Physics Electromagnetism - How to Solve: Coulomb’s Law & Superposition of Forces (IIT JEE Guide)

How to Solve: Coulomb’s Law & Superposition of Forces (IIT JEE Guide)

Introduction

Mastering Coulomb’s Law and superposition of forces unlocks 8–12 marks in IIT JEE (Main + Advanced) every year—enough to push you into the top 1%. Whether it’s calculating forces between charges, finding equilibrium positions, or solving complex multi-charge systems, this topic appears in both numerical and conceptual questions. Miss it, and you lose easy marks. Nail it, and you gain a reliable, high-scoring edge.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand:
1. Vector addition (resolving forces into components, using unit vectors).
2. Basic electrostatics (charge, conductors, insulators, quantization of charge).
3. Newton’s Third Law (forces between two charges are equal and opposite).

If any of these are shaky, pause and review them first.

KEY TERMS & FORMULAS

Key Terms

1. Point Charge: A charge concentrated at a single point (no size).
2. Coulomb’s Law: The force between two point charges.
3. Superposition Principle: The net force on a charge is the vector sum of forces from all other charges.
4. Permittivity of Free Space (ε₀): A constant (ε₀ = 8.85 × 10⁻¹² C²/N·m²). Given on exam sheet.

Formulas

1. Coulomb’s Law (Force between two point charges)

Formula: [ \vec{F}{12} = k \frac{q_1 q_2}{r ] - }^2} \hat{r}_{12F₁₂: Force on charge q₁ due to q₂ (vector). - k: Coulomb’s constant = ( \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{N·m²/C²} ). MEMORISE THIS. - q₁, q₂: Magnitudes of the charges (in Coulombs, C). - r₁₂: Distance between q₁ and q₂ (in meters, m). - r̂₁₂: Unit vector from q₁ to q₂ (direction of force).

MEMORISE THIS: Like charges repel, opposite charges attract.

2. Superposition Principle

Formula: [ \vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \dots ] - Fₙₑₜ: Net force on a charge due to all other charges. - F₁, F₂, …: Individual forces from each charge (calculated using Coulomb’s Law).

MEMORISE THIS: Forces add vectorially, not just numerically.

STEP-BY-STEP METHOD

Step 1: Draw the System

• Sketch all charges and their positions.
• Label distances and angles clearly.
• Assign a coordinate system (e.g., x-y axes).

Step 2: Identify the Target Charge

• Pick the charge on which you need to find the net force.
• Call this the "test charge".

Step 3: Calculate Individual Forces

• For each other charge, apply Coulomb’s Law:
• Find the distance (r) between the test charge and the other charge.
• Determine the direction of the force (repulsion/attraction).
• Write the force as a vector (magnitude + direction).

Step 4: Resolve Forces into Components

• Break each force into x and y components (or other axes if needed).
• Use trigonometry:
• ( F_x = F \cos \theta )
• ( F_y = F \sin \theta )

Step 5: Sum the Components

• Add all x-components to get ( F_{\text{net},x} ).
• Add all y-components to get ( F_{\text{net},y} ).

Step 6: Find the Net Force

• Combine components using the Pythagorean theorem: [ F_{\text{net}} = \sqrt{F_{\text{net},x}^2 + F_{\text{net},y}^2} ]
• Find the direction using: [ \theta = \tan^{-1} \left( \frac{F_{\text{net},y}}{F_{\text{net},x}} \right) ]

Step 7: Check Units and Signs

• Ensure all distances are in meters, charges in Coulombs.
• Verify signs: Repulsive forces are positive, attractive are negative (if using a coordinate system).

WORKED EXAMPLES

Example 1 – Basic (Two Charges)

Problem: Two charges, ( q_1 = +2 \, \mu C ) and ( q_2 = -3 \, \mu C ), are placed 0.5 m apart. Find the force on ( q_1 ) due to ( q_2 ).

Solution:
1. Draw the system: - ( q_1 ) (+) ← 0.5 m → ( q_2 ) (-)
2. Target charge: ( q_1 ).
3. Coulomb’s Law: [ F_{12} = k \frac{q_1 q_2}{r^2} = (9 \times 10^9) \frac{(2 \times 10^{-6})(-3 \times 10^{-6})}{(0.5)^2} ] [ F_{12} = 9 \times 10^9 \times \frac{-6 \times 10^{-12}}{0.25} = -2.16 \times 10^{-1} \, \text{N} ] - Negative sign → attractive force (toward ( q_2 )).
4. Direction: Along the line joining ( q_1 ) to ( q_2 ).
5. Final answer: ( \vec{F}_{12} = 0.216 \, \text{N} ) toward ( q_2 ).

What we did and why: - Applied Coulomb’s Law directly. - Negative sign confirmed attraction (opposite charges). - Converted μC to C (1 μC = 10⁻⁶ C).

Example 2 – Medium (Three Charges)

Problem: Three charges are placed as shown: - ( q_1 = +1 \, \mu C ) at (0, 0) - ( q_2 = +2 \, \mu C ) at (3, 0) - ( q_3 = -1 \, \mu C ) at (0, 4) Find the net force on ( q_1 ).

Solution:
1. Draw the system: - ( q_1 ) at origin. - ( q_2 ) 3 m to the right. - ( q_3 ) 4 m above.
2. Target charge: ( q_1 ).
3. Force due to ( q_2 ) (F₁₂): - Distance ( r_{12} = 3 \, \text{m} ). - ( F_{12} = k \frac{q_1 q_2}{r_{12}^2} = 9 \times 10^9 \times \frac{(1 \times 10^{-6})(2 \times 10^{-6})}{9} = 2 \times 10^{-3} \, \text{N} ). - Direction: Repulsive (away from ( q_2 )) → left along x-axis. - Vector: ( \vec{F}{12} = -2 \times 10^{-3} \, \hat{i} \, \text{N} ).
4. Force due to ( q_3 ) (F₁₃): - Distance ( r ). - ( F_{13} = k \frac{q_1 q_3}{r_{13}^2} = 9 \times 10^9 \times \frac{(1 \times 10^{-6})(-1 \times 10^{-6})}{16} = -5.625 \times 10^{-4} \, \text{N} ). - Negative sign → } = 4 \, \text{mattractive (toward ( q_3 )) → up along y-axis. - Vector: ( \vec{F}{13} = +5.625 \times 10^{-4} \, \hat{j} \, \text{N} ).
5. Net force: [ \vec{F}}} = \vec{F{12} + \vec{F} ] - Magnitude: [ F_{\text{net}} = \sqrt{(-2 \times 10^{-3})^2 + (5.625 \times 10^{-4})^2} = 2.08 \times 10^{-3} \, \text{N} ] - Direction: [ \theta = \tan^{-1} \left( \frac{5.625 \times 10^{-4}}{-2 \times 10^{-3}} \right) = 164.05^\circ \text{ (from +x-axis)} ]} = -2 \times 10^{-3} \, \hat{i} + 5.625 \times 10^{-4} \, \hat{j

What we did and why: - Used superposition to add forces from two charges. - Resolved forces into x and y components. - Calculated magnitude and direction of the net force.

Example 3 – Exam-Style (Equilibrium Problem)

Problem: Two charges ( q_1 = +4 \, \mu C ) and ( q_2 = +9 \, \mu C ) are fixed 0.5 m apart. Where should a third charge ( q_3 = +1 \, \mu C ) be placed so that it experiences zero net force?

Solution:
1. Draw the system: - ( q_1 ) and ( q_2 ) on a line, 0.5 m apart. - ( q_3 ) must be placed between them (since both are positive, forces can cancel).
2. Let ( q_3 ) be at distance ( x ) from ( q_1 ): - Then, distance from ( q_2 ) = ( 0.5 - x ).
3. Forces on ( q_3 ): - ( F_{31} = k \frac{q_3 q_1}{x^2} ) (repulsive, toward ( q_2 )). - ( F_{32} = k \frac{q_3 q_2}{(0.5 - x)^2} ) (repulsive, toward ( q_1 )).
4. Set net force to zero: [ F_{31} = F_{32} \implies \frac{q_1}{x^2} = \frac{q_2}{(0.5 - x)^2} ] [ \frac{4}{x^2} = \frac{9}{(0.5 - x)^2} ]
5. Solve for ( x ): - Take square roots: [ \frac{2}{x} = \frac{3}{0.5 - x} ] - Cross-multiply: [ 2(0.5 - x) = 3x \implies 1 - 2x = 3x \implies 1 = 5x \implies x = 0.2 \, \text{m} ]
6. Final answer: Place ( q_3 ) 0.2 m from ( q_1 ) (and 0.3 m from ( q_2 )).

What we did and why: - Recognized that equilibrium is possible only between the charges (for like charges). - Set forces equal and solved for distance. - Verified the solution by plugging back in.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP

Listen up—this is your last-minute lifeline.

Coulomb’s Law gives the force between two charges: ( F = k \frac{q_1 q_2}{r^2} ). Memorise k = 9 × 10⁹. Like charges repel, opposites attract.

For multiple charges, superposition is key: Find the force from each charge on your test charge, resolve into x and y components, then add them up. Never add magnitudes directly—forces are vectors.

For equilibrium problems, set forces equal and solve for distance. If charges are alike, the third charge must be between them. If they’re opposite, it can’t be on the line—think off-axis.

Common pitfalls? Wrong units, ignoring vectors, and misplacing equilibrium points. Double-check signs and directions.

You’ve got this. Now go solve those problems like a pro.