Physics - Electrodynamics and Optics - How to Solve: Electric Charges and Fields (Coulomb’s Law, Field Lines, Dipole) – NEET UG Physics Guide

How to Solve: Electric Charges and Fields (Coulomb’s Law, Field Lines, Dipole) – NEET UG Physics Guide

Introduction

Mastering Coulomb’s Law, electric field lines, and dipoles unlocks 5-7 direct questions in NEET Physics—worth 20+ marks—and helps you solve electrostatics problems in circuits, capacitors, and even biology (nerve impulses!). If you nail this, you’re one step closer to a top 100 rank.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand:
1. Basic electrostatics – Like charges repel, unlike charges attract.
2. Vector addition – Forces and fields are vectors; direction matters.
3. SI units – Charge (Coulomb, C), Force (Newton, N), Distance (meter, m).

KEY TERMS & FORMULAS

1. Coulomb’s Law

Formula: [ F = k \frac{|q_1 q_2|}{r^2} ] - ( F ) = Electrostatic force (N) - ( q_1, q_2 ) = Charges (C) - ( r ) = Distance between charges (m) - ( k ) = Coulomb’s constant = ( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 ) (MEMORISE THIS)

Direction: - Repulsive if ( q_1 ) and ( q_2 ) have the same sign. - Attractive if ( q_1 ) and ( q_2 ) have opposite signs.

2. Electric Field (E)

Formula: [ E = \frac{F}{q} ] - ( E ) = Electric field (N/C or V/m) - ( F ) = Force on test charge (N) - ( q ) = Test charge (C)

For a point charge: [ E = k \frac{|q|}{r^2} ] (MEMORISE THIS)

Direction: - Away from positive charges. - Toward negative charges.

3. Electric Field Lines

• Start on positive charges, end on negative charges.
• Never intersect.
• Density = Strength of field (closer lines = stronger field).

4. Electric Dipole

Definition: Two equal and opposite charges ((+q, -q)) separated by distance (2a).

Dipole Moment (( \vec{p} )): [ \vec{p} = q \times 2\vec{a} ] (MEMORISE THIS) - Direction: From negative to positive charge.

Electric Field on Axial Line (along dipole): [ E = \frac{2k \vec{p}}{r^3} ] (MEMORISE THIS)

Electric Field on Equatorial Line (perpendicular to dipole): [ E = \frac{k \vec{p}}{r^3} ] (MEMORISE THIS)

STEP-BY-STEP METHOD

Step 1: Identify the Problem Type

• Coulomb’s Law? → Force between two charges.
• Electric Field? → Field due to one or more charges.
• Dipole? → Field due to a dipole or torque on a dipole.

Step 2: Draw a Diagram

• Sketch charges, distances, and directions.
• Label all given values.

Step 3: Write Down Known & Unknown Quantities

• List ( q_1, q_2, r, k, E, F ) etc.
• Circle what you need to find.

Step 4: Choose the Right Formula

• Force? → Coulomb’s Law.
• Field due to point charge? → ( E = k \frac{|q|}{r^2} ).
• Field due to dipole? → Axial or equatorial formula.

Step 5: Plug in Values & Solve

• Check units (C, m, N).
• Calculate magnitude first, then assign direction.

Step 6: Verify Direction

• Repulsion/Attraction? → Use signs of charges.
• Field direction? → Away from positive, toward negative.

Step 7: Check for Vector Addition (if multiple charges)

• Break into x and y components.
• Add components separately.
• Find resultant using Pythagoras & trigonometry.

WORKED EXAMPLES

Example 1 – Basic (Coulomb’s Law)

Question: Two charges ( +3 \, \mu C ) and ( -5 \, \mu C ) are 0.2 m apart. Find the force between them.

Step 1: Problem type → Coulomb’s Law (force between two charges). Step 2: Diagram → Two charges, 0.2 m apart. Step 3: Known: - ( q_1 = +3 \times 10^{-6} \, C ) - ( q_2 = -5 \times 10^{-6} \, C ) - ( r = 0.2 \, m ) - ( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 ) Step 4: Formula → ( F = k \frac{|q_1 q_2|}{r^2} ) Step 5: Plug in: [ F = 9 \times 10^9 \times \frac{(3 \times 10^{-6})(5 \times 10^{-6})}{(0.2)^2} ] [ F = 9 \times 10^9 \times \frac{15 \times 10^{-12}}{0.04} ] [ F = 9 \times 10^9 \times 3.75 \times 10^{-10} ] [ F = 3.375 \, N ] Step 6: Direction → Attractive (opposite charges). Answer: ( 3.375 \, N ) (attractive).

What we did and why: - Used Coulomb’s Law because we needed force between two charges. - Converted ( \mu C ) to ( C ) (NEET expects SI units). - Calculated magnitude first, then assigned direction based on charge signs.

Example 2 – Medium (Electric Field Due to Multiple Charges)

Question: Two charges ( +4 \, \mu C ) and ( -6 \, \mu C ) are 0.3 m apart. Find the electric field at a point 0.4 m from ( +4 \, \mu C ) and 0.5 m from ( -6 \, \mu C ).

Step 1: Problem type → Electric field due to two charges. Step 2: Diagram → Two charges, point P between them. Step 3: Known: - ( q_1 = +4 \times 10^{-6} \, C ), ( r_1 = 0.4 \, m ) - ( q_2 = -6 \times 10^{-6} \, C ), ( r_2 = 0.5 \, m ) Step 4: Formula → ( E = k \frac{|q|}{r^2} ) for each charge. Step 5: Calculate ( E_1 ) and ( E_2 ): [ E_1 = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{(0.4)^2} = 2.25 \times 10^5 \, N/C ] (away from ( +4 \, \mu C )) [ E_2 = 9 \times 10^9 \times \frac{6 \times 10^{-6}}{(0.5)^2} = 2.16 \times 10^5 \, N/C ] (toward ( -6 \, \mu C )) Step 6: Vector addition → Both fields are in the same direction (from ( +4 ) to ( -6 )). [ E_{total} = E_1 + E_2 = 2.25 \times 10^5 + 2.16 \times 10^5 = 4.41 \times 10^5 \, N/C ] Answer: ( 4.41 \times 10^5 \, N/C ) (toward ( -6 \, \mu C )).

What we did and why: - Calculated field due to each charge separately. - Assigned directions (away from positive, toward negative). - Added vectors (since they were in the same line).

Example 3 – Exam-Style (Dipole in External Field)

Question: An electric dipole of moment ( 2 \times 10^{-8} \, C \, m ) is placed in a uniform electric field of ( 5 \times 10^4 \, N/C ). If the dipole is perpendicular to the field, find the torque acting on it.

Step 1: Problem type → Torque on a dipole in an electric field. Step 2: Diagram → Dipole ( \vec{p} ) perpendicular to ( \vec{E} ). Step 3: Known: - ( p = 2 \times 10^{-8} \, C \, m ) - ( E = 5 \times 10^4 \, N/C ) - ( \theta = 90^\circ ) (perpendicular) Step 4: Formula → ( \tau = pE \sin \theta ) (MEMORISE THIS) Step 5: Plug in: [ \tau = (2 \times 10^{-8}) \times (5 \times 10^4) \times \sin 90^\circ ] [ \tau = 10^{-3} \times 1 = 10^{-3} \, N \, m ] Answer: ( 10^{-3} \, N \, m ).

What we did and why: - Recognized dipole in external field → torque formula. - Used ( \sin 90^\circ = 1 ) (maximum torque when perpendicular). - No vector addition needed (only magnitude asked).

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 20-mark shortcut in NEET Physics. First, Coulomb’s Law: Force between two charges is ( F = k \frac{q_1 q_2}{r^2} ). Same signs repel, opposite signs attract. Second, electric field: ( E = k \frac{q}{r^2} )—direction is away from positive, toward negative. Third, dipole: Dipole moment ( p = q \times 2a ), field on axial line is ( \frac{2kp}{r^3} ), on equatorial line is ( \frac{kp}{r^3} ). Torque on dipole? ( \tau = pE \sin \theta ). Biggest mistake? Forgetting direction—always draw arrows! Exam trap? Three charges—break into x and y components. Now go crush it!