Unit 01: Electrostatics • Module 1.1
Coulomb's Law & Principle of Superposition
Crisp first-principles breakdown, interactive live vector mechanics, exam traps, and exemplar numericals.
Imagine two magnets floating on completely frictionless ice: push them closer, and repulsion blows up exponentially, not linearly. Electrostatic force operates on a rigid, invisible line of sight connecting two point charges. Adding a third charge into the system does not alter the mutual handshake between the first two—it simply adds its own independent force vector on top.
Coulomb's Law (Point Charges)
Force between stationary charges in vacuum along their line of center:
In medium of dielectric constant $K$: $F_{\text{med}} = \frac{F_{\text{vac}}}{K}$.
Superposition Principle
Net force is the vector sum of individual two-body pairwise forces:
Scalar addition of magnitudes without directional angles leads to 100% error.
1. Position vectors: Let point charges $q_1$ and $q_2$ be at position vectors $\vec{r}_1$ and $\vec{r}_2$. The displacement vector pointing from $q_2$ to $q_1$ is $\vec{r}_{12} = \vec{r}_1 - \vec{r}_2$.
2. Formulating Force: The electrostatic force exerted on $q_1$ by $q_2$ along the line of sight is:
3. Expanding unit vector: Since $\hat{r}_{12} = \frac{\vec{r}_{12}}{|\vec{r}_{12}|} = \frac{\vec{r}_1 - \vec{r}_2}{|\vec{r}_1 - \vec{r}_2|}$:
4. Action-Reaction symmetry: Inverting coordinates gives $(\vec{r}_2 - \vec{r}_1) = -(\vec{r}_1 - \vec{r}_2)$, proving Newton's 3rd Law:
Trap 1: The "Scalar Magnitude" Sum Trap
Students frequently compute forces $|F_1|$ and $|F_2|$ and add them directly ($F_1 + F_2$). Forces are vectors! Always resolve into orthogonal components ($\sum F_x \hat{i} + \sum F_y \hat{j}$) before finding the resultant.
Trap 2: Large Conducting Spheres vs Point Charges
Coulomb's Law strictly assumes point charges. For conducting spheres of finite radii, induction relocates charge centers. Like charges repel to opposite sides ($r_{\text{eff}} > r$), reducing force below Coulomb prediction.
Problem: Two fixed point charges $+4q$ and $+q$ are separated by distance $L$. At what position should a third charge $Q$ be placed so that it is in equilibrium?
Solution: Let $Q$ be at distance $x$ from $+4q$. For equilibrium:
$$\frac{1}{4\pi\varepsilon_0} \frac{4q \cdot Q}{x^2} = \frac{1}{4\pi\varepsilon_0} \frac{q \cdot Q}{(L-x)^2} \implies \frac{2}{x} = \frac{1}{L-x}$$
$$2(L - x) = x \implies 2L = 3x \implies x = \frac{2L}{3}$$
Punchline: Distance is $2L/3$ from $+4q$. The sign and magnitude of $Q$ cancel completely, having zero effect on equilibrium position.
Problem: Five equal point charges $+q$ are fixed at 5 vertices of a regular hexagon of side $a$. Find the net force on charge $+q_0$ at the center.
Solution: By geometric inversion symmetry, diametrically opposite pairs cancel pairwise:
$$\vec{F}_1 + \vec{F}_4 = 0, \quad \vec{F}_2 + \vec{F}_5 = 0$$
The single charge across the vacant vertex remains completely unbalanced. Distance from center to vertex in a regular hexagon equals side $a$:
$$\vec{F}_{\text{net}} = \frac{1}{4\pi\varepsilon_0} \frac{q \cdot q_0}{a^2} \quad \text{(directed directly toward the empty vertex)}$$
Scalar Force
$F = \frac{k |q_1 q_2|}{r^2}$
Dielectric Effect
$F_{\text{med}} = \frac{F_{\text{vac}}}{K}$
Electrostatic Constant
$9 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2$
Validity Boundary
Point charges, $r > 10^{-15}\text{ m}$