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Study NotePHY102

Electrostatics: Charge & Coulomb's Law

Charge Basics

  • The SI unit of charge is the Coulomb.
  • The magnitude of charge is the same regardless of sign (a +2 μC and −2 μC charge have equal magnitude).
  • Franklin's positive/negative labeling convention came from an arbitrary choice (the charge on a rubbed glass rod was called positive) — it has nothing to do with charges being "motionless."
  • Charge is conserved: a charge that disappears in one setting can, in principle, be accounted for turning up elsewhere.

Worked example: a −2.5 C point charge contains how many electrons? n=Q/e=2.5/(1.6×1019)1.56×1019n = Q/e = 2.5/(1.6\times10^{-19}) \approx 1.56\times10^{19} electrons.

Coulomb's Law

F=Kq1q2r2,K=14πε08.99×109 N⋅m2/C2F = \dfrac{Kq_1q_2}{r^2}, \qquad K = \dfrac{1}{4\pi\varepsilon_0} \approx 8.99\times10^9 \text{ N·m}^2/\text{C}^2

The units of Coulomb's constant K are N·m²/C² — don't confuse this with the units of permittivity ε₀ (C²/N·m²), which is the inverse combination.

Worked example — building up charge assembly one at a time: four charges (+1μC, +2μC, +3μC, +4μC) are placed one-by-one at the corners of a 1 cm square:

  • Bringing the first charge into otherwise-empty space costs zero work (nothing to push against yet).
  • Bringing the second charge to an adjacent corner (0.01 m away): W=Kq1q2r=9×109×1×106×2×106/0.01=1.8W = \dfrac{Kq_1q_2}{r} = 9\times10^9\times1\times10^{-6}\times2\times10^{-6}/0.01 = 1.8 J.
  • Each subsequent charge must do work against every charge already placed — sum the pairwise work against each existing charge (using the straight side-length or diagonal distance as appropriate) to get that step's total.
  • The grand total work to assemble the whole configuration is simply the sum of all the individual steps.

Force Between Multiple Charges: Work in Components

When a charge feels forces from two different directions (e.g. attraction toward one charge, repulsion from another at a right angle), find each force separately, then combine them as perpendicular vector components:

Fnet=Fx2+Fy2,θ=arctan(FyFx)F_{net} = \sqrt{F_x^2 + F_y^2}, \qquad \theta = \arctan\left(\dfrac{F_y}{F_x}\right)

Why This Matters for Your Exams

Multi-part "assemble N charges one at a time" problems look intimidating but are just Coulomb's law applied repeatedly — draw the geometry once, label every pairwise distance (sides vs diagonals), and each sub-question becomes a single plug-in calculation.

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