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A-Level Coulomb's law, electric field strength, potential and energy, uniform fields and capacitance

What the A-Level syllabus expects for Coulomb's law, electric field strength, potential and energy, uniform fields and capacitance, and how to practise it.

What the syllabus expects

How it's examined

Questions on this topic most often ask you to find, show, sketch. About 5% of the past-paper style questions in Rae's bank for this subject sit in this topic.

Worked examples

Example 1 (5 marks)

A straight wire of infinite length carries a uniform linear charge density λ. Apply Gauss's Law to find the electric field strength E at a perpendicular distance r from the wire, expressed in terms of λ. Set out your reasoning clearly.

Show the worked answer

By symmetry the field is radial and has constant magnitude at fixed r. Choose a coaxial Gaussian cylinder of radius r and length L about the wire. Flux passes only through the curved surface (the flat ends are parallel to E, so contribute zero): flux = E(2 pi r L). Charge enclosed = lambda L. Gauss's Law: flux = Q_enc / e0, so E(2 pi r L) = lambda L / e0. Hence E = lambda / (2 pi e0 r).

Example 2 (3 marks)

(c) Although air is normally an insulator, when the electric potential gradient becomes large enough sparks (arcing) can leap across the gap between the capacitor plates. The lowest voltage at which such arcing begins is the breakdown voltage Vb. The capacitor from part (b) is charged by a supply built to deliver a constant current of I = 5.0 μA. A student times a period T = 10 minutes 30 seconds over which N = 250 sparks take place. Take it that each spark fully discharges the capacitor and that the permittivity of air is essentially equal to the permittivity of free space. (i) Show that the breakdown voltage of air may be written as Vb = ITd / (N ε0 A).

Show the worked answer

For a parallel-plate capacitor of plate area A and separation d with air (permittivity ~ epsilon_0), C = epsilon_0*A/d. It arcs when charged to the breakdown voltage Vb, so the charge stored at breakdown is Q = C*Vb = (epsilon_0*A/d)*Vb. The supply delivers constant current I, so in the total time T it delivers charge I*T, producing N sparks (full discharges). The charge accumulated between successive sparks is therefore Q = I*T/N. Equating the two expressions for Q: (epsilon_0*A/d)*Vb = I*T/N. Rearranging: Vb = I*T*d/(N*epsilon_0*A).

Example 3 (2 marks)

For an LC circuit in which a 2.0 μF capacitor is charged to 60 V and then connected across an inductor of L = 10 mH: find the initial charge Q_0 stored on the capacitor.

Show the worked answer

Initial charge on the capacitor Q0 = C V = (2.0x10⁻⁶ F)(60 V) = 1.2x10⁻⁴ C = 120 microcoulomb.

More worked questions on this topic

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