Chapter 21: Coulomb’s Law

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Coulomb’s law governs the electric forces between charged particles, a fundamental principle in electrostatics with applications ranging from atomic interactions to charged particle propulsion in rockets. Building on the thermodynamic concepts from Chapters 18–20, this chapter introduces the principles of electrostatics. For JEE Main, JEE Advanced, and NEET students, mastering Coulomb’s law is essential, as it forms the basis for understanding electric fields, potentials, and interactions in physics problems. This chapter, Coulomb’s Law, covers electric charge and Coulomb’s law, vector nature of electric forces, superposition principle, and applications of Coulomb’s law, providing detailed explanations, derivations, solved examples, and practical applications to ensure conceptual clarity and problem-solving proficiency.

21.1 Electric Charge and Coulomb’s Law

Coulomb’s law describes the electric force between charged particles, laying the foundation for electrostatics, a critical topic for JEE/NEET problems.

Electric Charge

• Charge Types: Positive (protons) and negative (electrons). Like charges repel, unlike charges attract.
• Quantization: Charge is quantized, $q=ne$, where $e=1.6\times {10}^{-19}\text{C}$ (elementary charge), $n$ is an integer.
• Conservation: Charge is conserved in isolated systems (e.g., in reactions, total charge remains constant).
• SI Unit: Coulomb (C).

Coulomb’s Law

The force between two point charges $q_1$ and $q_2$ separated by distance $r$ in a vacuum is:

$$F=k\frac{|q_1{q}_2|}{r^2}$$

• $k$: Coulomb’s constant, $k=\frac{1}{4\pi {ϵ}_0}$, where ${ϵ}_0=8.85\times {10}^{-12}{\text{C}}^2/{\text{N·m}}^2$ (permittivity of free space). Numerically, $k\approx 9\times {10}^9{\text{N·m}}^2/{\text{C}}^2$.
• $F$: Magnitude of the force (N).
• Direction: Attractive if charges are unlike, repulsive if like, along the line joining the charges.

In a Medium

In a medium with permittivity $ϵ$, the force is reduced: $F=\frac{1}{4\pi ϵ}\frac{|q_1{q}_2|}{r^2}$, where $ϵ={ϵ}_r{ϵ}_0$, and ${ϵ}_r$ is the relative permittivity (dielectric constant). Thus, $F_{\text{medium}}=\frac{F_{\text{vacuum}}}{{ϵ}_r}$.

Properties

• Inverse Square Law: $F\propto \frac{1}{r^2}$.
• Sign of Force: Positive $F$ indicates repulsion, negative indicates attraction (in vector form).
• Point Charges: Assumes charges are point-like (size negligible compared to $r$).

Derivation: Coulomb’s Constant $k$
Coulomb’s law is an experimental result, but $k$ relates to ${ϵ}_0$, the permittivity of free space, defined in SI units. The force $F=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}$ ensures consistency with Gauss’s law and electromagnetic theory. Numerically, $k=\frac{1}{4\pi \times 8.85\times {10}^{-12}}\approx 9\times {10}^9{\text{N·m}}^2/{\text{C}}^2$.

Derivation: Force in a Medium
In a medium, electric fields are reduced due to polarization effects. The permittivity becomes $ϵ={ϵ}_r{ϵ}_0$, so $F=\frac{1}{4\pi ϵ}\frac{q_1{q}_2}{r^2}=\frac{1}{4\pi {ϵ}_r{ϵ}_0}\frac{q_1{q}_2}{r^2}=\frac{F_{\text{vacuum}}}{{ϵ}_r}$.

Derivation: Charge Quantization
Charge quantization arises from the discrete nature of electrons and protons: $q=ne$, where $e=1.6\times {10}^{-19}\text{C}$. This is a fundamental property confirmed by experiments like Millikan’s oil drop experiment.

Derivation: Electrostatic Force in Rocket Systems
In a charged particle propulsion system, two charges $q_1=2\times {10}^{-6}\text{C}$, $q_2=-3\times {10}^{-6}\text{C}$ are separated by $r=0.1\text{m}$. Force: $F=9\times {10}^9\times \frac{(2\times {10}^{-6})\times (3\times {10}^{-6})}{(0.1{)}^2}=5.4\text{N}$ (attractive), aiding propulsion (your interest, April 19, 2025).

Solved Example: A JEE Main problem involves two charges $q_1=+4\mu \text{C}$ and $q_2=-2\mu \text{C}$ separated by $r=0.2\text{m}$ in vacuum. Find the force magnitude.

• Solution:
  $q_1=4\times {10}^{-6}\text{C}$, $q_2=-2\times {10}^{-6}\text{C}$, $r=0.2\text{m}$, $k=9\times {10}^9{\text{N·m}}^2/{\text{C}}^2$:$$F=k\frac{|q_1{q}_2|}{r^2}=9\times {10}^9\times \frac{(4\times {10}^{-6})\times (2\times {10}^{-6})}{(0.2{)}^2}=9\times {10}^9\times \frac{8\times {10}^{-12}}{0.04}=1.8\text{N}$$
  • JEE Tip: Use magnitudes for $F$; the negative sign indicates attraction. Common error: Forgetting to convert $\mu \text{C}$ to C (${10}^{-6}$).

Solved Example: A NEET problem involves two charges $q_1=+5\text{nC}$ and $q_2=+5\text{nC}$ separated by $r=0.1\text{m}$ in vacuum. Find the force.

• Solution:
  $q_1=q_2=5\times {10}^{-9}\text{C}$, $r=0.1\text{m}$:$$F=9\times {10}^9\times \frac{(5\times {10}^{-9}{)}^2}{(0.1{)}^2}=9\times {10}^9\times \frac{25\times {10}^{-18}}{0.01}=2.25\times {10}^{-5}\text{N}$$Repulsive (like charges).
  • NEET Tip: Like charges repel; convert nC to C (${10}^{-9}$). Common error: Misinterpreting the direction of the force.

Solved Example: A JEE Advanced problem involves two charges in water (${ϵ}_r=80$), $q_1=+3\mu \text{C}$, $q_2=-6\mu \text{C}$, $r=0.5\text{m}$. Find the force.

• Solution:
  In vacuum: $F_{\text{vacuum}}=9\times {10}^9\times \frac{(3\times {10}^{-6})\times (6\times {10}^{-6})}{(0.5{)}^2}=0.648\text{N}$. In water: $F=\frac{F_{\text{vacuum}}}{{ϵ}_r}=\frac{0.648}{80}\approx 0.0081\text{N}$. Attractive.
  • JEE Tip: Medium reduces force by ${ϵ}_r$; direction depends on charge signs. Common error: Forgetting to divide by ${ϵ}_r$.

Solved Example: A JEE Main problem involves a charge of $+8\mu \text{C}$. How many electrons were removed? ($e=1.6\times {10}^{-19}\text{C}$)

• Solution:
  $q=ne$, $q=8\times {10}^{-6}\text{C}$: $n=\frac{q}{e}=\frac{8\times {10}^{-6}}{1.6\times {10}^{-19}}=5\times {10}^{13}$ electrons removed.
  • JEE Tip: Positive charge means electrons removed; use $e$ to find $n$. Common error: Assuming electrons were added.

Application: Coulomb’s law applies to atomic physics (electron-proton forces), capacitor design, and rocketry (e.g., charged particle propulsion, aligning with your interest, April 19, 2025).

21.2 Vector Nature of Electric Forces

Coulomb’s law is a vector law, requiring careful consideration of directions and components, a key skill for JEE/NEET electrostatics problems.

Vector Form of Coulomb’s Law

For charges $q_1$ and $q_2$ at positions ${\overrightarrow{r}}_1$ and ${\overrightarrow{r}}_2$, the force on $q_1$ due to $q_2$ is:

$${\overrightarrow{F}}_{1,2}=k\frac{q_1{q}_2}{r^2}{\hat{r}}_{12}$$

• ${\overrightarrow{r}}_{12}={\overrightarrow{r}}_1-{\overrightarrow{r}}_2$, $r=|{\overrightarrow{r}}_{12}|$, ${\hat{r}}_{12}=\frac{{\overrightarrow{r}}_{12}}{r}$ (unit vector from $q_2$ to $q_1$).
• ${\overrightarrow{F}}_{2,1}=-{\overrightarrow{F}}_{1,2}$ (Newton’s third law).

Force Components

In 2D or 3D, break the force into components:

• $x$-component: $F_x=F\cos \theta$.
• $y$-component: $F_y=F\sin \theta$.

Multiple Charges

For multiple charges, compute the vector force on each charge due to all others and resolve into components.

Derivation: Vector Form of Coulomb’s Law
Consider $q_1$ at ${\overrightarrow{r}}_1$ and $q_2$ at ${\overrightarrow{r}}_2$. The displacement vector ${\overrightarrow{r}}_{12}={\overrightarrow{r}}_1-{\overrightarrow{r}}_2$, with magnitude $r=|{\overrightarrow{r}}_{12}|$. The unit vector ${\hat{r}}_{12}=\frac{{\overrightarrow{r}}_1-{\overrightarrow{r}}_2}{|{\overrightarrow{r}}_1-{\overrightarrow{r}}_2|}$ points from $q_2$ to $q_1$. The force magnitude is $F=k\frac{|q_1{q}_2|}{r^2}$. The vector force on $q_1$ is ${\overrightarrow{F}}_{1,2}=F{\hat{r}}_{12}$ if $q_1{q}_2>0$ (repulsive), or $-{\hat{r}}_{12}$ if $q_1{q}_2<0$ (attractive). General form: ${\overrightarrow{F}}_{1,2}=k\frac{q_1{q}_2}{r^2}{\hat{r}}_{12}$, where the sign of $q_1{q}_2$ determines the direction.

Derivation: Force Components in 2D
For charges $q_1$ at $(x_1,y_1)$ and $q_2$ at $(x_2,y_2)$, ${\overrightarrow{r}}_{12}=(x_1-x_2,y_1-y_2)$, $r=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$. Unit vector: ${\hat{r}}_{12}=(\frac{x_1-x_2}{r},\frac{y_1-y_2}{r})$. Force: $\overrightarrow{F}=k\frac{q_1{q}_2}{r^2}{\hat{r}}_{12}$, so $F_x=k\frac{q_1{q}_2}{r^2}\cdot \frac{x_1-x_2}{r}$, $F_y=k\frac{q_1{q}_2}{r^2}\cdot \frac{y_1-y_2}{r}$.

Derivation: Net Force on a Charge in a System
For a charge $q_1$ with charges $q_2$ and $q_3$, compute ${\overrightarrow{F}}_{1,2}$ and ${\overrightarrow{F}}_{1,3}$ separately using vector form, then add: ${\overrightarrow{F}}_{\text{net}}={\overrightarrow{F}}_{1,2}+{\overrightarrow{F}}_{1,3}$. Resolve into components if needed.

Derivation: Electrostatic Steering in Rockets
In a rocket’s ion propulsion, a charge $q_1=1\times {10}^{-6}\text{C}$ at $(0,0)$ experiences forces from $q_2=2\times {10}^{-6}\text{C}$ at $(0.1,0)$ and $q_3=-3\times {10}^{-6}\text{C}$ at $(0,0.1)$. Net force components: $F_x=1.8\text{N}$, $F_y=-2.7\text{N}$, steering the particle (your interest, April 19, 2025).

Solved Example: A JEE Main problem involves charges $q_1=+3\mu \text{C}$ at $(0,0)$ and $q_2=-3\mu \text{C}$ at $(0.3,0)$. Find the force on $q_1$.

• Solution:
  $r=0.3\text{m}$, $F=9\times {10}^9\times \frac{(3\times {10}^{-6}{)}^2}{(0.3{)}^2}=0.9\text{N}$. ${\hat{r}}_{12}$ points from $q_2$ to $q_1$ (along negative x-axis): ${\hat{r}}_{12}=(-1,0)$. ${\overrightarrow{F}}_{1,2}=-0.9\hat{i}\text{N}$ (attractive).
  • JEE Tip: Use unit vectors to determine direction; negative sign for attraction. Common error: Ignoring direction, giving only magnitude.

Solved Example: A NEET problem involves $q_1=+2\mu \text{C}$ at $(0,0)$ and $q_2=+2\mu \text{C}$ at $(0,0.2)$. Find the force vector on $q_1$.

• Solution:
  $F=9\times {10}^9\times \frac{(2\times {10}^{-6}{)}^2}{(0.2{)}^2}=0.9\text{N}$. ${\hat{r}}_{12}=(0,1)$ (along positive y-axis). ${\overrightarrow{F}}_{1,2}=0.9\hat{j}\text{N}$ (repulsive).
  • NEET Tip: Like charges repel; specify direction using coordinates. Common error: Forgetting repulsion direction.

Solved Example: A JEE Advanced problem involves $q_1=+4\mu \text{C}$ at $(0,0)$, $q_2=-2\mu \text{C}$ at $(0.1,0)$, $q_3=+3\mu \text{C}$ at $(0,0.1)$. Find the net force on $q_1$.

• Solution:
  ${\overrightarrow{F}}_{1,2}=7.2\hat{i}\text{N}$ (attractive), ${\overrightarrow{F}}_{1,3}=-10.8\hat{j}\text{N}$ (repulsive). Net: ${\overrightarrow{F}}_{\text{net}}=7.2\hat{i}-10.8\hat{j}\text{N}$.
  • JEE Tip: Add forces as vectors; compute components separately. Common error: Adding magnitudes without direction.

Solved Example: A JEE Main problem involves $q_1=+5\mu \text{C}$ at $(0.2,0.2)$ and $q_2=-5\mu \text{C}$ at $(0,0)$. Find $\overrightarrow{F}$ on $q_1$.

• Solution:
  $r=\sqrt{(0.2{)}^2+(0.2{)}^2}=0.2\sqrt{2}$, $F=2.8125\text{N}$. ${\hat{r}}_{12}=(\frac{-0.2}{0.2\sqrt{2}},\frac{-0.2}{0.2\sqrt{2}})=(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$. $\overrightarrow{F}=-2.8125(\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j})\approx -1.99\hat{i}-1.99\hat{j}\text{N}$.
  • JEE Tip: Resolve unit vectors using coordinates; round components appropriately (April 14, 2025). Common error: Incorrect unit vector direction.

Application: Vector forces apply to charged particle dynamics, crystal structures, and rocketry (e.g., ion propulsion steering, aligning with your interest, April 19, 2025).

21.3 Superposition Principle

The superposition principle allows the calculation of net forces from multiple charges, a key concept for JEE/NEET electrostatics problems involving systems of charges.

Superposition Principle

The net force on a charge due to multiple charges is the vector sum of the forces from each charge:

$${\overrightarrow{F}}_{\text{net}}=\sum _i{\overrightarrow{F}}_i$$

Each ${\overrightarrow{F}}_i$ is calculated using Coulomb’s law, considering magnitude and direction.

Steps to Apply

1. Identify all charges and their positions.
2. Calculate the force on the target charge due to each other charge (vector form).
3. Resolve forces into components (if needed).
4. Sum the components to find the net force.
5. Compute magnitude and direction if required.

Equilibrium of Charges

A charge is in equilibrium if ${\overrightarrow{F}}_{\text{net}}=0$. For a system, solve for positions or charges where forces balance.

Derivation: Superposition Principle
Coulomb’s law applies to pairs of charges. For a charge $q_1$ with charges $q_2,q_3,\dots ,q_n$, the force due to each is independent (electric forces are linear). Thus, ${\overrightarrow{F}}_{1,\text{net}}={\overrightarrow{F}}_{1,2}+{\overrightarrow{F}}_{1,3}+\cdots +{\overrightarrow{F}}_{1,n}$, where each ${\overrightarrow{F}}_{1,i}=k\frac{q_1{q}_i}{r_{1i}^2}{\hat{r}}_{1i}$. This principle holds due to the linearity of Maxwell’s equations in electrostatics.

Derivation: Equilibrium Position
For three charges in a line, $q_1=+q$ at $x=0$, $q_2=+q$ at $x=a$, find where $q_3=-Q$ (at $x$, $0<x<a$) has ${\overrightarrow{F}}_{\text{net}}=0$. Forces: $F_{3,1}=k\frac{qQ}{x^2}$ (right), $F_{3,2}=k\frac{qQ}{(a-x{)}^2}$ (left). Set equal: $\frac{1}{x^2}=\frac{1}{(a-x{)}^2}$, $x=a-x$, $x=\frac{a}{2}$.

Derivation: Net Force in a Triangle
Charges $q_1=+q$, $q_2=+q$, $q_3=-q$ at vertices of an equilateral triangle (side $a$). Net force on $q_1$: ${\overrightarrow{F}}_{1,2}=k\frac{q^2}{a^2}{\hat{r}}_{12}$, ${\overrightarrow{F}}_{1,3}=-k\frac{q^2}{a^2}{\hat{r}}_{13}$. Use geometry to sum vectors, yielding ${\overrightarrow{F}}_{\text{net}}$ along symmetry axis.

Derivation: Ion Propulsion Force Balance
In a rocket ion engine, three charges align for propulsion: $q_1=+q$, $q_2=+q$, $q_3=-q$. Equilibrium ensures controlled motion, optimizing thrust (your interest, April 19, 2025).

Solved Example: A JEE Main problem involves $q_1=+2\mu \text{C}$ at $(0,0)$, $q_2=-3\mu \text{C}$ at $(0.2,0)$, $q_3=+4\mu \text{C}$ at $(0,0.2)$. Find the net force on $q_1$.

• Solution:
  ${\overrightarrow{F}}_{1,2}=-1.35\hat{i}\text{N}$, ${\overrightarrow{F}}_{1,3}=1.8\hat{j}\text{N}$. ${\overrightarrow{F}}_{\text{net}}=-1.35\hat{i}+1.8\hat{j}\text{N}$.
  • JEE Tip: Apply superposition by summing vector forces; compute components. Common error: Adding magnitudes without direction.

Solved Example: A NEET problem involves $q_1=+q$, $q_2=+q$ at $x=0$ and $x=1\text{m}$. Where does $q_3=-q$ experience zero force?

• Solution:
  At $x$ (between 0 and 1), $F_{3,1}=k\frac{q^2}{x^2}$ (right), $F_{3,2}=k\frac{q^2}{(1-x{)}^2}$ (left). Set equal: $x=0.5\text{m}$.
  • NEET Tip: Equilibrium requires forces to balance; solve for position. Common error: Placing $q_3$ outside the charges.

Solved Example: A JEE Advanced problem involves charges at vertices of a square (side $a$): $q_1=+q$, $q_2=+q$, $q_3=-q$, $q_4=-q$. Find the net force on $q_1$.

• Solution:
  Compute forces: ${\overrightarrow{F}}_{1,2}$, ${\overrightarrow{F}}_{1,3}$, ${\overrightarrow{F}}_{1,4}$. Sum vectors considering angles (45° for diagonals). Net force magnitude: $\sqrt{2}k\frac{q^2}{a^2}$.
  • JEE Tip: Use symmetry and vector addition; diagonals require angle resolution. Common error: Ignoring vector nature.

Solved Example: A JEE Main problem involves $q_1=+2q$ at $(0,0)$, $q_2=-q$ at $(a,0)$, $q_3=-q$ at $(0,a)$. Find $\overrightarrow{F}$ on $q_1$.

• Solution:
  ${\overrightarrow{F}}_{1,2}=k\frac{2q^2}{a^2}\hat{i}$, ${\overrightarrow{F}}_{1,3}=k\frac{2q^2}{a^2}\hat{j}$. ${\overrightarrow{F}}_{\text{net}}=k\frac{2q^2}{a^2}(\hat{i}+\hat{j})$.
  • JEE Tip: Symmetry simplifies components; net force at 45°. Common error: Incorrect signs for attraction/repulsion.

Application: Superposition applies to multi-particle systems, molecular forces, and rocketry (e.g., ion engine charge configurations, aligning with your interest, April 19, 2025).

21.4 Applications of Coulomb’s Law

Coulomb’s law is applied to calculate forces, equilibrium positions, and electric fields, providing practical problem-solving skills for JEE/NEET electrostatics.

Force Calculations

Use Coulomb’s law to find forces in systems of charges, often requiring vector addition (Sections 21.2–21.3).

Equilibrium Problems

Find positions where the net force on a charge is zero, often involving solving for distances or charges.

Electric Field Introduction

The electric field at a point due to a charge $q$ is:

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

For multiple charges, use superposition: ${\overrightarrow{E}}_{\text{net}}=\sum {\overrightarrow{E}}_i$.

Charge Distributions

For continuous distributions (e.g., line, ring), integrate over the charge elements: $d\overrightarrow{F}=k\frac{qdq}{r^2}\hat{r}$.

Derivation: Electric Field from Coulomb’s Law
The electric field $\overrightarrow{E}$ at a point is the force per unit charge: $\overrightarrow{E}=\frac{\overrightarrow{F}}{q_{\text{test}}}$. For a charge $q$, $\overrightarrow{F}=k\frac{q{q}_{\text{test}}}{r^2}\hat{r}$, so $\overrightarrow{E}=k\frac{q}{r^2}\hat{r}$. For multiple charges, sum vectorially.

Derivation: Equilibrium in a System
Three charges in a line: $q_1=+q$, $q_2=+q$, $q_3=-2q$. Find $q_3$’s position for equilibrium (as in Section 21.3). Generalize for non-linear systems using vector sums.

Derivation: Field Due to a Line Charge
A line charge with linear charge density $\lambda$ (C/m), length $L$. Field at point $P$ distance $d$ perpendicular to the midpoint: $dE=k\frac{dq}{r^2}$, $dq=\lambda dx$, integrate over the line, yielding $E=k\frac{\lambda L}{d\sqrt{d^2+(L/2{)}^2}}$.

Derivation: Rocket Ion Engine Field
In an ion engine, a line of charge (length $L$, $\lambda$) creates a field $E$ to accelerate ions, optimizing thrust (your interest, April 19, 2025).

Solved Example: A JEE Main problem involves finding the electric field at $(0,0.3)$ due to $q=+5\mu \text{C}$ at $(0,0)$.

• Solution:
  $r=0.3\text{m}$, $\overrightarrow{E}=9\times {10}^9\times \frac{5\times {10}^{-6}}{(0.3{)}^2}\hat{j}=5\times {10}^5\hat{j}\text{N/C}$.
  • JEE Tip: Field direction is along the line from charge to point; positive charge means field points away. Common error: Incorrect direction.

Solved Example: A NEET problem involves $q_1=+2\mu \text{C}$ at $(0.1,0)$, $q_2=-2\mu \text{C}$ at $(-0.1,0)$. Find $\overrightarrow{E}$ at $(0,0.2)$.

• Solution:
  $E_1=4.5\times {10}^5\hat{j}$, $E_2=-4.5\times {10}^5\hat{j}$. ${\overrightarrow{E}}_{\text{net}}=0$.
  • NEET Tip: Superposition for fields; symmetry can cancel components. Common error: Adding magnitudes without direction.

Solved Example: A JEE Advanced problem involves a ring of charge (radius $R$, total charge $Q$). Find $\overrightarrow{E}$ on the axis at distance $z$.

• Solution:
  $d{E}_z=k\frac{dq}{(R^2+z^2)}\cdot \frac{z}{\sqrt{R^2+z^2}}$, integrate: $E_z=k\frac{Qz}{(R^2+z^2{)}^{3/2}}$.
  • JEE Tip: Symmetry cancels radial components; integrate over the ring. Common error: Forgetting $z$-component projection.

Solved Example: A JEE Main problem involves finding the equilibrium position of $q_3=-q$ between $q_1=+q$ at $(0,0)$ and $q_2=+4q$ at $(a,0)$.

• Solution:
  At $x$, $k\frac{q^2}{x^2}=k\frac{4q^2}{(a-x{)}^2}$, $2x=a-x$, $x=\frac{a}{3}$.
  • JEE Tip: Equilibrium balances forces; solve for $x$. Common error: Incorrect force ratios.

Application: Applications include field calculations, capacitor design, and rocketry (e.g., electric fields in ion propulsion, aligning with your interest, April 19, 2025).

Summary and Quick Revision

• Electric Charge: Quantized: $q=ne$, $e=1.6\times {10}^{-19}\text{C}$. Conserved. Unit: Coulomb (C).
• Coulomb’s Law: $F=k\frac{|q_1{q}_2|}{r^2}$, $k=\frac{1}{4\pi {ϵ}_0}\approx 9\times {10}^9{\text{N·m}}^2/{\text{C}}^2$, ${ϵ}_0=8.85\times {10}^{-12}{\text{C}}^2/{\text{N·m}}^2$. In medium: $F=\frac{F_{\text{vacuum}}}{{ϵ}_r}$.
• Vector Form: ${\overrightarrow{F}}_{1,2}=k\frac{q_1{q}_2}{r^2}{\hat{r}}_{12}$, ${\hat{r}}_{12}=\frac{{\overrightarrow{r}}_1-{\overrightarrow{r}}_2}{|{\overrightarrow{r}}_1-{\overrightarrow{r}}_2|}$. Components: $F_x=F\cos \theta$, $F_y=F\sin \theta$.
• Superposition: ${\overrightarrow{F}}_{\text{net}}=\sum {\overrightarrow{F}}_i$. Equilibrium: ${\overrightarrow{F}}_{\text{net}}=0$.
• Applications: Electric field: $\overrightarrow{E}=k\frac{q}{r^2}\hat{r}$. Continuous distributions: integrate $d\overrightarrow{F}$ or $d\overrightarrow{E}$.
• JEE/NEET Tips: Use vector addition for forces/fields, convert units (e.g., $\mu \text{C}$ to C), apply superposition for multiple charges, verify significant figures (April 14, 2025), distinguish attraction/repulsion based on charge signs.
• SI Units: Charge (C), force (N), electric field (N/C), distance (m).

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