11.1 Electrostatics

11.1 Electrostatics

Quick Answer: Electrostatics treats charges at rest. Coulomb's law gives the force between point charges; the electric field and potential describe how a charge would affect others nearby; capacitors store charge and energy using C = Q/V.

For the PAF Aeronautical Engineer Initial physics paper, electrostatics is examined at FSc (intermediate) level: formulas, directions, unit awareness, and one- or two-step numerical problems. Treat every symbol carefully — a missing square on distance or a wrong series/parallel capacitor rule is a classic mark-loss pattern.

Electric Charge and Coulomb's Law

Electric charge is a scalar quantity measured in coulombs (C). The elementary charge is e ≈ 1.6 × 10⁻¹⁹ C. Charge is conserved and comes in two types conventionally labeled positive and negative.

Coulomb's law for the magnitude of the electrostatic force between two point charges in vacuum (or air, to good approximation):

$F = k \frac{|q_1 q_2|}{r^2}$

where $k = \dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \mathrm{N\cdot m^2/C^2}$ and $r$ is the separation. Direction: like charges repel, unlike charges attract, along the line joining the charges.

Worked example — Coulomb force. Two point charges +2 μC and −3 μC are 30 cm apart in air. Find the force magnitude and state attraction or repulsion.

$q_1 = 2 \times 10^{-6}\ \mathrm{C}$, $q_2 = 3 \times 10^{-6}\ \mathrm{C}$, $r = 0.30\ \mathrm{m}$.

$F = (9 \times 10^9)\frac{(2\times 10^{-6})(3\times 10^{-6})}{(0.30)^2} = (9 \times 10^9)\frac{6 \times 10^{-12}}{0.09} = 0.60\ \mathrm{N}$

Unlike signs → attractive force of 0.60 N.

Exam trap: keep $r$ in metres. Using 30 cm without converting gives an answer wrong by a factor of 10⁴.

Electric Field

The electric field at a point is the force per unit positive test charge:

$\vec{E} = \frac{\vec{F}}{q_0} \quad \Rightarrow \quad \vec{F} = q\vec{E}$

Unit: N/C (equivalently V/m). For a point charge $q$ in vacuum:

$E = k \frac{|q|}{r^2}$

Field lines leave positive charge and enter negative charge; denser lines mean stronger field. In a uniform field (ideal parallel plates), E is constant in magnitude and direction between the plates.

Worked example — field and force. A +4 μC charge produces a field at a point 20 cm away. Find E, then the force on a +1 μC test charge placed there.

$E = (9 \times 10^9)\frac{4 \times 10^{-6}}{(0.20)^2} = 9 \times 10^5\ \mathrm{N/C}$

$F = qE = (1 \times 10^{-6})(9 \times 10^5) = 0.90\ \mathrm{N}$

(away from the source charge, since both are positive).

Electric Potential and Potential Difference

Electric potential $V$ at a point is the work done by an external agent per unit positive charge in bringing that charge from infinity (reference $V_\infty = 0$) to the point, slowly, without kinetic energy change. For a point charge:

$V = k \frac{q}{r}$

Unit: volt (V) = J/C. Potential is a scalar — easier for multi-charge problems than vector fields: add algebraic potentials, then find field from geometry if needed.

Potential difference between two points:

$\Delta V = V_A - V_B = \frac{W}{q}$

In a uniform field of magnitude $E$ over separation $d$:

$V = Ed \quad \text{(magnitude relation)}$

Moving a charge $q$ through potential difference ΔV requires work $W = q\Delta V$ (sign depends on whether the field or an external agent does the work — exam questions usually ask magnitude or specify "against the field").

Worked example — potential. Find the potential due to a +5 μC charge at 50 cm, then the work to bring a +2 μC charge from infinity to that point.

$V = (9 \times 10^9)\frac{5 \times 10^{-6}}{0.50} = 9 \times 10^4\ \mathrm{V}$

$W = qV = (2 \times 10^{-6})(9 \times 10^4) = 0.18\ \mathrm{J}$

Capacitors

A capacitor stores separated charge. Capacitance:

$C = \frac{Q}{V}$

Unit: farad (F). Practical devices are μF, nF, or pF. For a parallel-plate capacitor (vacuum/air):

$C = \frac{\varepsilon_0 A}{d}$

Increasing plate area $A$ or decreasing separation $d$ increases C. Inserting a dielectric of constant κ multiplies capacitance: $C = \kappa\varepsilon_0 A/d$.

Energy stored in a charged capacitor:

$U = \frac{1}{2}CV^2 = \frac{1}{2}QV = \frac{Q^2}{2C}$

In parallel, voltage across each capacitor equals the supply; charges add. In series, charge on each is the same; voltages add.

Worked example — parallel plates and energy. A parallel-plate capacitor has A = 0.02 m², d = 1.0 mm, air dielectric. It is charged to 100 V. Find C and stored energy.

$C = \frac{(8.85 \times 10^{-12})(0.02)}{1.0 \times 10^{-3}} = 1.77 \times 10^{-10}\ \mathrm{F} = 177\ \mathrm{pF}$

$U = \frac{1}{2}(1.77 \times 10^{-10})(100)^2 \approx 8.85 \times 10^{-7}\ \mathrm{J}$

Worked example — series capacitors. Two capacitors 3 μF and 6 μF in series across 12 V. Find C_eq, charge, and voltage on each.

$\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2} \Rightarrow C_{eq} = 2\ \mu\mathrm{F}$

$Q = C_{eq}V = 2 \times 12 = 24\ \mu\mathrm{C}$

$V_3 = Q/C_3 = 24/3 = 8\ \mathrm{V},\quad V_6 = 24/6 = 4\ \mathrm{V}$

(Sum 12 V checks.) Smaller capacitance in series takes the larger share of voltage.

Aviation Context (Exam Awareness)

Fueling and hangar work involve static discharge risk: charge builds on airframe or personnel and can spark. Bonding and grounding equalize potential. You are not asked for aviation SOPs here, but linking "potential difference → discharge" helps retention for interview-style oral questions.

Section Checklist

• Convert μC ↔ C and cm ↔ m before substituting.
• Field is a vector; potential is a scalar.
• Capacitor series/parallel rules are the inverse of resistor rules.
• Energy formulas: prefer $½CV^2$ when V is known; $Q^2/(2C)$ when Q and C are known.