Electrostatics and Direct Current Circuits
Scope of the domain
This domain covers electric force and field, potential and capacitance, and direct-current circuits with resistance and internal resistance. The exam asks you to use Coulomb's law, field and potential relations, series and parallel rules, and Kirchhoff's laws to solve circuit problems.
Core rules
- Coulomb's law: F = kQ₁Q₂/r², with k = 8.99 × 10⁹ N m² C⁻²; force is along the line joining the charges and is repulsive for like charges.
- Electric field E = F/q; field lines point from positive to negative. In a uniform field E = V/d.
- Potential V = kQ/r; potential difference is work done per unit charge.
- Capacitance C = Q/V; energy stored = ½QV = ½CV². Capacitors in parallel add; in series the reciprocals add.
- Ohm's law V = IR at constant temperature; R = ρL/A. Series resistances add; for parallel, 1/R = 1/R₁ + 1/R₂.
- EMF ε = I(R + r); terminal p.d. V = ε − Ir. Power P = IV = I²R = V²/R.
- Kirchhoff: sum of currents at a junction is zero; sum of p.d.s around a loop equals the total EMF.
- An ammeter is connected in series with low resistance; a voltmeter in parallel with high resistance.
Worked example
A cell of EMF 12 V and internal resistance 0.50 Ω is connected to a 5.5 Ω resistor. Total resistance = 6.0 Ω, so I = 12 ÷ 6.0 = 2.0 A. Terminal p.d. = 12 − (2.0 × 0.50) = 11 V. Power dissipated in the external resistor = I²R = 4.0 × 5.5 = 22 W; power wasted internally = 4.0 × 0.50 = 2.0 W. The sum, 24 W, equals the total power εI = 12 × 2.0 = 24 W.
Traps that cost marks
- Ignoring internal resistance and quoting the EMF as the terminal p.d.
- Using the series rule for parallel resistors, or forgetting to invert the reciprocal at the end.
- Mixing mA and A, or cm and m, before substitution.
- Applying V = IR to the whole circuit without including r.
- Forgetting that adding resistors in parallel lowers total resistance.
- Treating kWh as a unit of power rather than energy.
- Reversing ammeter and voltmeter connections.