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No. 0426 November 2025.
EM fields and waves, EE2147: my electrostatics notes

From one force to a field: Coulomb’s law, added up

My electrostatics notes run from Coulomb’s law to one boxed line. As a vector, the law points along the line between two charges and falls off with the square of the distance between them. With many charges the forces on a test charge Q add, and Q comes out of the sum as a common factor. What is left depends only on where Q sits: that is the electric field.

F12 = Q1Q2 (r2 − r1) / 4πε0 |r2 − r1|³

F = (Q / 4πε0) Σk Qk (r − rk) / |r − rk|³⇒E(r) = F / Q

With 1/4πε0 ≈ 8.99 × 10⁹ N m² C⁻², from ε0 ≈ 8.854 × 10⁻¹² F m⁻¹, the field comes out in newtons per coulomb, the same unit as volts per metre. One line in my notes was marked TODO: F12 = −F21. Swapping the labels turns r2 − r1 into r1 − r2 and leaves the distance alone, so the force reverses and keeps its size. Newton’s third law falls out of the formula.

test point (0.30, 0.90) m|E| = 8.0 N/C|E₁| = 3.6, |E₂| = 6.9 N/C++1 nC−−1 nCE₁E₂E1 m
test point (0.30, 0.90) m|E| = 8.0 N/C++1 nC−−1 nCE₁E₂E1 m
Two charges, +1 nC and −1 nC, 2 m apart. At the test point the field of each charge, the thin arrows, adds up to the total, the thick one. All three share one scale, so the parallelogram is the sum. The field lines are traced from the same sum. Drag the test point. Computed on this page.

what I don’t understand yet

Gauss’s law. My notes have its heading and nothing under it yet. The course asks for Laplace’s and Poisson’s equations to be derived from it, and I want to reach them from this sum rather than quote them.

where this goes

The field is minus the gradient of a potential, E = −∇V, which is why entry 05 is about gradients.