NCERT Plus Two (Class 12) Physics • Chapter 1

Electric Charges and Fields

Charge conservation & quantisation, Coulomb's law & superposition, electric field lines, electric dipoles, electric flux & Gauss's law applications.

📄 Official NCERT Textbook PDF

Topic 1.2 – 1.4: Charge Conservation, Quantisation (q = ±ne) & Induction

Run a dry plastic comb through your hair on a breezy Kerala afternoon and hold it over tiny bits of torn notebook paper — magically, the paper leaps up and sticks! You did not invent new electricity out of thin air; rubbing simply shifted invisible packets of electrons from hair to comb, leaving one negative and the other positive.
QUANTISATIONq = ± n e where n = 1, 2, 3... & e = 1.6 × 10⁻¹⁹ C
CONSERVATIONNet charge of an isolated system never changes — charges only transfer!
1 COULOMB COUNT1 C equals about 6.25 × 10¹⁸ electrons! (SI unit: Coulomb, C)
🧠 Memory Trick: Think of charge like 1-rupee coins (e) — you can carry 5e or −12e, NEVER 2.5e! And in Induction, no touch needed — opposite charge stays!
💡 Simple Kid-Check: Two like charges (+ and +, or − and −) push each other away; unlike charges (+ and −) pull together. Unlike mass, charge can be positive, negative, or zero — yet total charge always balances like a bank ledger!
  • 1
    Additivity & Conservation: Total charge of a system is the simple algebraic sum qnet = q₁ + q₂ + ... (scalar!). In an isolated system, charge can be transferred or created in equal +/− pairs, keeping net charge constant.
  • 2
    Quantisation of Charge: Free charge always comes in integer multiples of elementary charge e = 1.602 × 10⁻¹⁹ C: q = ±ne. Only at huge macroscopic charges (like 1 μC ≈ 6 × 10¹² e) does granularity look continuous.
  • 3
    Charging by Induction (No Touch!): Bring a negatively charged rod near a neutral metal sphere on an insulating stand → electrons flee to the far side; earth the far side so they drain away; break earth first, then remove rod → the sphere stays purely positively charged!
🧒 Super-Easy Example: Can a body hold a charge of 4.8 × 10⁻¹⁹ C? Divide by e: n = (4.8 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) = 3 (whole integer → YES!). What about 2.4 × 10⁻¹⁹ C? n = 1.5 → IMPOSSIBLE!
Classic Kerala HSE 1-mark trap: (1) Always verify n = q / e is an exact integer before accepting any charge value! (2) When charging a sphere positively by induction using a negative rod, you MUST disconnect the earthing wire BEFORE removing the charged rod, otherwise electrons rush right back from Earth! (3) Charge is invariant (does not change with speed), unlike relativistic mass!

Topic 1.5 – 1.6: Coulomb's Law & Superposition of Multiple Charges

Imagine two tiny magnets pushing or pulling each other across your study table — bring them twice as close and the push suddenly gets FOUR times stronger! Charles Coulomb discovered the exact same inverse-square law for electric charges, except electric forces can BOTH attract and repel!
COULOMB'S LAWF = 14πε₀ |q₁ q₂|r²  |  k = 9 × 10⁹ N·m²/C²
MEDIUM EFFECTFmed = Fvacεᵣ (Water εᵣ ≈ 80 cuts force 80×!)
SUPERPOSITIONF₁ = F₁₂ + F₁₃ + ... (Pairwise vector sum; third charge doesn't disturb F₁₂!)
🧠 Memory Trick: Coulomb is Newton's twin! F ∝ q₁q₂ / r² — double r ⇒ force drops to ¼; put charges in water (K = 80) ⇒ force drops 80×!
💡 Simple Kid-Check: Two charges talk along the straight line joining them. Even if ten noisy classmates join the room, the individual force between charge 1 and charge 2 refuses to change — you simply add all force arrows head-to-tail!
  • 1
    Scalar & Vector Form: Force acts along line joining charges: F₂₁ = 14πε₀ q₁ q₂r² r̂₁₂. Note F₁₂ = −F₂₁ (obeys Newton's 3rd Law even for unequal charges!).
  • 2
    Permittivity & Dielectric Constant: Vacuum permittivity ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻². Inside any medium of relative permittivity εᵣ (or dielectric constant K), force drops by factor K: Fm = F₀ / K.
  • 3
    Principle of Superposition: Net force on any charge equals the vector sum of individual two-body Coulomb forces from every other charge. For equal forces F at angle θ: Fnet = √(F₁² + F₂² + 2F₁F₂ cosθ).
🧒 Super-Easy Example: Two charges repel with 36 N in air at distance r. (a) Move them to 3r → force drops by 3² = 9 times to 36/9 = 4 N. (b) Keep at r inside glass (K = 6) → force drops to 36/6 = 6 N!
Watch three classic traps: (1) If charge +10 μC faces +1 μC, students think the bigger charge pushes harder! WRONG! Both experience EXACTLY EQUAL and opposite force (Newton's 3rd Law, ratio 1 : 1!). (2) Always verify Coulomb vector direction (attraction vs repulsion) BEFORE adding components — do NOT drop signs blindly! (3) Remember ε₀ unit is C² N⁻¹ m⁻² (inverse of k).

Topic 1.7 – 1.8: Electric Field, Field Lines & Continuous Charge Density

Think of a bright campfire on a chilly hill: even before you stretch your hand near it, the fire sets up an invisible warm zone around itself! Similarly, any charge creates an invisible Electric Field everywhere around it, waiting to push any tiny test charge that walks in.
ELECTRIC FIELDE = Fq₀  |  Point charge: E = k Qr² (Unit: N/C or V/m)
FIELD LINE RULESStart at +, end at − • NEVER intersect • NEVER form closed loops!
3 CHARGE DENSITIESLinear λ = Q/L (C/m) • Surface σ = Q/A (C/m²) • Volume ρ = Q/V (C/m³)
🧠 Memory Trick: Field lines are like one-way traffic streets: crowded lines = strong E; tangent = direction of E; two streets NEVER cross (otherwise 2 directions at 1 spot!).
💡 Simple Kid-Check: Test charge q₀ → 0 is kept super tiny so it behaves like a polite spy — it measures the host charge's field without pushing the host charge out of place!
  • 1
    Electric Field Vector: E = limq₀→0 (F/q₀). Points radially away from +Q and toward −Q. Force on any charge q placed in field E is simply F = qE!
  • 2
    4 Key Properties of Field Lines: (a) Tangent gives direction of E. (b) Never cross each other. (c) Electrostatic field lines never form closed loops (because electrostatic force is conservative!). (d) Always hit a conductor surface at 90°; zero lines exist inside a conductor!
  • 3
    Continuous Charge Distribution: When billions of electrons pack on a wire, sheet, or sphere, use dq = λ dl, dq = σ dA, or dq = ρ dV to sum fields smoothly.
🧒 Super-Easy Example: An electron (q = −1.6 × 10⁻¹⁹ C) enters a downward field E = 500 N/C. Force magnitude is F = |q|E = 8 × 10⁻¹⁷ N, directed UPWARD (opposite to E because charge is negative!).
Top 2-mark Kerala HSE board question: Why can two electric field lines NEVER intersect each other? Answer: If they crossed, a test charge placed at the intersection would have two different tangents (two directions of electric field) at the same point, which is physically impossible! Also remember: electrostatic field lines never form closed loops (unlike magnetic field lines!).

Topic 1.10 – 1.11: Electric Dipole Moment, Axial/Equatorial Field & Torque

Look at a simple water molecule (H₂O): oxygen pulls electrons slightly closer than hydrogen, leaving one side slightly negative and the other slightly positive! That tiny equal-and-opposite pair separated by a microscopic gap is an Electric Dipole — nature's miniature electric compass needle!
DIPOLE MOMENTp = q · (2a) pointing from −q to +q (Unit: C·m)
FAR-FIELD (r ≫ a)Axial: Eax = 2kpr³ (∥ p)  |  Eq: Eeq = kpr³ (−∥ p)
UNIFORM E-FIELDNet Force = 0! Torque τ = p × E ⇒ τ = pE sinθ
🧠 Memory Trick: Eaxial = 2 × Eequatorial at same large distance r! Point charge falls as 1/r², while a dipole falls faster as 1/r³!
💡 Simple Kid-Check: Inside a uniform electric field, +q gets pulled right by +qE and −q gets pulled left by −qE. Those equal opposite pulls cancel (net force = 0, so it doesn't fly sideways!), yet they twist the bar like turning a bicycle handlebar — that twist is Torque!
  • 1
    Direction Rule: Dipole moment p always points from −q toward +q (opposite to internal field!).
  • 2
    Axial vs Equatorial Field (Derivations!):
    • Along axis: Eaxial = 14πε₀ 2pr(r² − a²)² ≈ 2kpr³ (for r ≫ a), along p.
    • Along perpendicular bisector: Eeq = 14πε₀ p(r² + a²)3/2 ≈ kpr³, opposite to p!
  • 3
    Torque in Uniform Field: τ = (qE) · (2a sinθ) = pE sinθ (τ = p × E). Maximum torque τmax = pE at θ = 90°! In a non-uniform field, BOTH net force AND torque act!
🧒 Super-Easy Example: A dipole with p = 4 × 10⁻⁹ C·m sits at 30° to a uniform field E = 5 × 10⁴ N/C. Torque = pE sin30° = (4 × 10⁻⁹)(5 × 10⁴)(0.5) = 10⁻⁴ N·m!
Two classic Kerala HSE traps: (1) Students forget dipole moment vector p runs from −q to +q, NOT + to −! (2) At θ = 0° (p ∥ E), torque is 0 and equilibrium is STABLE; at θ = 180° (antiparallel), torque is ALSO 0, yet equilibrium is UNSTABLE (tiny nudge flips it right around)! (3) Remember in a uniform field, net force is ALWAYS ZERO (only in non-uniform fields does a dipole experience net force!).

Topic 1.9 & 1.13: Electric Flux & Gauss's Law

Imagine holding a butterfly net under steady monsoon rain! Hold the ring flat facing the rain and maximum water pours through; tilt the ring sideways edge-on to the rain and ZERO drops pass through! Electric Flux simply counts how many electric field arrows pierce through a surface.
ELECTRIC FLUXΦ = E · A = E A cosθ (Scalar! Unit: N·m²/C or V·m)
AREA VECTOR AAlways points perpendicular (outward normal) to the surface!
GAUSS'S LAWΦtotal = ∮ E · dA = qenclosedε₀
🧠 Memory Trick: Gauss's Law cares ONLY about WHO IS INSIDE the closed bag (qin), NOT the bag's size, shape, or outside charges!
💡 Simple Kid-Check: Put a glowing bulb (+q) inside a tiny football or a giant cinema tent — the EXACT same amount of light streams out of both! Outside lamps send rays into one wall and straight out the far wall, adding zero net flux!
  • 1
    Angle θ Trap-Check: θ is the angle between E and the normal vector n̂ (perpendicular to surface), NOT the surface itself! So surface perpendicular to E means θ = 0° ⇒ Φ = EA (max); surface parallel to E means θ = 90° ⇒ Φ = 0!
  • 2
    Gauss's Law Formula: Total electric flux out of ANY closed Gaussian surface equals 1/ε₀ times the net charge enclosed inside: Φ = qenc / ε₀.
  • 3
    What About Charges Outside? Every field line from an outside charge enters and exits the closed surface, contributing 0 net flux! Note: E on the surface includes ALL charges (inside + outside), while total Φ depends ONLY on qin!
🧒 Super-Easy NCERT Classic: A point charge +q sits at the center of a cube of side a. Total flux out of all 6 faces = q/ε₀. By symmetry, flux through ONE face = q / (6ε₀)! If the charge sits inside an electric dipole (+q and −q enclosed), qnet = 0 ⇒ Φ = 0!
Watch three giant exam traps: (1) If a question says "the plane of a loop makes 30° with E", then the angle with the normal is θ = 90° − 30° = 60°, so use cos 60° = 0.5, NOT cos 30°! (2) Doubling the radius of a Gaussian sphere around a charge DOES NOT CHANGE electric flux (Φ = q/ε₀ stays identical!). (3) Enclosing a dipole gives net flux 0 because qin = +q − q = 0.

Topic 1.14: Applications of Gauss's Law (Wire, Sheet & Spherical Shell)

Trying to integrate Coulomb's law over an infinite sheet or giant spherical shell takes pages of messy calculus! Using Gauss's symmetry trick — wrapping a tidy cylinder, pillbox, or sphere around the charge — collapses the whole derivation into just three algebra lines!
1. INFINITE WIRE (λ)E = λ2πε₀ r (Falls as 1/r, cylindrical symmetry!)
2. INFINITE SHEET (σ)E = σ2ε₀ (CONSTANT! Independent of distance r!)
3. SPHERICAL SHELL (R)Outside (r ≥ R): E = kQr² • Inside (r < R): E = 0!
🧠 Memory Trick: Watch the power of r drop as dimension grows: Point (1/r²) → Line Wire (1/r¹) → Flat Sheet (1/r⁰ = constant)! And inside a hollow charged shell, E = 0!
💡 Simple Kid-Check: Inside a uniform charged spherical shell, pulls from nearby walls and huge far-away walls cancel out everywhere inside! That is why sitting inside a metal car keeps you safe during lightning (electrostatic shielding)!
  • 1
    Infinitely Long Thin Straight Wire (λ): Wrap a coaxial cylinder of radius r, length l. Flat end-caps have E ⊥ dA ⇒ Φ = 0; curved wall gives E(2πrl) = λl / ε₀ ⇒ E = λ / (2πε₀r).
  • 2
    Uniformly Charged Infinite Plane Sheet (σ): Wrap a symmetric pillbox cutting through with two end faces of area A: 2EA = σA / ε₀ ⇒ E = σ / (2ε₀) — totally independent of distance! (Between two opposite sheets +σ and −σ, fields add to σ / ε₀ inside and 0 outside!)
  • 3
    Thin Spherical Shell (Radius R, Charge Q):
    • Outside (r > R): Gaussian sphere encloses Q → E(4πr²) = Q/ε₀ ⇒ E = Q / (4πε₀r²) (acts as if all charge sits at centre!).
    • On surface (r = R): E = σ / ε₀.
    • Inside (r < R): qenclosed = 0 ⇒ E = 0!
🧒 Super-Easy Comparison: A single thin sheet of charge σ produces E = σ/(2ε₀). Place a +σ sheet parallel to a −σ sheet: between them both point same way → σ/(2ε₀) + σ/(2ε₀) = σ/ε₀ (that's the exact parallel-plate capacitor field in Ch 2!).
Watch these top 3 DHSE mistakes: (1) Students confuse σ/(2ε₀) (single thin non-conducting sheet) with σ/ε₀ (outside a charged conductor or between opposite plates)! (2) Inside a thin charged spherical shell, Electric Field E = 0 everywhere inside, even though Electric Potential V is NOT zero (we'll see V = kQ/R constant in Ch 2!). (3) Always draw the Gaussian cylinder/pillbox clearly for full 3-mark derivation credit!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Is a charge of 5.8 × 10⁻¹⁸ C possible on an isolated conductor? Justify.

No. By quantisation q = ne, n = (5.8 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹) = 36.25. Since n is not an integer, this charge cannot exist.

Q2 (Kerala HSE Say • 2 Marks): State any two properties of electric field lines. Why can two electric field lines never intersect?

Properties: (1) They originate from + charges and end on − charges without forming closed loops. (2) Tangent gives direction of E.
No intersection: If two lines crossed, a test charge at the point would experience two directions of E simultaneously, which is impossible.

Q3 (Kerala HSE March • 2 Marks): Write the expression for torque acting on an electric dipole in a uniform electric field. Identify the angles for stable and unstable equilibrium.

τ = p × E ⇒ τ = pE sinθ (net force = 0).
• Stable equilibrium: θ = 0° (p ∥ E, minimum energy −pE).
• Unstable equilibrium: θ = 180° (antiparallel, maximum energy +pE).

Q4 (Kerala HSE March • 3 Marks): Derive an expression for the electric field at a point on the axial line of an electric dipole.

Consider dipole −q and +q separated by 2a; point P at distance r from centre along axis.
E₊ = kq/(r − a)² (along p);  E₋ = kq/(r + a)² (opposite to p).
Eaxial = E₊ − E₋ = kq [1/(r − a)² − 1/(r + a)²] = kq · 4ar / (r² − a²)².
Since p = 2qa: Eaxial = 2pr4πε₀(r² − a²)². For r ≫ a, Eaxial = 2kpr³.

Q5 (Kerala HSE Model • 2 Marks): State Gauss's law in electrostatics. A point charge of +10 μC sits 5 cm directly above the centre of a 10 cm × 10 cm square. Find the electric flux through the square.

Gauss's Law: Total flux through a closed surface is Φ = qenc / ε₀.
Imagine a cube of side 10 cm enclosing the charge right at its centre (5 cm from each of 6 identical faces).
Flux through one square face = q/(6ε₀) = (10 × 10⁻⁶) / (6 × 8.854 × 10⁻¹²) ≈ 1.88 × 10⁵ N·m²/C.

Q6 (Kerala HSE March • 3 Marks): Using Gauss's law, derive the electric field due to an infinitely long straight uniformly charged wire of linear charge density λ.

Choose a coaxial Gaussian cylinder of radius r and length l.
• End caps: E ⊥ dA ⇒ Φends = 0.
• Curved surface: E ∥ dA and uniform ⇒ Φ = E(2πrl).
• By Gauss's Law: E(2πrl) = qenc/ε₀ = λl/ε₀ ⇒ E = λ / (2πε₀r) radially outward (for λ > 0).

Q7 (Kerala HSE Say • 3 Marks): Apply Gauss's law to find the electric field (i) outside and (ii) inside a uniformly charged thin spherical shell of radius R and charge Q. Sketch E vs r.

(i) Outside (r > R): Concentric Gaussian sphere of radius r encloses Q ⇒ E(4πr²) = Q/ε₀ ⇒ E = Q/(4πε₀r²).
(ii) Inside (r < R): Gaussian sphere encloses qenc = 0 ⇒ E = 0.
Graph: E = 0 from 0 to R, jumps to max kQ/R² at r = R, then decays as 1/r².

Q8 (NCERT / DHSE Numerical • 2 Marks): Two point charges +4 μC and +9 μC are 20 cm apart in air. Find where a third charge q₀ must be placed on the line joining them so it stays in equilibrium.

Equilibrium point lies between them at distance x from +4 μC:
k(4)q₀ / x² = k(9)q₀ / (20 − x)² ⇒ 2/x = 3/(20 − x) ⇒ 3x = 40 − 2x ⇒ 5x = 40 ⇒ x = 8 cm from +4 μC.