amazon.co.ukA-Level Physics revision: Coulomb's law, electric field strength, potential, charged particle motion, capacitance, energy stored, and exponential capacitor charge/discharge with the time constant - AQA 7408 3.7.3-3.7.4, OCR A H556 6.1-6.2, Edexcel 9PH0 Topic 7.
Coulomb's law: F = q1q2/(4πε0r^2) - the force between two point charges is proportional to each charge and to the inverse square of their separation; like charges repel, unlike attract (AQA 7408 3.7.3.1; Edexcel 9PH0 Topic 7; OCR A H556 6.2). ε0 = 8.85 × 10^-12 F/m (permittivity of free space); in a medium, replace with ε = ε0εr. The inverse-square form matches gravity (F = GMm/r^2) - same geometry (field lines spreading over spheres of area 4πr^2) - but charge has two signs, so electric forces attract or repel while gravity only attracts, and the electric force between subatomic particles is vastly stronger than gravity.
Electric field strength is force per unit positive charge: E = F/q, in N/C (= V/m). For a point charge, substitute Coulomb's law: E = Q/(4πε0r^2) - radially outward from a positive charge, inward for negative. Field lines point in the direction of the force on a positive test charge; denser lines mean stronger field. Between two parallel plates the field is uniform: E = V/d (derivation: work done moving charge q between plates = qV; also W = Eqd; equate: Eqd = qV, so E = V/d) - field lines parallel, evenly spaced, from the positive to the negative plate.
Two parallel plates 5.0 mm apart are connected to a 200 V supply. Find E, and the force and acceleration on an electron between them (m = 9.11 × 10^-31 kg).
Line 1: E = V/d = 200/5.0 × 10^-3 = 4.0 × 10^4 V/m (M1 A1: uniform field).
Line 2: F = eE = 1.60 × 10^-19 × 4.0 × 10^4 = 6.4 × 10^-15 N (M1 A1), towards the positive plate.
Line 3: a = F/m = 6.4 × 10^-15/9.11 × 10^-31 = 7.0 × 10^15 m/s^2 (A1) - enormous compared with g.
Electric potential at a point is the work done per unit charge bringing a small positive test charge from infinity: for a point charge Q, V = Q/(4πε0r) (AQA 7408 3.7.3.3). Positive near positive charges (repulsive - you do positive work pushing in), negative near negative charges, zero at infinity. Potential energy of charge q: E_p = qV = q1q2/(4πε0r). Moving a charge between two points: ΔW = qΔV - work done equals charge times potential difference; the pd is the energy per coulomb.
Field strength is the potential gradient: E = -dV/dr - the field points down the potential slope. Graphically, V against r for a point charge is a curve falling as 1/r from large values at small r; the gradient at any r gives -E. Equipotential surfaces are spheres of constant V around a point charge (parallel planes between plates): moving a charge along one does no work (ΔV = 0), and field lines cross them at right angles - the two diagram facts always tested. Between parallel plates, equipotentials are evenly spaced planes because E is uniform.
Comparison table with gravity (standard six-mark synthesis): force inverse-square in both; field strength force per unit charge/mass; potential zero at infinity in both; E/g = -dV/dr in both; gravity always attractive (V_g = -GM/r always negative) while electric potential has either sign; electric field can be shielded by conductors, gravity cannot.
Find the potential 0.30 m from a +5.0 μC point charge, and the work done moving a +2.0 μC charge from 0.50 m to 0.30 m.
Line 1: V = Q/(4πε0r) = (5.0 × 10^-6)/(4π × 8.85 × 10^-12 × 0.30) (M1) = 1.5 × 10^5 V (A1).
Line 2: V at 0.50 m: = 9.0 × 10^4 V (same expression, r = 0.50) (A1).
Line 3: ΔV = 1.5 × 10^5 - 9.0 × 10^4 = 6.0 × 10^4 V (M1: potential rises moving towards the positive charge).
Line 4: ΔW = qΔV = 2.0 × 10^-6 × 6.0 × 10^4 = 0.12 J (A1: work done against the repulsive force).
A charged particle between parallel plates behaves exactly like a projectile (AQA 7408 3.7.3.2; OCR A H556 6.2): constant velocity parallel to the plates (no force that way) and constant acceleration perpendicular (force qE). The path is a parabola. Standard analysis: enter with speed v parallel to the plates; acceleration a = qE/m = qV/(md) perpendicular; time between plates t = L/v (length of plates over horizontal speed); vertical deflection y = ½at^2 = qVL^2/(2mdv^2).
Work-energy route: a particle accelerated through a pd V gains kinetic energy qV: ½mv^2 = qV, so v = √(2qV/m) - the standard electron-gun calculation (used for the de Broglie and specific-charge questions too). Between deflecting plates, the exit velocity has components: parallel v unchanged, perpendicular v_y = at.
An electron enters at 2.0 × 10^7 m/s midway between plates 5.0 mm apart, 200 V, of length 40 mm. Find its vertical deflection on exit.
Line 1: E = V/d = 200/5.0 × 10^-3 = 4.0 × 10^4 V/m (M1 A1).
Line 2: a = eE/m = 1.60 × 10^-19 × 4.0 × 10^4/9.11 × 10^-31 = 7.0 × 10^15 m/s^2 (M1 A1).
Line 3: t = L/v = 0.040/2.0 × 10^7 = 2.0 × 10^-9 s (M1: time from horizontal motion).
Line 4: y = ½at^2 = ½ × 7.0 × 10^15 × (2.0 × 10^-9)^2 = 1.4 × 10^-2 m = 14 mm (A1) - more than half the plate gap, so the electron would hit a plate; check the geometry.
A capacitor stores charge on two plates separated by an insulator (AQA 7408 3.7.4.1; OCR A H556 6.1; Edexcel 9PH0 Topic 7). Capacitance is charge stored per unit pd: C = Q/V, in farads (1 F = 1 C/V); practical values are μF (10^-6) or pF (10^-12). For a parallel-plate capacitor with plate area A, separation d and dielectric of relative permittivity εr: C = Aε0εr/d - larger plates and closer separation store more charge (εr for air ≈ 1, ceramics and oxides much larger).
Dielectric behaviour: the insulator's molecules polarise in the field - negative ends shift towards the positive plate - reducing the field inside for the same charge, so a lower pd is needed for the same Q and C increases by εr.
Energy stored derivation: as the capacitor charges, the pd rises linearly from 0 to V while charge builds to Q, so the average pd during charging is V/2. Energy stored = ∫V dQ = ½QV - the area under the Q-V graph. Substituting Q = CV gives the useful forms W = ½CV^2 and W = ½Q^2/C.
Combinations: capacitors in parallel share the pd and add their charges: C_total = ΣC (opposite to resistors); in series each stores the same charge and the pds add: 1/C_total = Σ1/C - series capacitors always give less than the smallest capacitance. Remember by which quantity is common: charge common in series, pd common in parallel.
Capacitors of 100 μF and 300 μF are connected in series across a 24 V supply. Find the charge stored, the pd across each, and the total energy stored.
Line 1: Series: 1/C_total = 1/100 + 1/300 (μF) = 0.0133, C_total = 75 μF (M1 A1).
Line 2: Series charge is common: Q = C_total V = 75 × 10^-6 × 24 = 1.8 × 10^-3 C (M1 A1).
Line 3: Pds: V_1 = Q/C_1 = 1.8 × 10^-3/100 × 10^-6 = 18 V; V_2 = 6 V - sum 24 V, checking Kirchhoff's loop rule (A1: verification).
Line 4: Total energy W = ½C_total V^2 = ½ × 75 × 10^-6 × 24^2 = 2.2 × 10^-2 J (A1) - less than a single 100 μF at 24 V would store (½CV^2 = 2.9 × 10^-2 J), because series capacitance is smaller.
Discharging a capacitor C through a resistor R (AQA 7408 3.7.4.2-3.7.4.3; OCR PAG8; Edexcel CP9): the pd drives current I = V/R, which falls as the charge (and pd) falls - an exponential decay.
Derivation of Q = Q0e^(-t/RC): at any moment Q = CV and the discharge current I = -dQ/dt = V/R = Q/(RC). This is a differential equation: dQ/dt = -Q/(RC). Separating variables and integrating: ∫dQ/Q = -∫dt/(RC), so ln(Q/Q0) = -t/RC, giving Q = Q0e^(-t/RC). Non-calculus version: in each small time step the charge loses a constant fraction proportional to itself - the defining property of exponential decay (like radioactive decay).
Time constant: τ = RC - the time for Q, V or I to fall to 1/e ≈ 37% of the initial value (set t = RC: Q = Q0/e). After 2RC: 13.5%; after 3RC: 5%; effectively discharged after ~5RC. Charging: Q = Q0(1 - e^(-t/RC)) - rises to 63% after one time constant, asymptotically to full. Half-life: t_½ = RC ln 2 = 0.693RC - the time to halve, constant like radioactivity.
Practical (required practical): charge a capacitor, discharge through a known R, and record V (or Q, via a data logger or multimeter) at timed intervals. Plot ln V against t: a straight line of gradient -1/RC, so C = -1/(R × gradient) - the log-graph method averages random error and tests the exponential model (a curve on a linear V-t plot, straight on ln V-t). Errors: multimeter resistance in parallel with R (use a high-impedance meter or data logger), timing resolution, capacitor tolerance (often ±20% - compare with the marked value using percentage difference).
A capacitor discharges through 47 kΩ. Its pd falls from 9.0 V to 3.3 V in 5.0 s.
Line 1: 3.3/9.0 = 0.37 ≈ 1/e, so t = 5.0 s is one time constant: RC = 5.0 s (M1 A1: recognition of 37%).
Line 2: C = RC/R = 5.0/47 × 10^3 = 1.06 × 10^-4 F ≈ 106 μF (M1 A1).
Line 3: Log-graph check: ln(9.0/3.3) = 1.0, gradient = 1.0/5.0 = 0.20 s^-1 = 1/RC (A1: consistent).