1.1 Introduction — Why Should You Care?

Ever rubbed a balloon on your hair and watched it stick to the wall? Or got a tiny shock touching a metal door in winter? That’s not magic — that’s electrostatics.

This whole chapter is basically answering one question: What happens when charges exist near each other? Simple question. Powerful answers.

1.2 Electric Charge — The Basics

So what even IS charge? Think of it as a property — like how some people are naturally “magnetic” in personality. Matter has this property called electric charge, and it comes in exactly two flavours:

• Positive — carried by protons
• Negative — carried by electrons

And you already know the rule from childhood: like poles repel, unlike attract. Same thing here.

• SI unit: Coulomb (C)
• Charge of 1 electron: −1.6 × 10⁻¹⁹ C (tiny, right?)

Real talk: You’ll never use “1 Coulomb” in practice — it’s massive. We mostly work in microcoulombs (μC) or nanocoulombs (nC).

1.3 Conductors vs Insulators

Here’s a simple way to think about it:

Conductors = highways for electrons. Charges move freely. (Copper, Silver, Iron)

Insulators = blocked roads. Electrons are stuck. (Rubber, Glass, Plastic)

That’s why electricians wear rubber gloves — rubber doesn’t let current pass through. Simple, life-saving physics!

There’s also a middle ground — semiconductors (like Silicon) — but that’s a story for Chapter 14. 😄

1.4 Three Properties of Charge — Don’t Skip These!

These three come up in exams more than you’d expect.

1. Additivity Just add them up algebraically.

+3C and −1C together? Total = +2C. Easy.

2. Conservation Charge is never created or destroyed — only transferred.

When you rub a comb on hair, electrons move from hair to comb. Hair becomes +ve, comb becomes −ve. No charge was born or killed — just relocated!

3. Quantisation This one’s interesting. Charge doesn’t come in any random amount. It always comes in multiples of e = 1.6 × 10⁻¹⁹ C.

$$q = ne$$

So you can have 1e, 2e, 3e worth of charge — but never 1.5e. Nature doesn’t do fractions here.

1.5 Coulomb’s Law — The Star of the Chapter ⭐

Okay, this is THE formula. Learn it, love it.

$$F = k\frac{q_1 q_2}{r^2}$$

• F = force between charges (Newtons)
• q₁, q₂ = the two charges
• r = distance between them
• k = 9 × 10⁹ N·m²/C²

Also written as: $k = \frac{1}{4\pi\varepsilon_0}$, where ε₀ = 8.85 × 10⁻¹² C²/N·m²

What does this tell us?

• More charge → more force ✅
• More distance → less force (and it drops as r² — so double the distance, force becomes 1/4th!) ✅

Notice it looks exactly like Newton’s Law of Gravitation? That’s not a coincidence — both are inverse square laws. Physics loves patterns!

1.6 Multiple Charges — Superposition Principle

What if there are 3, 4, or 10 charges? Do they all mess with each other simultaneously?

Yes — but here’s the beautiful part: each pair of charges interacts independently. You just find the force from each pair and add them as vectors.

$$\vec{F1} = \vec{F{12}} + \vec{F{13}} + \vec{F{14}} + …$$

Imagine three people pulling you in different directions. The net force on you is the vector sum of all three pulls. Same idea!

1.7 Electric Field — Thinking Smarter

Here’s a smarter way to think about forces between charges. Instead of always asking “what does charge A do to charge B?”, we ask:

“What does charge A do to the space around it?”

That’s the Electric Field — it’s the force per unit positive test charge at any point.

$$\vec{E} = \frac{\vec{F}}{q_0}$$

For a point charge Q:

$$E = \frac{kQ}{r^2}$$

• Unit: N/C (or V/m — same thing)
• It’s a vector — direction matters
• Positive charge? Field points away from it
• Negative charge? Field points toward it

Why do we use a “test charge”? Because we want to measure the field without disturbing it. So we use a tiny, tiny positive charge — just enough to feel the field, not change it.

1.8 Electric Field Lines — Visualising the Invisible

You can’t see electric fields. So physicists invented field lines — imaginary lines that show you where the field is and how strong it is.

Rules (exam-important!):

• Start at +ve charge, end at −ve charge
• Never cross each other (two directions at one point = impossible)
• Denser lines = stronger field
• Always perpendicular to conductor surfaces
• Do NOT form closed loops

Drawing tip: Single +ve charge = lines going outward like a sun 🌞. Single −ve charge = lines coming inward like a drain 🌀. Dipole = lines going from + to −.

1.9 Electric Flux — Counting Field Lines

Flux is basically asking: how many field lines pass through this surface?

$$\Phi = EA\cos\theta$$

• E = electric field strength
• A = area of surface
• θ = angle between field and the normal to surface

Unit: N·m²/C

• θ = 0° → maximum flux (field hits surface head-on)
• θ = 90° → zero flux (field runs parallel to surface, nothing passes through)

Think of it like rain on a window. Window facing the rain = gets soaked (max flux). Window parallel to rain = stays dry (zero flux). 🌧️

1.10 Electric Dipole — Two Charges, One System

A dipole = one +q and one −q, separated by distance 2a.

The dipole moment tells you the strength and direction of the dipole:

$$\vec{p} = q \times 2a$$

Direction: always from −q to +q (remember this — it’s opposite to what you’d expect!)

Field at different positions:

• Axial point (on the axis): $E = \frac{2kp}{r^3}$
• Equatorial point (on perpendicular bisector): $E = \frac{kp}{r^3}$

So axial field = 2 × equatorial field at the same distance. Note it!

Water (H₂O) is a natural dipole — the oxygen pulls electrons more strongly, making one end slightly negative. That’s why water dissolves almost everything. Chemistry meets physics! 🔬

1.11 Dipole in a Uniform Field — What Happens?

Put a dipole in an external electric field E. What does it do?

It experiences a torque that tries to align it with the field:

$$\tau = pE\sin\theta$$

• θ = 0° → τ = 0 → stable equilibrium (dipole aligned with field)
• θ = 90° → τ = pE → maximum torque
• θ = 180° → τ = 0 → unstable equilibrium (dipole anti-aligned)

Potential Energy: $$U = -pE\cos\theta$$

It’s exactly like a compass needle in Earth’s magnetic field — it always tries to align itself. The dipole does the same in an electric field!

1.12 Continuous Charge Distribution — Real World Charges

Point charges are great for theory. But real objects have charge spread across them. So we have three types:

To find the total field, you integrate over all tiny elements — each contributing a little bit of field, and you add them all up (superposition again!).

1.13 Gauss’s Law — The Elegant Shortcut 🏆

This is arguably the most powerful tool in this chapter. Ready?

Gauss’s Law says:

The total electric flux through any closed surface = total charge inside ÷ ε₀

$$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}$$

The closed surface you choose is called a Gaussian surface — it’s imaginary, and you pick its shape to make your life easier.

What’s beautiful about this?

• Charges outside the surface? They contribute zero net flux
• Only what’s inside matters

Imagine a closed room with a fan inside. No matter where the fan is placed inside the room, the total airflow out of the room is the same — it only depends on the fan’s power, not its position. That’s Gauss’s Law in spirit! 🌬️

1.14 Applications of Gauss’s Law — Where It Gets Powerful

Three standard applications. All three are board exam favourites. Learn them cold.

Application 1: Infinite Line Charge

(Use a cylindrical Gaussian surface)

$$\boxed{E = \frac{\lambda}{2\pi\varepsilon_0 r}}$$

Field decreases as 1/r — not 1/r² like a point charge. Different geometry, different result!

Application 2: Infinite Plane Sheet of Charge

(Use a pill-box shaped Gaussian surface)

$$\boxed{E = \frac{\sigma}{2\varepsilon_0}}$$

Shocking result: field is constant, independent of distance! Move closer or farther from an infinite sheet — field doesn’t change. Wild, right?

Application 3: Uniformly Charged Spherical Shell

This one has two cases:

• Outside the shell (r > R): $$E = \frac{Q}{4\pi\varepsilon_0 r^2}$$ Behaves exactly like a point charge sitting at the centre!
• Inside the shell (r < R): $$\boxed{E = 0}$$

Zero field inside! This is one of the most elegant results in physics.

This is why you’re safe inside a car during lightning ⚡ — the metal body is like a shell, and the electric field inside is zero. This is called a Faraday Cage. Your microwave oven works on the same principle — it keeps the microwaves inside!

📋 Formula Sheet — Stick This Somewhere

🎯 Last-Minute Exam Tips

• Gauss’s Law — at least one question is guaranteed. Know all 3 applications with diagrams.
• Field lines never cross — if an MCQ says they do, it’s wrong.
• Dipole formulas — axial = 2 × equatorial. Don’t mix them up.
• Inside a conductor, E = 0 — always. This is a direct consequence of Gauss’s Law.
• For Coulomb’s Law in a medium, replace ε₀ with ε₀εᵣ (εᵣ = relative permittivity).
• Draw diagrams wherever possible — examiners love them and you get method marks!

🧠 MCQs — Electric Charges and Fields

(25 Questions | Board + JEE Level)

Q1. The SI unit of electric field intensity is:

• (a) N·m²/C
• (b) N/C ✅
• (c) C/N
• (d) J/C

Q2. Which of the following is NOT a property of electric field lines?

• (a) They start from positive charge
• (b) They end at negative charge
• (c) They can intersect each other ✅
• (d) They are perpendicular to conductor surface

Q3. The charge on an electron is:

• (a) +1.6 × 10⁻¹⁹ C
• (b) −1.6 × 10⁻¹⁹ C ✅
• (c) −1.6 × 10⁻²⁹ C
• (d) +1.6 × 10⁻²⁹ C

Q4. If the distance between two charges is doubled, the Coulomb force becomes:

• (a) Double
• (b) Half
• (c) Four times
• (d) One-fourth ✅

Q5. Electric flux through a closed surface depends on:

• (a) Shape of the surface
• (b) Size of the surface
• (c) Charges inside the surface ✅
• (d) Charges outside the surface

Q6. The electric field inside a uniformly charged spherical shell is:

• (a) Maximum at centre
• (b) Equal to field outside
• (c) Zero ✅
• (d) Infinite

Q7. Quantisation of charge means:

• (a) Charge is always positive
• (b) Charge exists in multiples of e ✅
• (c) Charge can be any real number
• (d) Charge is always conserved

Q8. The direction of electric dipole moment is:

• (a) From +q to −q
• (b) From −q to +q ✅
• (c) Perpendicular to the dipole axis
• (d) Along the equatorial line

Q9. The torque on a dipole in a uniform electric field is maximum when angle θ is:

• (a) 0°
• (b) 45°
• (c) 90° ✅
• (d) 180°

Q10. Electric field due to an infinite plane sheet of charge is:

• (a) Proportional to distance
• (b) Inversely proportional to distance
• (c) Independent of distance ✅
• (d) Inversely proportional to square of distance

Q11. The value of Coulomb’s constant k is:

• (a) 9 × 10⁶ N·m²/C²
• (b) 9 × 10⁹ N·m²/C² ✅
• (c) 8.85 × 10⁻¹² N·m²/C²
• (d) 6.67 × 10⁻¹¹ N·m²/C²

Q12. A Faraday cage works because:

• (a) It absorbs all electric charges
• (b) Electric field inside a conductor is zero ✅
• (c) It repels all charges
• (d) It is made of insulating material

Q13. The ratio of axial field to equatorial field of a dipole at the same distance is:

• (a) 1:2
• (b) 2:1 ✅
• (c) 1:1
• (d) 4:1

Q14. Which surface is used as Gaussian surface for an infinite line charge?

• (a) Spherical
• (b) Planar
• (c) Cylindrical ✅
• (d) Conical

Q15. If total charge enclosed in a Gaussian surface is zero, then:

• (a) E = 0 everywhere on the surface
• (b) Net flux through the surface is zero ✅
• (c) No charge exists outside
• (d) The surface must be spherical

Q16. Two charges +q and −q are placed at distance 2a. The dipole moment is:

• (a) q/2a
• (b) 2qa ✅
• (c) qa
• (d) q/a

Q17. Electric field lines in a uniform electric field are:

• (a) Curved and diverging
• (b) Parallel and equidistant ✅
• (c) Converging toward centre
• (d) Circular

Q18. The unit of electric flux is:

• (a) N/C
• (b) N·m²/C ✅
• (c) C/m²
• (d) V·m

Q19. A charge of 3μC and −3μC are placed 10cm apart. This system is called:

• (a) Conductor
• (b) Insulator
• (c) Electric dipole ✅
• (d) Gaussian surface

Q20. Which law gives the relationship between electric flux and enclosed charge?

• (a) Coulomb’s Law
• (b) Ohm’s Law
• (c) Gauss’s Law ✅
• (d) Faraday’s Law

Q21. The electric field at the equatorial point of a dipole is directed:

• (a) Same as dipole moment
• (b) Opposite to dipole moment ✅
• (c) Perpendicular to dipole moment
• (d) At 45° to dipole moment

Q22. When a dipole is in stable equilibrium in a uniform field, the angle between p and E is:

• (a) 90°
• (b) 180°
• (c) 45°
• (d) 0° ✅

Q23. The electric field due to an infinite line charge varies as:

• (a) 1/r²
• (b) 1/r ✅
• (c) r
• (d) r²

Q24. Conservation of charge means:

• (a) Charge can be created in pairs
• (b) Total charge of an isolated system remains constant ✅
• (c) Charge always stays positive
• (d) Charge cannot be transferred

Q25. Outside a uniformly charged spherical shell, the field behaves as if:

• (a) Charge is spread on the surface
• (b) All charge is concentrated at the centre ✅
• (c) There is no charge
• (d) Field is uniform everywhere

📝 Important Questions — Board Exam Style

1 Mark Questions

Q1. State Coulomb’s Law in electrostatics.

Ans. The electrostatic force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. F = kq₁q₂/r²

Q2. What is meant by quantisation of electric charge?

Ans. Electric charge always exists in integral multiples of the elementary charge e (= 1.6 × 10⁻¹⁹ C). So q = ne, where n is an integer.

Q3. Why do electric field lines never intersect?

Ans. If two field lines intersected, there would be two directions of electric field at that point — which is physically impossible. Hence they never cross.

Q4. What is a Gaussian surface?

Ans. A Gaussian surface is an imaginary closed surface chosen to apply Gauss’s Law. Its shape is chosen based on the symmetry of the charge distribution.

Q5. Define electric dipole moment. Give its SI unit.

Ans. Electric dipole moment p = q × 2a, where 2a is the separation between charges. It is directed from −q to +q. SI unit: C·m

2 Mark Questions

Q6. State and explain the superposition principle for electric forces.

Ans. The net force on a charge due to multiple charges is the vector sum of individual forces due to each charge, calculated independently. $$\vec{F1} = \vec{F{12}} + \vec{F{13}} + \vec{F{14}} + …$$ Each pair interacts independently — presence of other charges doesn’t affect their interaction.

Q7. Distinguish between conductors and insulators with examples.

Ans.

• Conductors: Allow free movement of charges. Electrons move freely. Examples: Copper, Silver, Iron.
• Insulators: Do not allow charge flow. Electrons are tightly bound. Examples: Rubber, Glass, Plastic.

Q8. What is electric flux? Write its SI unit and formula.

Ans. Electric flux is the total number of electric field lines passing through a given surface area. $$\Phi = EA\cos\theta$$ SI unit: N·m²/C. It is maximum when field is perpendicular to surface (θ = 0°) and zero when parallel (θ = 90°).

Q9. Write two properties of electric field lines.

Ans.

• Field lines start from positive charges and end at negative charges.
• Field lines never intersect each other.
• Denser field lines indicate stronger electric field.
• They are always perpendicular to the surface of a conductor.

(Any two with explanation)

Q10. What happens when an electric dipole is placed in a uniform electric field?

Ans. The dipole experiences a torque τ = pE sinθ that tries to align it with the field. It does not experience a net translational force (since field is uniform). At θ = 0°, torque is zero — stable equilibrium. At θ = 90°, torque is maximum.

3 Mark Questions

Q11. Derive the expression for electric field at an axial point of an electric dipole.

Ans.

Consider a dipole with charges +q and −q separated by 2a. Let P be a point on the axial line at distance r from the centre.

Field at P due to +q: $E_+ = \frac{kq}{(r-a)^2}$ (toward P, away from +q)

Field at P due to −q: $E_- = \frac{kq}{(r+a)^2}$ (toward −q, away from P)

Net field: $$E{axial} = E+ – E_- = kq\left[\frac{1}{(r-a)^2} – \frac{1}{(r+a)^2}\right]$$

Simplifying: $$E_{axial} = \frac{2kpr}{(r^2-a^2)^2}$$

For r >> a: $$\boxed{E_{axial} = \frac{2kp}{r^3}}$$

Direction: same as dipole moment p.

Q12. State Gauss’s Law and use it to find the electric field due to an infinitely long straight wire of linear charge density λ.

Ans.

Gauss’s Law: Total electric flux through a closed surface = Q_enc/ε₀

$$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}$$

For infinite line charge: Choose a cylindrical Gaussian surface of radius r and length l coaxial with the wire.

• Flux through curved surface = E × 2πrl
• Flux through flat ends = 0 (E ⊥ area vector)
• Charge enclosed = λl

Applying Gauss’s Law: $$E \times 2\pi rl = \frac{\lambda l}{\varepsilon_0}$$

$$\boxed{E = \frac{\lambda}{2\pi\varepsilon_0 r}}$$

Direction: radially outward for positive λ.

Q13. Using Gauss’s Law, show that electric field inside a uniformly charged spherical shell is zero.

Ans.

Consider a spherical shell of radius R with total charge Q. To find field at point P inside (r < R):

Choose a spherical Gaussian surface of radius r < R centred at the shell’s centre.

By symmetry, E is uniform over this surface.

Charge enclosed inside this Gaussian surface = 0 (all charge is on the shell, outside our surface)

Applying Gauss’s Law: $$\oint E \cdot dA = \frac{Q_{enc}}{\varepsilon_0} = \frac{0}{\varepsilon_0} = 0$$

Since area ≠ 0: $$\boxed{E = 0 \text{ inside the shell}}$$

This is why a Faraday cage protects you — no electric field can exist inside a conducting shell!

5 Mark Questions

Q14. (a) State Gauss’s Law. (b) Apply it to find electric field due to: (i) infinite plane sheet (ii) spherical shell (inside and outside).

Ans.

(a) Gauss’s Law: $$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}$$

(b)(i) Infinite Plane Sheet (surface charge density σ):

Choose a pill-box (cylinder) Gaussian surface with flat faces of area A on both sides of the sheet.

• Flux through two flat faces = 2EA
• Flux through curved side = 0
• Charge enclosed = σA

$$2EA = \frac{\sigma A}{\varepsilon_0}$$ $$\boxed{E = \frac{\sigma}{2\varepsilon_0}}$$

Field is uniform and independent of distance!

(b)(ii) Spherical Shell:

Outside (r > R): Spherical Gaussian surface of radius r. $$E \times 4\pi r^2 = \frac{Q}{\varepsilon_0}$$ $$\boxed{E = \frac{Q}{4\pi\varepsilon_0 r^2} = \frac{kQ}{r^2}}$$ Behaves like a point charge!

Inside (r < R): Q_enc = 0 $$\boxed{E = 0}$$