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State Board · Class 12 · Physics

Lesson 8: Electrostatics

7 concepts · 56 practice questions with answers · Free to read

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Gauss's law and symmetric charge distributions

From daily life

A charged balloon sets up an electric field around it; the number of field lines depends on the charge.

Simple definition

Electric flux Φ = ∮ E · dA. Gauss's law links the flux to the enclosed charge.

Gaussian surfaces : sphere, cylinder, pillboxΦ = q ÷ ε₀E = λ ÷ 2π ε₀ rE = σ ÷ 2ε₀

In the figure : A sphere, cylinder and pillbox — the Gaussian surface fits the symmetry.

Rule 1: Gauss's law — The total electric flux through a closed surface is Φ = q_enclosed ÷ ε₀ (ε₀ = 8.85 × 10⁻¹² C² N⁻¹m⁻²). Charges outside do not matter; neither does the shape of the surface.

  • Easy : q = 8.85 nC inside ⇒ Φ = 8.85 × 10⁻⁹ ÷ 8.85 × 10⁻¹² = 1000 N m²/C.
  • Medium : A surface with +3 µC and −3 µC inside ⇒ Φ = 0.
  • Exam (CET) : A charge q at the centre of a cube ⇒ flux through one face? (1) q/6ε₀ (2) q/ε₀ (3) q/3ε₀ (4) q/24ε₀ → answer 1). Trick : 6 equal faces.

Rule 2: Sphere, line and sheet — Spherical shell : outside E = kq ÷ r², inside 0. Infinite line : E = λ ÷ (2π ε₀ r). Infinite sheet : E = σ ÷ 2ε₀ (independent of distance).

  • Easy : λ = 2 µC/m, r = 0.1 m : E = 3.6 × 10⁵ N/C.
  • Medium : σ = 4 µC/m² : E = 2.26 × 10⁵ N/C.
  • Exam (CET) : Distance from a line doubled ⇒ E? (1) halves (2) quarter (3) doubles (4) same → answer 1). Trick : E ∝ 1/r.

Rule 3: Method of using Gauss's law — (1) Identify the symmetry, (2) choose a Gaussian surface on which E is constant and normal, (3) Φ = E × area, (4) find q_enclosed, (5) solve for E.

  • Easy : Sphere : area 4πr²; E 4πr² = q/ε₀.
  • Medium : Line : curved area of a cylinder 2πrl.
  • Exam (CET) : Best Gaussian surface? (1) one on which E is constant (2) any (3) a cube (4) a small one → answer 1)

Derivation : E of an infinite sheet = σ ÷ 2ε₀

  • Step 1 : By symmetry E is normal to the sheet and equal on both sides.
  • Step 2 : Take a pillbox Gaussian surface with end area A; no flux through the curved part.
  • Step 3 : Φ = E A + E A = 2 E A.
  • Step 4 : q_enclosed = σ A ⇒ 2 E A = σ A ÷ ε₀ ⇒ E = σ ÷ 2ε₀.
Remember

Φ = q/ε₀; E_line = λ/2πε₀r; E_sheet = σ/2ε₀; inside a shell 0.

Common mistake / CET trap

Φ depends on enclosed charge only; E depends on all charges. Sheet E has 2ε₀ (a conductor surface has σ/ε₀ — do not mix).

Shortcut

Memory clue : "sphere r², line r, sheet r⁰" (E ∝ 1/r², 1/r, constant).

Spot the difference

DistributionEUnit / dimensions
Point/sphere (outside)kq/r²N/C; [M L T⁻³ A⁻¹]
Infinite lineλ/(2πε₀r)N/C
Infinite sheetσ/2ε₀N/C
Inside a shell0N/C

Practice questions

Easy
  1. 1State Gauss's law.
  2. 2E inside a shell?
  3. 3Does the sheet's E depend on distance?
Medium
  1. 4q = 17.7 nC : Φ?
  2. 5σ = 8.85 µC/m² : E (sheet)?
  3. 6λ = 1 µC/m, r = 0.2 m : E?
Challenging
  1. 7Spherical shell Q, radius R : E at r > R?
  2. 8Flux through one face of a cube, q = 12 nC?
Show answers (try them yourself first)
  1. 1Φ = q_enclosed/ε₀
  2. 2Zero
  3. 3No
  4. 4Φ = 2000 N m²/C
  5. 5E = 5 × 10⁵ N/C
  6. 6E = 8.99 × 10⁴ N/C
  7. 7E = kQ/r²
  8. 8Φ = 226 N m²/C

Electric potential and potential energy

From daily life

Like the height of a hill, electric potential is a "height"; a positive charge slides from high to low potential.

Simple definition

Electric potential is the work needed to bring a unit positive charge from infinity.

Potential of a point charge V = kq/rV (q>0)V (q<0)VrE = −dV ÷ dr

In the figure : V is positive and falls with r for +q; negative for −q.

Rule 1: Potential — V = k q ÷ r (point charge), unit V = J/C. Many charges : algebraic sum. W = q (V_B − V_A); E = −dV/dr.

  • Easy : q = 2 nC, r = 0.3 m : V = 9 × 10⁹ × 2 × 10⁻⁹ ÷ 0.3 = 60 V.
  • Medium : Moving 1 µC through 50 V : W = 10⁻⁶ × 50 = 5 × 10⁻⁵ J.
  • Exam (CET) : Uniform E = 200 V/m; ΔV over 0.05 m? (1) 10 V (2) 4000 V (3) 0.25 V (4) 100 V → answer 1). Trick : ΔV = E d.

Rule 2: Potential energy — For two charges U = k q₁ q₂ ÷ r. Like signs ⇒ U positive (repulsion); unlike ⇒ negative. For three charges U = sum over pairs.

  • Easy : 2 µC and 3 µC, r = 0.5 m : U = 0.108 J.
  • Medium : +q and −q : U negative (a bound system).
  • Exam (CET) : r halved ⇒ U? (1) doubles (2) halves (3) four times (4) same → answer 1)

Rule 3: Potential difference and E — For a uniform E : V = E d. E is the negative gradient of V and points from high to low potential. 1 eV = 1.6 × 10⁻¹⁹ J.

  • Easy : E = 500 V/m, d = 4 cm ⇒ V = 20 V.
  • Medium : An electron accelerated through 20 V ⇒ KE = 20 eV = 3.2 × 10⁻¹⁸ J.
  • Exam (CET) : 1 eV = ? (1) 1.6 × 10⁻¹⁹ J (2) 1.6 J (3) 9 × 10⁹ J (4) 1 J → answer 1)

Derivation : V = kq ÷ r

  • Step 1 : E of a point charge = k q ÷ r²; V = work to bring +1 C from infinity to r.
  • Step 2 : V = −∫(∞→r) E dr = −∫ k q ÷ r² dr.
  • Step 3 : = −k q [−1 ÷ r] from ∞ to r = k q ÷ r.
  • Step 4 : V = k q ÷ r (V = 0 at infinity).
Remember

V = kq/r; U = kq₁q₂/r; W = qΔV; V = Ed; 1 eV = 1.6 × 10⁻¹⁹ J.

Common mistake / CET trap

V is a scalar (algebraic sum), E a vector. Keep signs in U; r is always positive.

Shortcut

Memory clue : "V is a scalar — add directly".

Spot the difference

QuantityFormulaUnit / dimensions
Vkq/rV; [M L² T⁻³ A⁻¹]
Ukq₁q₂/rJ
WqΔVJ
E−dV/drV/m

Practice questions

Easy
  1. 1Unit of potential?
  2. 2V at infinity?
  3. 3Relation of E and V?
Medium
  1. 4q = 5 nC, r = 0.9 m : V?
  2. 5E = 1000 V/m, d = 2 cm : V?
  3. 61 µC through 100 V : work?
Challenging
  1. 7+3 µC and −3 µC, r = 0.3 m : U?
  2. 8Three +1 µC charges at the corners of a 1 m triangle : U?
Show answers (try them yourself first)
  1. 1V (J/C)
  2. 2Zero
  3. 3E = −dV/dr
  4. 4V = 50 V
  5. 5V = 20 V
  6. 6W = 10⁻⁴ J
  7. 7U = -0.27 J
  8. 8U = 3 × 9 × 10⁹ × 10⁻¹² ÷ 1 = 0.027 J

Equipotential surfaces, charge systems and the dipole in a field

From daily life

Water molecules are dipoles, so they turn in an electric field.

Simple definition

An electric dipole is +q and −q separated by 2a; its dipole moment is p = q × 2a.

Electric dipole : axial and equatorial E+−pE_axialE_equatorialp = q × 2a

In the figure : On the axis E is along p (2kp/r³); on the equator it is opposite (kp/r³).

Live figure

Equipotential surfaces, charge systems and the dipole in a field

In the app you can move this figure with your finger.

Open in the app

Rule 1: Equipotential surfaces — Surfaces of constant V. E is always normal to them; moving a charge on one needs no work. For a point charge : concentric spheres; for uniform E : parallel planes.

  • Easy : Moving a charge 5 m along the surface : W = 0.
  • Medium : Two equipotential surfaces never intersect.
  • Exam (CET) : Angle between E and an equipotential surface? (1) 90° (2) 0° (3) 45° (4) 180° → answer 1)

Rule 2: Electric dipole — Two equal and opposite charges 2a apart : p = q (2a) (from −q to +q). On the axis E = 2kp ÷ r³ (along p); on the equator E = kp ÷ r³ (opposite to p).

  • Easy : q = 1 µC, 2a = 2 cm ⇒ p = 2 × 10⁻⁸ C m.
  • Medium : r = 0.5 m : E_axial = 2880 N/C; E_equatorial = 1440 N/C.
  • Exam (CET) : E_axial : E_equatorial (same r) = ? (1) 2 : 1 (2) 1 : 2 (3) 1 : 1 (4) 4 : 1 → answer 1)

Rule 3: Dipole in a field — In a uniform E : τ = p E sin θ, U = − p E cos θ. Stable equilibrium at θ = 0° (minimum U), unstable at 180°. The net force is zero.

  • Easy : p = 2 × 10⁻⁸, E = 10⁴ N/C, θ = 30° : τ = 1 × 10⁻⁴ N m.
  • Medium : θ = 0° ⇒ U = −p E; θ = 180° ⇒ +p E; the work between them is 2 p E.
  • Exam (CET) : When is τ maximum? (1) θ = 90° (2) 0° (3) 180° (4) 45° → answer 1)

Derivation : τ = p E sin θ

  • Step 1 : Each charge feels F = q E; they are equal and opposite ⇒ no net force but a couple.
  • Step 2 : Arm of the couple = 2a sin θ.
  • Step 3 : τ = F × arm = q E × 2a sin θ.
  • Step 4 : p = q 2a ⇒ τ = p E sin θ (vector : τ = p × E).
Remember

p = q·2a; E_axis = 2kp/r³; E_equator = kp/r³; τ = pE sinθ; U = −pE cosθ.

Common mistake / CET trap

p points from −q to +q. Axial E is twice the equatorial E. θ is the angle between p and E.

Shortcut

Memory clue : "axis double, equator single".

Spot the difference

LocationEDirection
Axial2kp/r³along p
Equatorialkp/r³opposite to p
τ in a fieldpE sin θ—
U in a field−pE cos θ—

Practice questions

Easy
  1. 1Unit of p?
  2. 2W on an equipotential surface?
  3. 3θ for stable equilibrium?
Medium
  1. 4q = 2 µC, 2a = 1 cm : p?
  2. 5p = 10⁻⁸, E = 10⁵, θ = 90° : τ?
  3. 6θ 0° → 90° : change in U (pE = 4 × 10⁻⁴ J)?
Challenging
  1. 7Axial E = 800 N/C ⇒ equatorial E (same r)?
  2. 8Why is the net force on a dipole zero in a uniform field?
Show answers (try them yourself first)
  1. 1C m
  2. 2Zero
  3. 30°
  4. 4p = 2 × 10⁻⁸ C m
  5. 5τ = 10⁻³ N m
  6. 6ΔU = +4 × 10⁻⁴ J
  7. 7400 N/C
  8. 8The forces on the two charges are equal and opposite

Conductors, electrostatic shielding and dielectrics

From daily life

A phone loses signal in a metal lift — metal blocks external fields.

Simple definition

A conductor has free charges; a dielectric has none.

Conductor cavity : E = 0 inside, E normal at surfaceE = 0E_ext

In the figure : E = 0 in the cavity; the outside E is normal to the surface.

Rule 1: Electrostatics in a conductor — In equilibrium, inside a conductor E = 0, the whole body is equipotential, charge sits only on the surface and E at the surface is normal; just outside E = σ ÷ ε₀.

  • Easy : σ = 2 µC/m² : E just outside = 2.26 × 10⁵ N/C.
  • Medium : Inside a conducting sphere V equals the surface value (constant).
  • Exam (CET) : E inside a conductor = ? (1) 0 (2) σ/ε₀ (3) σ/2ε₀ (4) infinite → answer 1)

Rule 2: Electrostatic shielding — The field inside a hollow conductor is unaffected by outside fields — so a car is safe in lightning and instruments are kept in metal boxes.

  • Easy : External E = 10⁵ N/C ⇒ E in the cavity = 0.
  • Medium : A charge in the cavity does produce a field outside; earthing the box removes it.
  • Exam (CET) : Why is a car safe in lightning? (1) shielding (2) rubber tyres (3) glass (4) speed → answer 1)

Rule 3: Dielectrics — A non-conductor polarises in a field and reduces it : E = E₀ ÷ K (K = dielectric constant, K > 1, unitless). For air K ≈ 1.

  • Easy : E₀ = 6000 N/C, K = 3 ⇒ E = 2000 N/C.
  • Medium : Force F = F₀/K (K = 80 in water ⇒ force 80 times smaller).
  • Exam (CET) : Unit of K? (1) none (2) N/C (3) F (4) C → answer 1)

Derivation : E near a conductor surface = σ ÷ ε₀

  • Step 1 : Small pillbox at the surface; flux only through the outer end (E = 0 inside).
  • Step 2 : Φ = E A.
  • Step 3 : q_enclosed = σ A.
  • Step 4 : E A = σ A ÷ ε₀ ⇒ E = σ ÷ ε₀.
Remember

Inside E = 0; outside σ/ε₀; E = E₀/K.

Common mistake / CET trap

Do not mix σ/ε₀ (conductor) with σ/2ε₀ (sheet); K > 1.

Shortcut

Memory clue : "conductor — zero inside".

Spot the difference

PropertyConductorDielectric
E inside0E₀/K
Chargesfreebound
Responsecharges flowmolecules polarise

Practice questions

Easy
  1. 1E in a conductor?
  2. 2What is K?
  3. 3An example of shielding?
Medium
  1. 4σ = 4 µC/m² : E outside?
  2. 5E₀ = 9000, K = 3 : E?
  3. 6Force in air 12 N; in a medium K = 4?
Challenging
  1. 7Water K = 80 : why does salt dissolve easily in water?
  2. 8V in the cavity of a hollow conductor?
Show answers (try them yourself first)
  1. 1Zero
  2. 2Dielectric constant
  3. 3A car / Faraday cage
  4. 4E = 4.52 × 10⁵ N/C
  5. 5E = 3000 N/C
  6. 6F = 3 N
  7. 7Force 80 times smaller ⇒ ions separate
  8. 8Same as the surface (constant)

Capacitors and the parallel plate capacitor

From daily life

A camera flash stores energy in a capacitor and releases it at once.

Simple definition

A capacitor is two conductors with a vacuum or dielectric between; capacitance C = Q ÷ V.

Parallel plate capacitor+Q−QdC = ε₀A ÷ d

In the figure : Uniform E between the plates; C = ε₀A/d.

Rule 1: Capacitance — C = Q ÷ V, unit farad (F = C/V), dimensions [M⁻¹L⁻²T⁴A²]. C depends on geometry and medium, not on Q or V.

  • Easy : Q = 20 µC, V = 10 V ⇒ C = 2 µF.
  • Medium : C = 5 µF, V = 12 V ⇒ Q = 60 µC.
  • Exam (CET) : 1 F = ? (1) 1 C/V (2) 1 V/C (3) 1 J/C (4) 1 N/C → answer 1)

Rule 2: Parallel plate capacitor — C = ε₀ A ÷ d (air); with a dielectric K : C = K ε₀ A ÷ d. Increase A or reduce d ⇒ C grows.

  • Easy : A = 0.02 m², d = 1 mm : C = 1.77 × 10⁻¹⁰ F = 0.177 nF.
  • Medium : Filled with K = 5 : C = 0.885 nF.
  • Exam (CET) : d doubled ⇒ C? (1) halves (2) doubles (3) four times (4) same → answer 1)

Rule 3: Battery connected / removed — Battery connected : V fixed; Q = CV. Battery removed : Q fixed; inserting K gives V = V₀/K, E = E₀/K, C = K C₀.

  • Easy : Battery connected, K = 4 : Q becomes four times.
  • Medium : Battery removed, V₀ = 100 V, K = 4 ⇒ V = 25 V.
  • Exam (CET) : Battery removed, K inserted; energy? (1) falls K times (2) rises K times (3) same (4) zero → answer 1)

Derivation : C = ε₀ A ÷ d

  • Step 1 : Plate charge density σ = Q ÷ A; between plates E = σ ÷ ε₀ = Q ÷ ε₀A.
  • Step 2 : V = E d = Q d ÷ ε₀ A.
  • Step 3 : C = Q ÷ V ⇒ C = ε₀ A ÷ d.
Remember

C = Q/V; C = ε₀A/d; K times with a dielectric.

Common mistake / CET trap

C does not depend on Q or V. Battery-connected and isolated cases differ.

Shortcut

Memory clue : "connected — V fixed, removed — Q fixed".

Spot the difference

QuantityFormulaUnit / dimensions
CQ/VF; [M⁻¹L⁻²T⁴A²]
Plate Cε₀A/dF
With dielectricKε₀A/dF
ε₀—C²N⁻¹m⁻²

Practice questions

Easy
  1. 1Unit of C?
  2. 2What does C depend on?
  3. 3Where does K appear?
Medium
  1. 4Q = 40 µC, V = 8 V : C?
  2. 5C = 3 µF, V = 20 V : Q?
  3. 6d halved : C?
Challenging
  1. 7A = 0.04 m², d = 2 mm : C?
  2. 8Battery removed, V₀ = 60 V, K = 3 : V?
Show answers (try them yourself first)
  1. 1Farad
  2. 2Geometry and medium
  3. 3K C₀
  4. 45 µF
  5. 560 µC
  6. 6Doubles
  7. 7C = 1.77 × 10⁻¹⁰ F
  8. 8V = 20 V

Capacitor combinations and stored energy

From daily life

To get a large capacitance, small capacitors are connected in parallel.

Simple definition

Series = one after another; parallel = both ends shared.

Series and parallel capacitors1/C = 1/C₁ + 1/C₂C = C₁ + C₂ + …

In the figure : Series : sum of reciprocals; parallel : direct sum.

Rule 1: Series — 1/C = 1/C₁ + 1/C₂ + …; same Q on each, V divides. For two : C = C₁C₂ ÷ (C₁ + C₂).

  • Easy : 2 µF and 3 µF in series : C = 1.2 µF.
  • Medium : Three equal 6 µF in series ⇒ 2 µF.
  • Exam (CET) : In series C is always? (1) less than the smallest (2) more (3) the sum (4) same → answer 1)

Rule 2: Parallel — C = C₁ + C₂ + …; same V, Q divides.

  • Easy : 2 µF and 3 µF in parallel : C = 5 µF.
  • Medium : V = 10 V : Q₁ = 20 µC, Q₂ = 30 µC.
  • Exam (CET) : In parallel C? (1) the sum (2) smaller (3) product (4) zero → answer 1)

Rule 3: Stored energy — U = ½ C V² = ½ Q V = Q² ÷ 2C. Energy density u = ½ ε₀ E².

  • Easy : C = 2 µF, V = 100 V ⇒ U = 1 × 10⁻² J.
  • Medium : V doubled ⇒ U four times.
  • Exam (CET) : U = ? (1) ½CV² (2) CV² (3) CV (4) ½CV → answer 1)

Derivation : U = ½ C V²

  • Step 1 : Work to add charge dq : dW = V dq = (q ÷ C) dq.
  • Step 2 : U = ∫ from 0 to Q (q ÷ C) dq.
  • Step 3 : = Q² ÷ 2C.
  • Step 4 : Q = CV ⇒ U = ½ C V².
Remember

Series 1/C = Σ1/Cᵢ; parallel ΣCᵢ; U = ½CV²; u = ½ε₀E².

Common mistake / CET trap

Series/parallel formulae are opposite to those for resistors.

Shortcut

Memory clue : "capacitors are reversed : parallel adds".

Spot the difference

ConnectionCV / Q
Series1/C = Σ1/Cᵢsame Q
ParallelC = ΣCᵢsame V
Energy½CV²J

Practice questions

Easy
  1. 1Q in series?
  2. 2V in parallel?
  3. 3Formula for U?
Medium
  1. 44 µF and 12 µF in series : C?
  2. 53 µF, 5 µF, 2 µF in parallel : C?
  3. 6C = 10 µF, V = 50 V : U?
Challenging
  1. 76 µF and 3 µF in series on 90 V : V across each?
  2. 8V doubled, same C : ratio of U?
Show answers (try them yourself first)
  1. 1Same
  2. 2Same
  3. 3½CV²
  4. 4C = 3 µF
  5. 5C = 10 µF
  6. 6U = 1.25 × 10⁻² J
  7. 730 V on 6 µF, 60 V on 3 µF
  8. 84 : 1

Displacement current and the Van de Graaff generator

From daily life

Combing hair shows static charge; a Van de Graaff is its large version.

Simple definition

Displacement current is the current due to a changing electric field.

Van de Graaff generator+ + +belt

In the figure : The belt carries charge into the sphere; it collects on the outer surface.

Rule 1: Displacement current — Current due to a changing electric flux : I_d = ε₀ dΦ_E/dt. While a capacitor charges, I_d between the plates equals the conduction current I_c.

  • Easy : I_c = 2 A ⇒ I_d between the plates = 2 A.
  • Medium : E constant ⇒ I_d = 0.
  • Exam (CET) : Cause of I_d? (1) changing Φ_E (2) free charge (3) battery (4) air → answer 1)

Rule 2: Ampere–Maxwell law — ∮ B · dl = μ₀ (I_c + I_d). I_d keeps current continuous and makes electromagnetic waves possible.

  • Easy : I_c in the wire and I_d between the plates — the total is continuous.
  • Medium : In a vacuum I_c = 0, only I_d.
  • Exam (CET) : Credit for I_d? (1) Maxwell (2) Faraday (3) Newton (4) Gauss → answer 1)

Rule 3: Van de Graaff generator — A belt carries charge into a hollow sphere; it collects on the outer surface. Maximum V = E_max × r (air breaks down near 3 × 10⁶ V/m).

  • Easy : r = 0.5 m ⇒ V = 1.5 × 10⁶ V.
  • Medium : r doubled ⇒ maximum V doubles.
  • Exam (CET) : E inside the sphere? (1) 0 (2) maximum (3) σ/ε₀ (4) infinite → answer 1)

Derivation : I_d = I_c between the plates

  • Step 1 : Plate charge q; E = q ÷ ε₀ A; Φ_E = q ÷ ε₀.
  • Step 2 : dΦ_E/dt = (1 ÷ ε₀) dq/dt.
  • Step 3 : I_d = ε₀ dΦ_E/dt = dq/dt = I_c.
  • Step 4 : Hence I_d = I_c.
Remember

I_d = ε₀ dΦ_E/dt; I_d = I_c; V_max = E_max r.

Common mistake / CET trap

I_d is not a flow of real charges; it comes from a changing field.

Shortcut

Memory clue : "changing field = current".

Spot the difference

CurrentSourceWhere
I_cfree chargesconductors
I_dchanging Φ_Evacuum/dielectric

Practice questions

Easy
  1. 1Formula for I_d?
  2. 2What did Maxwell add?
  3. 3What does a Van de Graaff produce?
Medium
  1. 4r = 0.2 m : maximum V?
  2. 5I_c = 3 A while charging : I_d?
  3. 6I_d when E is constant?
Challenging
  1. 7Why is V constant inside the sphere?
  2. 8Importance of I_d in a vacuum?
Show answers (try them yourself first)
  1. 1ε₀ dΦ_E/dt
  2. 2Displacement current
  3. 3High potential
  4. 4V = 6 × 10⁵ V
  5. 53 A
  6. 6Zero
  7. 7E = 0 ⇒ V constant
  8. 8For electromagnetic waves

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
GaussΦ = q/ε₀N m²/C
PotentialV = kq/r; W = qΔVV; [ML²T⁻³A⁻¹]
Dipolep = q2a; τ = pE sinθC m; N m
ConductorE inside = 0; outside σ/ε₀N/C
CapacitorC = Q/V = ε₀A/dF; [M⁻¹L⁻²T⁴A²]
Combinationseries 1/C; parallel ΣCF
EnergyU = ½CV²; u = ½ε₀E²J; J/m³
I_dε₀ dΦ/dtA

🎯 Most important points for the exam

  • Φ = q/ε₀; sheet σ/2ε₀, line λ/2πε₀r.
  • V = kq/r scalar; E = −dV/dr.
  • Dipole : axial 2kp/r³, equatorial kp/r³.
  • E = 0 in a conductor; shielding.
  • C = ε₀A/d; K times.
  • Series 1/C; parallel adds.
  • U = ½CV²; battery on — V fixed, off — Q fixed.
Study mantra

Spot the symmetry → choose the surface → formula → convert µ, n, p to SI → answer.