conjugate to split Zp into real and imaginary parts.
Zp = R₂·(−jXC)/(R₂ − jXC) = (R₂·XC² − jR₂²·XC)/(R₂² + XC²)
Zp = 14.83 − j8.757 Ω
Total impedance
Z = 40 + 14.83 − j8.757 = 54.83 − j8.757 Ω
|Z| = √(54.83² + 8.757²) = 55.52 Ω
Answer
|Z| = 55.52 Ω
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