Pancyprian Physics study materials
Pancyprian Physics study resources

Pancyprian Physics: Electricity and Magnetism: Electrostatics, DC Circuits, Fields and Induction

Pancyprian Physics · Work through your exam, one topic at a time.

All guide sections

Electricity and Magnetism: Electrostatics, DC Circuits, Fields and Induction

This domain links three strands that the exam often joins in one question: electrostatic fields and potentials, DC circuit analysis with real sources, and magnetic forces with electromagnetic induction. You are expected to calculate, but also to justify directions using field lines, Lenz's law and conservation of energy.

Electrostatics

Coulomb's law: F = kQ₁Q₂/r², with k = 8.99 × 10⁹ N·m²/C². Field strength E = F/q, and between parallel plates E = V/d is uniform. Potential V = kQ/r is a scalar, so potentials add algebraically, while fields add as vectors. Work done moving a charge is W = qV. A capacitor stores Q = CV and energy ½QV; inserting a dielectric increases C.

DC circuits

Ohm's law V = IR holds for ohmic conductors at constant temperature. Resistors in series add; in parallel, 1/R = Σ1/Rᵢ and the total is smaller than the smallest. Kirchhoff's junction rule is charge conservation; the loop rule is energy conservation. A real source has emf ε and internal resistance r, so the terminal voltage is V = ε − Ir and the current is I = ε/(R + r). Power: P = VI = I²R = V²/R.

Magnetism and induction

A current-carrying wire in a field feels F = BIL sin θ; a moving charge feels F = qvB, giving circular motion of radius r = mv/qB. Magnetic flux is Φ = BA cos θ, where θ is measured from the normal to the area. Faraday's law gives ε = −N dΦ/dt, and Lenz's law fixes the direction: the induced current opposes the change producing it. Transformers obey V_s/V_p = N_s/N_p.

Worked example

A battery of emf 12 V and internal resistance 0.50 Ω is connected to a 5.5 Ω resistor. Find the current, terminal voltage and power wasted internally. I = ε/(R + r) = 12/6.0 = 2.0 A. Terminal voltage V = ε − Ir = 12 − (2.0)(0.50) = 11 V. Internal power = I²r = (4.0)(0.50) = 2.0 W.

Traps that cost marks

  • Believing the field inside a charged conductor is large; it is zero in electrostatic equilibrium.
  • Setting E = 0 wherever V = 0, which is false at the midpoint between two equal like charges.
  • Forgetting to convert cm, cm² or mA to SI before substituting.
  • Using the angle with the plane rather than with the normal in Φ = BA cos θ.
  • Reversing the direction of the induced current, or omitting the minus sign that encodes Lenz's law.