Review / Chapter 1 — Electrostatics

Coulomb's Law & Superposition

Chapter 1 — Electrostatics

Electric charge comes in two signs, is conserved, and is quantized in units of \(e = 1.6\times 10^{-19}\) C. The whole of electrostatics is built from one empirical statement: the force between two point charges. Everything else — fields, potentials, Gauss's law — follows from Coulomb's law plus the superposition principle.

The formula

For a point charge \(q_1\) at position \(\vec{r}_1\) and \(q_2\) at \(\vec{r}_2\), the force on \(q_2\) due to \(q_1\) is

\[\vec{F}_{2} = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r_{21}^2}\,\hat{r}_{21},\]

where \(\vec{r}_{21} = \vec{r}_2 - \vec{r}_1\) points from the source to the target and \(\hat{r}_{21}\) is its unit vector. With the sign convention built in, like charges repel (force along \(\hat{r}_{21}\)) and opposite charges attract.

The constant is \(k = 1/(4\pi\epsilon_0) \approx 8.99\times 10^9\) N·m\(^2\)/C\(^2\), with \(\epsilon_0 = 8.854\times 10^{-12}\) C\(^2\)/(N·m\(^2\)).

Key properties

Three features jump out and shape everything downstream:

  1. Linear in each charge. Doubling either charge doubles the force. This is what makes superposition work.
  2. Inverse square. \(|\vec F|\propto 1/r^2\). We'll later see via Gauss's law that this is forced by the fact that we live in 3D.
  3. Central. The force points along the line between the charges. Symmetry demands this for point charges in empty space.

Coulomb's law also obeys Newton's 3rd law: swap the labels and you get \(\vec F_1 = -\vec F_2\).

Superposition

The force on a target charge due to a collection of sources is the vector sum of the pairwise Coulomb forces:

\[\vec F_{\text{on }q} = \sum_i \frac{1}{4\pi\epsilon_0}\frac{q\, q_i}{|\vec r - \vec r_i|^2}\,\hat{r}_i,\quad \hat r_i = \frac{\vec r - \vec r_i}{|\vec r - \vec r_i|}.\]

The presence of a third charge doesn't alter the force between the first two. Superposition relies on Coulomb's law being linear in the charges — if it went like \(q_1^2 q_2^2\) the principle would fail instantly.

Two point charges with the Coulomb force vector between them.
Force on \(q_2\) from \(q_1\): along \(\hat r_{21}\), falling off as \(1/r^2\).

Practice Problems

Problem 1easy
Two protons are separated by \(1.0\times 10^{-10}\) m (roughly an atomic scale). Find the magnitude of the Coulomb force between them.
Hint
\(F = kq^2/r^2\) with \(q = e\).
Solution

\(F = (9\times 10^9)(1.6\times 10^{-19})^2/(10^{-10})^2 = 9\times 10^9 \cdot 2.56\times 10^{-38}/10^{-20} \approx 2.3\times 10^{-8}\) N.

Answer: \(\approx 2.3\times 10^{-8}\) N, repulsive.

Problem 2easy
Charge \(+q\) sits at \((0,0)\) and another \(+q\) at \((a,0)\). A third charge \(+q\) is placed at \((a/2, 0)\). What is the net force on the middle charge?
Hint
The two outer charges pull the middle one in opposite directions with equal magnitudes.
Solution

By symmetry the forces cancel: each outer charge pushes the middle charge away with the same magnitude \(kq^2/(a/2)^2\), but in opposite directions.

Answer: \(\vec F = 0\) (unstable equilibrium).

Problem 3medium
Charges \(+q, +q, +q\) sit at the vertices of an equilateral triangle of side \(a\). Find the magnitude of the force on one of the vertices.
Hint
Two equal forces at \(60°\) to each other. Add components.
Solution

Each source exerts \(F_0 = kq^2/a^2\) on the target. The angle between them (as seen from the target) is \(60°\), so they each contribute a component \(F_0 \cos 30° = F_0 \sqrt{3}/2\) along the symmetry axis, and the perpendicular components cancel. Net \(F = 2 F_0 \cos 30° = \sqrt{3}\, kq^2/a^2\).

Answer: \(\sqrt{3}\,kq^2/a^2\), directed away from the centroid.

Problem 4medium
A charge \(+Q\) is fixed at the origin and a charge \(-q\) (\(q>0\)) is placed at \((d,0)\). Where on the \(x\)-axis (with \(x>d\)) could a third charge \(+q'\) be placed so that the net force on it from the first two is zero?
Hint
Outside the negative charge, the two forces oppose only if you pick the right region. Set the magnitudes equal.
Solution

At \(x > d\), the \(+Q\) pushes \(+q'\) further right (repel), the \(-q\) pulls it left (attract). Set \(kQq'/x^2 = kqq'/(x-d)^2\), so \((x-d)/x = \sqrt{q/Q}\), giving \(x = d/(1 - \sqrt{q/Q})\), provided \(Q > q\).

Answer: \(x = d/(1-\sqrt{q/Q})\), requires \(Q>q\).