NCERT Class 12 Physics • Unit I

Electric Charges and Fields

📄 Source PDF: leph101.pdf (NCERT Class 12 Physics, Chapter 1)

A focused NEET Faculty Exam Pack — exact concepts, a complete formula bank, trap-avoidance shortcuts, an interactive field-vs-distance graph, and 10 exam-level MCQs with the fastest solving methods. No long derivations.

1 The NEET Cheat Sheet — Core Concepts & Exceptions

The simple idea

Some objects carry an invisible property called charge. Charges of the same kind push each other apart, opposite kinds pull together. The push/pull at any point is what we call the electric field. The whole chapter is just rules for finding that field.

Properties of Electric Charge

Quantisation of charge

💡 Charge comes only in whole packets — never half an electron. The smallest packet is the charge of one electron (e).

Every charge is q = ± n·e, where n ∈ ℤ and e = 1.602 × 10⁻¹⁹ C.

📌 Charge of an α-particle = +2e; a Cl⁻ ion = −e.

⚠ Fails at quark level (±⅓ e, ±⅔ e), but quarks are never free — so quantisation holds for all observable charges.

Conservation of charge

💡 Charge is never created or destroyed — it only moves from one object to another (like a thread of beads being passed around).

Total charge of an isolated system stays constant in any process — even nuclear reactions.

📌 Beta decay: n → p + e⁻ + ν̄. Net charge before (0) = after (+1 − 1 + 0) = 0. ✓

Additivity

💡 If a body has many tiny charges sitting on it, the body's total charge is just their sum — with signs.

Qtotal = q₁ + q₂ + q₃ + … (algebraic sum).

📌 +5 μC plus −3 μC on the same body ⇒ net +2 μC.

Conductors, insulators & charging

Conductor vs Insulator

💡 In conductors (metals, salt water) charges can walk around freely. In insulators (plastic, glass) they're stuck where they are.

Conductors: free electrons → very low resistance. Insulators: bound electrons → resist motion.

⚠ A charged conductor at equilibrium has all extra charge on its outer surface and E = 0 inside.

3 ways to charge a body

💡 You can give an object charge by rubbing (friction), by touching (conduction), or by simply bringing a charged thing nearby (induction).

Friction (rubbing) · Conduction (direct contact) · Induction (no contact — separates charges, then ground).

📌 Earthing a conductor in induction removes the unwanted same-sign charge; left with opposite sign.

Coulomb's law — the foundation

Coulomb's law

💡 Two charges push/pull each other harder if they're bigger, and much weaker if farther apart (squared!).

F = k·q₁q₂/r², where k = 1/(4πε₀) ≈ 9 × 10⁹ N m²/C²; ε₀ = 8.854 × 10⁻¹² C²/N·m².

📌 The force is along the line joining the two charges — attractive for opposite signs, repulsive for like signs.

⚠ Valid for point charges at rest in vacuum. In a medium of dielectric constant K: Fmed = Fvac/K.

Superposition principle

💡 If many charges are around, the total force on one of them is just the vector sum of forces from each, one at a time.

Fnet = F₁ + F₂ + F₃ + … (vectors).

📌 Always resolve into x and y components first, then add.

Electric field & field lines

Electric field E

💡 An invisible arrow at every point — telling you which way and how strongly a +1 C test charge would be pushed.

E = F / q₀ (q₀ → 0); SI unit: N/C = V/m.

Field of a point charge: E = kq/r², radial (outward for +, inward for −).

Field-line rules

💡 Field lines are visual maps of the electric field — like wind-direction streaks.

  • Start on + charges, end on charges (or at infinity).
  • Never cross (single E direction at each point).
  • Density of lines ∝ field strength.
  • Tangent to a field line = direction of E.
  • Continuous in free space (don't form closed loops in electrostatics).

Positive point charge

Lines radiate outward.

Dipole (+q, −q)

From + to −, curved outside.

Two like (+, +)

Lines repel; neutral point midway.

Electric dipole

Dipole moment & field

💡 Two equal & opposite charges a tiny distance apart act together like a single "arrow" called the dipole moment, pointing from − to +.

p = q · 2a; SI unit: C·m. Direction: −q → +q.

Axial (end-on) field: E = (2kp)/r³ — along p.

Equatorial (broadside): E = kp/r³ — anti-parallel to p.

⚠ Both formulas assume r >> a (short / point dipole).

Dipole in uniform field

💡 Place a dipole in a uniform field and it rotates to align with the field — like a compass needle.

Torque τ = p × E ⇒ |τ| = pE sinθ. Stable equilibrium at θ = 0°; unstable at 180°.

Potential energy U = −p·E = −pE cosθ. Net force = 0 in uniform field.

Electric flux & Gauss's law

Electric flux ϕE

💡 Imagine field lines piercing a surface. Count the lines passing through — that "count" is the flux.

ϕE = ∫ E · dA = EA cosθ (uniform field). SI unit: V·m or N·m²/C.

📌 If E is parallel to surface (θ = 90°): ϕ = 0.

Gauss's law

💡 Total field lines piercing any closed surface depend only on the charge it encloses — nothing else inside or outside the box matters.

∮ E · dA = qenc / ε₀ — for any closed surface.

📌 Charge q at centre of a cube: flux through cube = q/ε₀; through one face = q/(6ε₀).

⚠ Charge outside the surface contributes zero net flux (in = out).

Exceptions, limits & common errors

2 The Formula & Approximation Bank

Every formula NEET tests, grouped by topic — with the time-saving notes beside each. A one-line plain summary sits above each group.

Coulomb force & field

💡 Two big "musts": k = 9 × 10⁹ in vacuum, and force/field always shrink as 1/r².

FormulaNotes / Approximation
F = (1/4πε₀)·q₁q₂/r² = kq₁q₂/r²k = 9 × 10⁹ N m²/C² (vacuum); ε₀ = 8.854 × 10⁻¹² C²/N·m².
Fmed = Fvac / KK = relative permittivity (dielectric constant) of the medium.
E = F / q₀ = kq/r² (point charge)Direction: away from +q, toward −q.
q = ± n·e ; e = 1.6 × 10⁻¹⁹ CQuantisation; integer multiples only.
1 C ≈ 6.25 × 10¹⁸ electron chargesHandy for "number of electrons" type Qs.

Continuous charge distributions

💡 When charge is spread out, "shrink" the distribution to a tiny piece dq, find its dE, then add (integrate).

FormulaNotes
Line: λ = dq/dl ; E = λ/(2πε₀ r)Infinite straight line. Radial; falls as 1/r.
Surface: σ = dq/dA ; E = σ/(2ε₀)Infinite plane sheet. Independent of distance.
Volume: ρ = dq/dVUse Gauss for high-symmetry volumes.
Two parallel sheets (opposite signs): Ebetween = σ/ε₀; outside = 0Useful in capacitor / Q5-style problems.

Electric dipole

💡 A dipole's field falls off faster than a single charge (1/r³ versus 1/r²) — because the + and − partially cancel far away.

FormulaNotes
p = q · 2a (from −q to +q)SI: C·m. Vector.
Eaxial = (2kp)/r³End-on; same direction as p; twice equatorial.
Eequatorial = kp/r³Broadside; opposite to p.
Egeneral = (kp/r³)·√(1 + 3cos²θ)θ measured from the dipole axis.
τ = pE sinθ ; U = −pE cosθTorque & potential energy in uniform field.

Flux & Gauss's Law applications

💡 Pick a "Gaussian surface" so that E is either perpendicular & constant (then EA = q/ε₀) or parallel (then contributes zero).

FormulaNotes
ϕE = EA cosθ ; ∮ E·dA = qenc/ε₀Closed surface only; charge outside contributes 0.
Point charge at cube centre: ϕtotal = q/ε₀ ; per face = q/(6ε₀)At a corner: per face touching = 0; total = q/(8ε₀).
Solid sphere (uniform ρ): Ein = kQr/R³ ; Eout = kQ/r²Inside: linear in r; outside: as point charge at centre.
Spherical shell: Ein = 0 ; Esurface = kQ/R² ; Eout = kQ/r²Inside any hollow conductor / shell, field is zero.
Conductor surface: E = σ/ε₀Twice a sheet because field exists on one side only.

Useful constants

k = 9 × 10⁹ε₀ = 8.854 × 10⁻¹²e = 1.6 × 10⁻¹⁹ CNA = 6.022 × 10²³1 μC = 10⁻⁶ C1 nC = 10⁻⁹ C

📈 Interactive Field-vs-Distance Graph

The simple idea

Different shapes of charge give different "falloff" laws for the electric field. A point charge falls as 1/r², a line as 1/r, and a sheet doesn't fall at all. Pick a configuration and watch the graph.

E at r: 3.60×10⁴ N/C
Falloff: 1/r²
Region: outside

Sliders update the curve in real time. Notice: sheet is a flat line (E independent of r), line falls as 1/r, point as 1/r², and shell jumps from 0 (inside) to kQ/R² (surface) then falls.

3 The "Trap Avoidance" & Shortcut Guide

The simple idea

These are the exact spots where students lose easy marks. Each card shows the trap, then the quick fix.

1

Coulomb's law in a medium

Wrong: Using F = kq₁q₂/r² with k = 9 × 10⁹ inside a medium with dielectric constant K.
Right: Fmedium = Fvacuum / K. Replace ε₀ by ε = Kε₀ in the denominator.

⚡ For water K ≈ 80 — the force becomes 80 times weaker, not the same.

2

"Inside vs Outside" of a charged sphere/shell

A favourite NEET twist. Use this table:

RegionUniform shellSolid sphere (uniform ρ)
Inside (r < R)E = 0E = kQr/R³ (linear in r)
At surfaceE = kQ/R²E = kQ/R²
Outside (r > R)E = kQ/r²E = kQ/r²

⚡ Inside a hollow conductor / shell: E = 0, V = constant = kQ/R. NEET classics.

3

Flux through a cube — corner trick

Wrong: A charge at a cube's corner still gives ϕ = q/ε₀.
Right: Imagine 8 such cubes meeting at the corner — by symmetry each gets q/(8ε₀).

At face centre: ϕ = q/(2ε₀). At edge midpoint: ϕ = q/(4ε₀). At centre: ϕ = q/ε₀.

4

Dipole — axial vs equatorial

  • Axial field is twice equatorial: Eaxial = 2·Eequatorial (at same r).
  • Both fall as 1/r³, faster than a single point charge.
  • On axial line E is parallel to p; on equatorial line E is anti-parallel to p.
  • Net force on a dipole in uniform field = 0; only torque.

⚡ Memory hook: "A"xial means "A"head, "2x" the strength.

5

Sheet vs Conductor surface — the factor of 2

Infinite charged sheet (insulator): E = σ/(2ε₀) — field on both sides.
Conductor surface: E = σ/ε₀ — field on one side only (inside conductor E=0).

⚡ The "2" disappears for a conductor because all the flux has only one side to go through.

6

Superposition signs

  • Always work with vectors: resolve each force into x and y, then add.
  • For three equal charges at corners of an equilateral triangle (side a), force on any one = √3·kq²/a², directed outward along the bisector.
  • For four equal charges at corners of a square, force on each charge ≠ zero — there is a net outward push along the diagonal.
7

Common numerical shortcuts

Memorise these to save 30+ seconds per problem:

kq²/a² @ 1 μC, 1 m ⇒ 9 × 10⁻³ N kq/r² @ 1 μC, 1 m ⇒ 9000 N/C σ/(2ε₀) ≈ 56.5·σ kN/C if σ in μC/m² 1 N/C = 1 V/m

4 High-Yield Practice MCQs

The simple idea

Try each question first, then open the solution to see the quickest route — using the formulas and shortcuts from above, not long textbook working.

10 NEET-level questions with numerically close options — 2 theory, 2 Coulomb, 3 field/dipole, 3 Gauss/flux. Tap "Show fastest solution".

TheoryQ1. Which of the following is NOT a correct property of electric field lines?

  1. They start from +ve and end on −ve charges.
  2. They can cross each other in regions of strong field.
  3. The tangent at any point gives the direction of E at that point.
  4. Their density at a point is proportional to the magnitude of E.
Show fastest solutionAnswer: (b)

Two crossing lines would give two directions for E at the crossing point — impossible. So field lines never cross. The other three are standard properties.

TheoryQ2. A point charge +q is placed at the centre of a cubical Gaussian surface. The flux through one face is:

  1. q/ε₀
  2. q/(2ε₀)
  3. q/(6ε₀)
  4. q/(8ε₀)
Show fastest solutionAnswer: (c) q/(6ε₀)

Total flux through the closed cube = q/ε₀. By symmetry six faces share it equally ⇒ q/(6ε₀) per face.

CoulombQ3. Two point charges +1 μC and +1 μC are placed 1 m apart in vacuum. The force on each is:

  1. 9 × 10⁻³ N (repulsive)
  2. 9 × 10⁻⁶ N (attractive)
  3. 9 × 10⁹ N (repulsive)
  4. 9 N (repulsive)
Show fastest solutionAnswer: (a)

F = kq₁q₂/r² = 9 × 10⁹ × (10⁻⁶)² / 1² = 9 × 10⁻³ N. Like charges → repulsive.

CoulombQ4. Two point charges +4q and +q lie on a line at distance d. The point on the line joining them where the net field is zero is at distance from +4q:

  1. d/3
  2. 2d/3
  3. d/2
  4. d/4
Show fastest solutionAnswer: (b) 2d/3

For E = 0 between the charges: k(4q)/x² = kq/(d−x)² ⇒ 2(d−x) = x ⇒ x = 2d/3. Closer to the smaller charge, as expected.

Field / DipoleQ5. An electric dipole of moment p is placed at angle θ with a uniform field E. Torque on it is:

  1. pE
  2. pE cosθ
  3. pE sinθ
  4. 0
Show fastest solutionAnswer: (c) pE sinθ

τ = p × E ⇒ magnitude = pE sinθ. Maximum at θ = 90°; zero at θ = 0° or 180°.

Field / DipoleQ6. At a point on the axial line of a short dipole (moment p) at distance r (r >> a), the field is E. The field at the same distance on the equatorial line is:

  1. E
  2. E/2
  3. 2E
  4. E/4
Show fastest solutionAnswer: (b) E/2

Eaxial = 2kp/r³ and Eequatorial = kp/r³ ⇒ ratio 2 : 1. So equatorial = E/2.

Field / DipoleQ7. Two large parallel sheets carry surface charge densities +σ and −σ. The field in the region between the sheets is:

  1. σ/(2ε₀)
  2. σ/ε₀
  3. 2σ/ε₀
  4. 0
Show fastest solutionAnswer: (b) σ/ε₀

Each sheet alone gives σ/(2ε₀). Between, the two fields add (opposite signs of σ, fields in same direction) ⇒ σ/ε₀. Outside, they cancel.

Gauss / FluxQ8. A charge of 8 μC is placed at the centre of a sphere of radius 2 m. The total electric flux out of the sphere is:

  1. 9 × 10⁵ V·m
  2. 9 × 10⁴ V·m
  3. 9 × 10⁵ V·m²/C
  4. (8/ε₀) μV·m
Show fastest solutionAnswer: (a)

ϕ = qenc/ε₀ = (8 × 10⁻⁶)/(8.854 × 10⁻¹²) ≈ 9.04 × 10⁵ V·m. Note: independent of radius.

Gauss / FluxQ9. A point charge q is placed at one corner of a cube. The total electric flux through the cube is:

  1. q/ε₀
  2. q/(2ε₀)
  3. q/(8ε₀)
  4. q/(6ε₀)
Show fastest solutionAnswer: (c) q/(8ε₀)

Surround the corner with 8 identical cubes; by symmetry, each gets q/8 of the total flux. Total enclosed (over 8 cubes) = q ⇒ each cube has q/(8ε₀).

Gauss / FluxQ10. A solid metallic sphere of radius R carries a charge Q. Field at distance r = R/2 from its centre is:

  1. kQ/R²
  2. kQr/R³
  3. 0
  4. kQ/r²
Show fastest solutionAnswer: (c) 0

Inside a metallic (conductor) sphere, all charge sits on the surface and the field inside is zero. Don't confuse with a solid insulator (uniform ρ) where Ein = kQr/R³.

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