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L-15: Electric Charge and Electric Field

Physics — Class 12 · NIOS Code 312 · Module 5 · Source: 312_Physics_Eng_Lesson15.pdf

Electric Charge and Electric Field — Electrostatics

Electrostatics is the branch of physics dealing with electric charges at rest — their properties, forces, fields, and the behaviour of surrounding space. Electrical energy powers nearly every modern device; this lesson lays the foundation for electricity and magnetism.

NIOS objectives: properties of charge; quantisation and conservation; Coulomb's law; electric field and field lines; dipole and dipole moment; Gauss's theorem; fields of point charge, line, shell, and plane sheet; Van de Graaff generator.

15.1 Frictional Electricity

Amber rubbed with fur attracted light objects — the word electric comes from Greek electron (amber). Rubbing a hard rubber rod with fur or a glass rod with silk produces charges that interact at a distance.

  • Like charges repel; unlike charges attract.
  • Franklin's convention: glass → positive; rubber → negative.
  • Charging by conduction (touching) or induction (near + earth).
Fig 15.1 — Like Charges Repel; Unlike Attract rubber rod + + repulsion (a) like charges repel glass rod + attraction (b) unlike charges attract
Fig 15.1 — Charged rods show repulsion (like) or attraction (unlike)

15.1.1 Conservation of Charge

When glass is rubbed with silk, the rod gains +q and silk gains −q of equal magnitude. Total charge of the isolated system is unchanged — charge is neither created nor destroyed, only transferred. Electrons move from glass to silk (or from fur to rubber).

15.1.2 Quantisation of Charge

Millikan (1909) showed charge always appears as an integral multiple of the fundamental charge e on an electron:

Q = N e
Q = total charge on body (C)
N = integer (1, 2, 3, …)
e = 1.6×10⁻¹⁹ C (electron charge)
A body cannot have 2.5e or 6.4e — charge is quantised.
  • Only two kinds of charge exist: positive and negative.
  • Charge is conserved and quantised.
  • Atom is neutral: equal electrons (−e) and protons (+e); neutron has no charge.

15.2 Coulomb's Law

Force between two stationary point charges q₁ and q₂ separated by distance r:

  • ∝ product q₁q₂; ∝ 1/r² (inverse-square law, like gravitation).
  • Along the line joining charges; repulsive for like, attractive for unlike.
  • Valid for point charges only; acts at a distance.
F = k q₁q₂ / r²
F = magnitude of force (N)
q₁, q₂ = charges (C)
r = separation (m)
Vector form: F₁₂ = k (q₁q₂/r²) r̂₁₂
F = (1/4πε₀) × q₁q₂ / r²
In vacuum/free space: k = 1/(4πε₀) = 9×10⁹ N·m²·C⁻²
ε₀ = 8.85×10⁻¹² C²·N⁻¹·m⁻² (permittivity of free space)
In medium: k = 1/(4πε); F_medium = F_vacuum × (ε₀/ε)

One coulomb: two 1 C charges 1 m apart experience ~10¹⁰ N — enormous compared to everyday forces. Action-reaction: F₁₂ = −F₂₁.

15.2.2 Principle of Superposition

For multiple charges, net force on q₁ is the vector sum of individual Coulomb forces:

F = F₁₂ + F₁₃ + F₁₄ + …
Calculate each pairwise force using Coulomb's law, then add vectorially.
Same principle applies to electric fields.

Example 15.2: Two 6.0×10⁻¹⁰ C charges 2.0 m apart → F = 81×10⁻¹¹ N.

15.3 Electric Field

Faraday introduced the electric field to explain action at a distance. Field at a point is force per unit positive test charge:

E = F / q₀ = k q / r² r̂
E = electric field (N·C⁻¹) — vector
q₀ = infinitesimal test charge (must not disturb source)
+q → field radially outward; −q → radially inward
Force on charge q in field: F = qE

Superposition: E = E₁ + E₂ + E₃ + … = Σ k qᵢ/rᵢ² r̂ᵢ. At centroid of equilateral triangle with equal +q at corners, E = 0 by symmetry.

15.3.1 Electric Dipole

Two equal and opposite charges ±q separated by small distance 2l form a dipole (e.g. H₂O).

p = q × 2l
p = dipole moment (C·m) — vector
Direction: from −q to +q along the dipole axis
|p| = 2ql
E_end-on = 2p / (4πε₀ r³)
End-on (axial) position — point on dipole axis, r ≫ l
Field parallel to p; magnitude twice the broad-on value
∝ 1/r³
E_broad-on = p / (4πε₀ r³)
Broad-on (equatorial) position — on perpendicular bisector, r ≫ l
Field antiparallel to p on one side
∝ 1/r³

15.3.2 Dipole in Uniform Field

In uniform E, forces on ±q form a couple tending to align the dipole with E:

τ = p E sin θ  |  τ = p × E
τ = torque (N·m)
θ = angle between p and E
τ = 0 when θ = 0 (aligned); maximum τ = pE at θ = 90°
In non-uniform field: net force ≠ 0 also.
Fig 15.13 — Dipole in Uniform Electric Field E → +q −q 2l qE qE θ couple τ = pE sin θ tends to align dipole with E
Fig 15.13 — Equal and opposite forces on dipole create aligning torque

15.3.3 Electric Lines of Force (Field Lines)

Fictitious lines depicting field direction and strength. Tangent at any point gives E direction; density ∝ field strength.

  • Start from +q radially outward to infinity; end on −q from infinity.
  • Dipole: lines from +q terminate on −q.
  • Two field lines never cross.
  • Two equal +q: field zero at midpoint P.
Fig 15.14–15.15 — Electric Field Lines +q +q: radial outward −q −q: inward radial + dipole: + to −
Fig 15.14–15.15 — Field lines for point charges and a dipole

15.4 Electric Flux and Gauss's Law

Electric flux through small area element: Δφ = E · Δs. For a closed Gaussian surface enclosing charge q:

Φ_E = q / ε₀
Φ_E = net electric flux through closed surface (N·m²·C⁻¹)
q = total enclosed charge
Gaussian surface is imaginary — need not be a real physical surface

15.4.1 Point Charge

E = q / (4πε₀ r²)
Spherical Gaussian surface of radius r centred on +q
Recovers Coulomb's law: F = q₀E
Fig 15.17 — Gaussian Sphere Around Point Charge +q E Δs Gaussian surface Φ = E·4πr² = q/ε₀ → E = q/(4πε₀r²)
Fig 15.17 — Spherical Gaussian surface for a point charge at centre

15.4.2 Long Line Charge

Uniform linear charge density λ (C·m⁻¹). Cylindrical Gaussian surface of radius r, length l:

E = λ / (2πε₀ r)
λ = charge per unit length
Field ∝ 1/r (not 1/r²)
Flat end caps contribute zero flux (E ⊥ Δs)

15.4.3 Uniformly Charged Spherical Shell

Outside (r > R): E = Q / (4πε₀ r²)
Entire charge Q acts as if at centre — same for hollow shell or solid conducting sphere (charge on outer surface)
Inside (r < R): E = 0
Enclosed charge Q_enclosed = 0 inside hollow shell
Field jumps at r = R then falls as 1/r² outside

15.4.4 Infinite Plane Sheet of Charge

Uniform surface charge density σ (C·m⁻²). Cylindrical Gaussian surface piercing the sheet:

E = σ / (2ε₀)
Field perpendicular to sheet; same magnitude on both sides
Independent of distance from the sheet
Factor 2: flux through both circular caps
         GAUSS'S LAW — FIELD SUMMARY
         ============================
    Point charge     :  E = q / (4πε₀ r²)
    Infinite line    :  E = λ / (2πε₀ r)
    Shell outside    :  E = Q / (4πε₀ r²)
    Shell inside     :  E = 0
    Plane sheet      :  E = σ / (2ε₀)

15.5 Van de Graaff Generator

Electrostatic machine producing potentials of a few million volts (up to ~20 MV). Named after Robert J. van de Graaff.

  • Large hollow metallic sphere S on insulating stand; rubber/silk belt on pulleys P₁ (motor-driven) and P₂ (at sphere centre).
  • Comb C₁ at ~10⁴ V positive; comb C₂ connected to inner surface of sphere.
  • Corona discharge near sharp needle points ionises air; belt carries + charge to C₂.
  • Charge accumulates on outer surface of S; earthed chamber T at high pressure reduces leakage.
  • Used to accelerate ion beams for nuclear reaction studies.

Quick Revision

  • Q = Ne; charge conserved and quantised; e = 1.6×10⁻¹⁹ C.
  • Coulomb: F = kq₁q₂/r²; k = 9×10⁹ N·m²·C⁻²; superposition for many charges.
  • Field: E = F/q₀ = kq/r²; F = qE.
  • Dipole: p = 2ql; end-on E = 2p/(4πε₀r³); broad-on E = p/(4πε₀r³); τ = pE sin θ.
  • Gauss: Φ = q/ε₀; shell inside E = 0; sheet E = σ/(2ε₀).
  • Van de Graaff: corona discharge + belt → megavolt sphere.
20 cards · click any card to flip
Two kinds of charge
Only positive and negative exist. Franklin: glass rubbed with silk → positive; rubber with fur → negative.
Like vs unlike charges
Like charges repel; unlike charges attract each other.
Conservation of charge
Total charge of an isolated system is constant. Rubbing transfers electrons — charge is neither created nor destroyed.
Quantisation of charge
Q = Ne where N is an integer and e = 1.6×10⁻¹⁹ C. No fractional multiples of e exist on a body.
Coulomb's law
F = kq₁q₂/r² along line joining charges. k = 9×10⁹ N·m²·C⁻² = 1/(4πε₀). Inverse-square law.
ε₀ (permittivity)
ε₀ = 8.85×10⁻¹² C²·N⁻¹·m⁻². In a medium, force reduces by factor ε₀/ε; εᵣ = ε/ε₀ > 1.
Superposition (forces)
Net force on a charge = vector sum of Coulomb forces from all other charges: F = F₁₂ + F₁₃ + …
Electric field definition
E = F/q₀ (N·C⁻¹). Force per unit positive test charge. Test charge must be vanishingly small.
Field of point charge
E = kq/r² r̂. +q → outward; −q → inward. F on charge q in field: F = qE.
Electric dipole moment
p = q × 2l (C·m). Vector from −q to +q along dipole axis. Example: water molecule.
Dipole field — end-on
On axis (r ≫ l): E = 2p/(4πε₀r³). Twice the broad-on value; parallel to p.
Dipole field — broad-on
On equatorial plane (r ≫ l): E = p/(4πε₀r³). Antiparallel to p on one side.
Torque on dipole
τ = pE sin θ = p × E. Aligns dipole with uniform E. Max at θ = 90°.
Electric field lines
Tangent gives E direction. Density ∝ field strength. Never cross. +q: outward; −q: inward; dipole: + to −.
Gauss's law
Φ_E = q_enclosed/ε₀. Net flux through any closed Gaussian surface equals enclosed charge divided by ε₀.
Line charge field
E = λ/(2πε₀r). Uniform linear density λ. Field falls as 1/r (cylindrical symmetry).
Spherical shell field
Outside: E = Q/(4πε₀r²) as point charge at centre. Inside hollow shell: E = 0.
Plane sheet field
E = σ/(2ε₀). Perpendicular to sheet; magnitude independent of distance from sheet.
Charging methods
Friction (rubbing), conduction (direct contact), and induction (near charged body + earthing).
Van de Graaff generator
Belt + corona discharge at combs transfers charge to hollow sphere. Potentials up to ~20 MV for ion acceleration.

Q1. The quantisation of charge is expressed as:

Q2. The value of Coulomb constant k in SI units is:

Q3. Coulomb force between two point charges varies with distance r as:

Q4. Electric field E at a point is defined as:

Q5. The dipole moment p has direction:

Q6. Electric field on the axis of a short dipole (end-on, r ≫ l) is:

Q7. Torque on a dipole in uniform electric field E is:

Q8. Gauss's law states that net flux through a closed surface equals:

Q9. Electric field inside a uniformly charged hollow spherical shell is:

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

Q = N e
F = k q₁q₂/r²
F = q₁q₂/(4πε₀r²)
F_net = F₁₂ + F₁₃ + …
E = F/q₀ = kq/r² r̂
F = qE
p = q(2l) = 2ql
E_axial = 2p/(4πε₀r³)
E_eq = p/(4πε₀r³)
τ = pE sin θ
Φ_E = q_enclosed/ε₀
E = q/(4πε₀r²)
E = λ/(2πε₀r)
Shell: E = Q/(4πε₀r²) (r > R)
Shell: E = 0 (r < R)
E = σ/(2ε₀)

1. Formulas & Definitions

Full Ch 15 study guide — charge, Coulomb, electric field, dipole, and Gauss's law.

Q = N e

Definition: Total charge on a body is an integer multiple of elementary charge e.

Derivation

Millikan oil-drop experiment (1909) — charge quantisation.

Variables

Q (C) · N = 1, 2, 3… · e = 1.6×10⁻¹⁹ C

Why it works

You cannot have fractional e — explains why rubbed objects gain discrete amounts of charge.

Historical context

Robert Millikan measured e precisely; Nobel Prize 1923.

Deep understanding

Conservation: total Q of isolated system constant. Atom neutral: N_e⁻ = N_p⁺.

2. Diagrams & Visuals

Q = N × 1.6×10⁻¹⁹ C

Color-coded visual · step-by-step breakdown below

  1. Count excess/deficit electrons N
  2. Q = Ne (sign from electron/proton excess)
  3. Check N is integer

3. Solved Examples

Basic

Q: Body loses 5 electrons.

Solution: Q = +5e

Answer: 8×10⁻¹⁹ C

Intermediate

Q: Q = −3.2×10⁻¹⁹ C.

Solution: N = 2

Answer: 2 electrons gained

Advanced

Q: Q = 4.8×10⁻¹⁹ C possible?

Solution: N=3, yes

Answer: Integer multiple

Exam

Q: Quantisation means?

Solution: Q = Ne only

Answer: Sec 15.1.2

F = k q₁q₂/r²

Definition: Coulomb force between two stationary point charges.

Derivation

Inverse-square law; k = 1/(4πε₀). Vector: F₁₂ = k(q₁q₂/r²)r̂₁₂.

Variables

F (N) · q₁,q₂ (C) · r (m) · k = 9×10⁹ N·m²·C⁻²

Why it works

Electrostatic analogue of gravity — like charges repel, unlike attract.

Historical context

Charles-Augustin de Coulomb (1785) with torsion balance.

Deep understanding

Point charges only. F₁₂ = −F₂₁ (third law). 1 C at 1 m → ~10¹⁰ N!

2. Diagrams & Visuals

F ∝ q₁q₂/r²

Color-coded visual · step-by-step breakdown below

  1. Convert to SI: C, m
  2. F = k|q₁q₂|/r² magnitude
  3. Direction: repel if same sign
  4. Vector sum if multiple charges

3. Solved Examples

Basic

Q: q₁=q₂=1 μC, r=0.1 m.

Solution: F=9×10⁻¹ N

Answer: 0.9 N repulsion

Intermediate

Q: Ex 15.2: 6×10⁻¹⁰ C, r=2 m.

Solution: F=81×10⁻¹¹ N

Answer: 8.1×10⁻¹⁰ N

Advanced

Q: Double r?

Solution: F quarters

Answer: F ∝ 1/r²

Exam

Q: Coulomb valid for?

Solution: Point charges at rest

Answer: Sec 15.2

F = q₁q₂/(4πε₀r²)

Definition: Coulomb law in terms of permittivity of free space ε₀.

Derivation

k = 1/(4πε₀); ε₀ = 8.85×10⁻¹² C²·N⁻¹·m⁻².

Variables

ε₀ = permittivity of vacuum

Why it works

Standard form in Gauss's law and field derivations.

Historical context

ε₀ links electrostatic force to field formulation.

Deep understanding

In medium: F_med = F_vac × (ε₀/ε). Dielectric reduces force.

2. Diagrams & Visuals

k = 1/(4πε₀) = 9×10⁹ ε₀ = 8.85×10⁻¹²

Color-coded visual · step-by-step breakdown below

  1. Use ε₀ = 8.85×10⁻¹²
  2. F = q₁q₂/(4πε₀r²)
  3. Same numeric result as k form

3. Solved Examples

Basic

Q: Value of k?

Solution: 1/(4πε₀)

Answer: 9×10⁹ N·m²/C²

Intermediate

Q: ε₀ units?

Solution: C²·N⁻¹·m⁻²

Answer: Farad per metre scale

Advanced

Q: In water ε≈80ε₀?

Solution: F reduced ~80×

Answer: ε₀/ε factor

Exam

Q: SI value k?

Solution: 9×10⁹

Answer: Memorise

F_net = F₁₂ + F₁₃ + …

Definition: Principle of superposition for forces on a charge.

Derivation

Vector sum of pairwise Coulomb forces from all other charges.

Variables

Each F_ij from Coulomb law

Why it works

Real systems have many charges — add forces, not scalars blindly.

Historical context

Same superposition as waves and fields — linearity of electrostatics.

Deep understanding

Resolve into x, y (or 3D) components before adding. Equilateral triangle equal +q at centre: F=0.

2. Diagrams & Visuals

ΣF

Color-coded visual · step-by-step breakdown below

  1. Draw all charges
  2. Find F from each other charge on target
  3. Add vectors component-wise
  4. Magnitude and direction of net F

3. Solved Examples

Basic

Q: Two equal F at 180°.

Solution: Cancel if equal

Answer: F_net=0

Intermediate

Q: Two equal F at 90°.

Solution: F_net=√2 F

Answer: Vector add

Advanced

Q: Equilateral +q corners, centre?

Solution: Symmetry → F=0

Answer: Sec 15.3

Exam

Q: Superposition applies to?

Solution: F and E

Answer: Sec 15.2.2

E = F/q₀ = kq/r² r̂

Definition: Electric field — force per unit positive test charge.

Derivation

E defined by source charge q; F = q₀E on test charge q₀.

Variables

E (N·C⁻¹) · q₀ must not disturb source

Why it works

Field explains action at a distance — charge alters space around it.

Historical context

Michael Faraday introduced field concept (~1830s).

Deep understanding

+q → E outward; −q → E inward. E = Σ k qᵢ/rᵢ² r̂ᵢ for many charges.

2. Diagrams & Visuals

E radial out

Color-coded visual · step-by-step breakdown below

  1. Find F on small +q₀
  2. E = F/q₀
  3. Or E = kq/r² along r̂ from q
  4. Direction: + test charge would move along E

3. Solved Examples

Basic

Q: F=0.5 N on 0.1 C test.

Solution: E=5 N/C

Answer: 5 N/C

Intermediate

Q: q=2 μC, r=0.2 m.

Solution: E=kq/r²

Answer: 4.5×10⁵ N/C

Advanced

Q: Negative source q?

Solution: E points inward

Answer: Toward charge

Exam

Q: E units?

Solution: N·C⁻¹

Answer: Same as V/m later

F = qE

Definition: Force on charge q placed in electric field E.

Derivation

Definition of E rearranged; valid for test charge in any field.

Variables

F (N) · q (C) · E (N/C)

Why it works

Once E is known, find force on any charge without recomputing from source.

Historical context

Bridges field concept to mechanics problems.

Deep understanding

F parallel to E for +q; opposite for −q. Dipole in uniform E: torque, not net force.

2. Diagrams & Visuals

E → F=qE

Color-coded visual · step-by-step breakdown below

  1. Field E at location
  2. Charge q (+ or −)
  3. F = qE (vector)
  4. Direction from sign of q

3. Solved Examples

Basic

Q: E=10³ N/C, q=2 mC.

Solution: F=2 N

Answer: Along E

Intermediate

Q: q=−5 μC, E east.

Solution: F west

Answer: Opposite to E

Advanced

Q: E=0 at midpoint two +q?

Solution: F=0 on test charge

Answer: Symmetry

Exam

Q: Relation F and E?

Solution: F = qE

Answer: Sec 15.3

p = q(2l) = 2ql

Definition: Electric dipole moment — two equal opposite charges ±q separated by 2l.

Derivation

p vector from −q to +q; magnitude p = 2ql.

Variables

p (C·m) · l = half-separation

Why it works

Models polar molecules (H₂O); key to dipole field and torque.

Historical context

Dipole approximation when r ≫ l for distant field calculation.

Deep understanding

Direction: −q → +q. Units C·m.

2. Diagrams & Visuals

2l p →

Color-coded visual · step-by-step breakdown below

  1. Identify q and separation 2l
  2. p = 2ql
  3. Direction − to +
  4. Use for field and torque formulas

3. Solved Examples

Basic

Q: q=2 nC, l=1 mm.

Solution: p=4×10⁻¹²

Answer: 4 nC·m

Intermediate

Q: p=3×10⁻¹² C·m, q=1 nC.

Solution: 2l=3 mm

Answer: l=1.5 mm

Advanced

Q: Zero dipole moment?

Solution: No separation or q=0

Answer: Monopole only

Exam

Q: p direction?

Solution: −q to +q

Answer: Sec 15.3.1

E_axial = 2p/(4πε₀r³)

Definition: Electric field on dipole axis (end-on), r ≫ l.

Derivation

Superposition of ±q fields; leading term ∝ p/r³.

Variables

r = distance from dipole centre along axis

Why it works

Axial field twice equatorial — common exam comparison.

Historical context

Far-field dipole approximation (r ≫ 2l).

Deep understanding

Falls as 1/r³ (not 1/r²). Parallel to p on axis.

2. Diagrams & Visuals

E_end-on ∝ 2p/r³

Color-coded visual · step-by-step breakdown below

  1. Confirm on-axis point, r ≫ l
  2. E = 2p/(4πε₀r³)
  3. Direction parallel to p
  4. Compare with broad-on

3. Solved Examples

Basic

Q: Same p, double r.

Solution: E÷8

Answer: 1/r³ dependence

Advanced

Q: Axial vs equatorial same r?

Solution: Axial = 2× equatorial

Answer: Factor 2

Exam

Q: Dipole field far away?

Solution: ∝ 1/r³

Answer: Not 1/r²

Intermediate

Q: p=10⁻¹², r=0.1 m.

Solution: Use formula

Answer: Compute E magnitude

E_eq = p/(4πε₀r³)

Definition: Electric field on dipole equatorial plane (broad-on), r ≫ l.

Derivation

Field antiparallel to p on one side of bisector.

Variables

r = distance from centre on perpendicular bisector

Why it works

Half the end-on value at same r.

Historical context

Equatorial = broad-on position in NIOS text.

Deep understanding

On equatorial line, E ⊥ dipole axis but vector antiparallel to p.

2. Diagrams & Visuals

E_eq ∝ p/r³

Color-coded visual · step-by-step breakdown below

  1. Point on perpendicular bisector
  2. r ≫ l
  3. E = p/(4πε₀r³)
  4. Half of axial value

3. Solved Examples

Basic

Q: E_axial = 8 N/C same dipole, r.

Solution: E_eq=4 N/C

Answer: Half

Intermediate

Q: p triples?

Solution: E triples

Answer: Linear in p

Advanced

Q: At centre r=0?

Solution: Formula invalid

Answer: Need r ≫ l

Exam

Q: Broad-on field?

Solution: p/(4πε₀r³)

Answer: Sec 15.3.1

τ = pE sin θ

Definition: Torque on dipole in uniform electric field E.

Derivation

Couple from forces qE on ±q; τ = p×E magnitude pE sinθ.

Variables

τ (N·m) · θ = angle between p and E

Why it works

Dipole aligns with field — basis of polar molecule behaviour.

Historical context

Fig 15.13 — maximum torque at θ=90°.

Deep understanding

τ=0 when aligned (θ=0). Non-uniform E also exerts net force on dipole.

2. Diagrams & Visuals

θ τ = pE sinθ

Color-coded visual · step-by-step breakdown below

  1. Angle θ between p and E
  2. τ = pE sinθ
  3. Max at θ=90°
  4. Zero when parallel

3. Solved Examples

Basic

Q: p=10⁻¹², E=10⁶, θ=90°.

Solution: τ=10⁻⁶ N·m

Answer: Maximum

Intermediate

Q: θ=30°.

Solution: τ=½ pE

Answer: Half of max

Advanced

Q: θ=0?

Solution: τ=0

Answer: Stable alignment

Exam

Q: Torque zero when?

Solution: θ=0 or 180°

Answer: Sec 15.3.2

Φ_E = q_enclosed/ε₀

Definition: Gauss's law — net flux through closed surface equals enclosed charge over ε₀.

Derivation

∮ E·dA = q_enclosed/ε₀ for any closed Gaussian surface.

Variables

Φ_E (N·m²·C⁻¹) · q = net enclosed charge

Why it works

Powerful shortcut for symmetric charge distributions.

Historical context

Carl Friedrich Gauss — cornerstone of electrostatics.

Deep understanding

Gaussian surface is imaginary. Enclosed q — not outside charges.

2. Diagrams & Visuals

Φ = q/ε₀

Color-coded visual · step-by-step breakdown below

  1. Choose symmetric Gaussian surface
  2. Find q inside only
  3. Φ = q/ε₀
  4. Relate Φ to E for E solution

3. Solved Examples

Basic

Q: q=ε₀ C enclosed.

Solution: Φ=1

Answer: 1 N·m²/C

Intermediate

Q: Charge outside surface?

Solution: Not in q_enclosed

Answer: Flux from outside can be zero net

Advanced

Q: Why Gaussian sphere for point q?

Solution: E constant on sphere

Answer: Easy Φ integral

Exam

Q: Gauss law needs?

Solution: Closed surface

Answer: Sec 15.4

E = q/(4πε₀r²)

Definition: Electric field of isolated point charge (Gauss / Coulomb).

Derivation

Spherical Gaussian surface: E·4πr² = q/ε₀.

Variables

q (C) · r (m)

Why it works

Fundamental field — building block via superposition.

Historical context

Recovers F = q₀E = Coulomb force.

Deep understanding

+q outward, −q inward. 1/r² fall-off.

2. Diagrams & Visuals

E∝1/r²

Color-coded visual · step-by-step breakdown below

  1. Spherical symmetry
  2. Gauss: E(4πr²)=q/ε₀
  3. E = q/(4πε₀r²)
  4. Direction radial

3. Solved Examples

Basic

Q: q=1 μC, r=0.3 m.

Solution: E≈10⁵ N/C

Answer: Outward if +q

Intermediate

Q: Double r?

Solution: E÷4

Answer: Inverse square

Advanced

Q: Two +q field at midpoint?

Solution: E=0 by symmetry

Answer: Fig 15.15 idea

Exam

Q: Point charge field?

Solution: q/(4πε₀r²)

Answer: Gauss case 1

E = λ/(2πε₀r)

Definition: Field of infinite uniformly charged line (linear density λ).

Derivation

Cylindrical Gaussian surface length l: E(2πrl)=λl/ε₀.

Variables

λ (C/m) · r = perpendicular distance

Why it works

Falls as 1/r — slower than point charge.

Historical context

Wire, lightning channel, long charged rod approximations.

Deep understanding

End caps contribute zero flux (E ⊥ caps). Infinite line assumption.

2. Diagrams & Visuals

E = λ/(2πε₀r)

Color-coded visual · step-by-step breakdown below

  1. Cylindrical symmetry
  2. Gaussian cylinder radius r
  3. E(2πrl) = λl/ε₀
  4. E = λ/(2πε₀r)

3. Solved Examples

Basic

Q: λ=2 μC/m, r=0.1 m.

Solution: E=3.6×10⁵

Answer: ~3.6×10⁵ N/C

Intermediate

Q: vs point charge fall-off?

Solution: 1/r not 1/r²

Answer: Line charge

Advanced

Q: Double distance?

Solution: E halves

Answer: E ∝ 1/r

Exam

Q: Line charge λ units?

Solution: C·m⁻¹

Answer: Sec 15.4.2

Shell: E = Q/(4πε₀r²) (r > R)

Definition: Field outside uniformly charged spherical shell (or conducting sphere).

Derivation

All charge Q acts as if at centre for r > R.

Variables

Q = total charge · R = shell radius

Why it works

Hollow shell outside looks like point charge at centre.

Historical context

Classic Gauss application — conducting sphere charge on surface.

Deep understanding

Solid conductor: charge resides on outer surface; same outside field.

2. Diagrams & Visuals

r > R

Color-coded visual · step-by-step breakdown below

  1. Point outside shell r > R
  2. Enclosed q = Q
  3. E = Q/(4πε₀r²)
  4. As point charge at centre

3. Solved Examples

Basic

Q: Q=10 μC shell R=0.1 m, r=0.5 m.

Solution: Same as point Q at 0.5 m

Answer: Use point formula

Intermediate

Q: Inside vs outside?

Solution: Inside E=0

Answer: Outside 1/r²

Advanced

Q: Charge on conductor?

Solution: Surface only

Answer: Electrostatic equilibrium

Exam

Q: Shell outside field?

Solution: Q/(4πε₀r²)

Answer: Sec 15.4.3

Shell: E = 0 (r < R)

Definition: Electric field inside hollow charged spherical shell.

Derivation

Gaussian sphere inside: q_enclosed = 0 → E = 0.

Variables

r < R inside cavity

Why it works

Faraday cage effect — no field in void of hollow conductor.

Historical context

Surprising result from Gauss — tested in many exams.

Deep understanding

Field jumps at r=R then falls as 1/r² outside. Not true if off-centre charge inside cavity with inner conductor.

2. Diagrams & Visuals

E=0 inside

Color-coded visual · step-by-step breakdown below

  1. Gaussian surface inside shell
  2. No enclosed charge
  3. Φ=0 → E=0 everywhere inside
  4. Valid for hollow shell

3. Solved Examples

Basic

Q: Inside charged shell field?

Solution: Zero

Answer: E=0

Intermediate

Q: At r=R from inside?

Solution: Discontinuity

Answer: Jumps to outside value

Advanced

Q: Solid insulating sphere uniform ρ?

Solution: Different — E≠0 inside

Answer: Not hollow shell case

Exam

Q: Hollow shell interior?

Solution: E=0

Answer: Gauss favourite

E = σ/(2ε₀)

Definition: Electric field near infinite uniformly charged plane sheet.

Derivation

Cylindrical Gaussian surface through sheet: flux 2EA = σA/ε₀.

Variables

σ (C/m²) surface charge density

Why it works

Remarkable: field independent of distance from sheet.

Historical context

Idealisation for large plates — parallel plate capacitor builds on this.

Deep understanding

Same magnitude both sides; perpendicular to sheet. Factor 2 from two faces.

2. Diagrams & Visuals

E = σ/(2ε₀)

Color-coded visual · step-by-step breakdown below

  1. Infinite sheet σ
  2. Gaussian pillbox through sheet
  3. 2EA = σA/ε₀
  4. E = σ/(2ε₀) each side

3. Solved Examples

Basic

Q: σ=8.85 μC/m².

Solution: E=500

Answer: 500 N/C

Intermediate

Q: Move farther from sheet?

Solution: E unchanged

Answer: Distance independent!

Advanced

Q: Two parallel sheets +σ, −σ?

Solution: Add fields

Answer: Capacitor preview

Exam

Q: Plane sheet field?

Solution: σ/(2ε₀)

Answer: Sec 15.4.4

5. Special Features & Extras

Complete study guide for Electric Charge and Electric Field.

Exam Tips & Tricks

  • Q = Ne — charge always integer multiple of e.
  • Coulomb: F ∝ q₁q₂/r²; k = 9×10⁹; use vector superposition.
  • Field: E = F/q₀; then F = qE on any charge.
  • Dipole far field: axial = equatorial; both ∝ 1/r³.
  • Gauss summary: point 1/r² · line 1/r · shell inside 0 · sheet constant.
  • Shell outside: treat as point charge Q at centre.

Common Student Mistakes

  • Adding Coulomb forces as scalars without direction
  • Using dipole 1/r² instead of 1/r³ at large r
  • Forgetting shell interior E = 0
  • Including charges outside Gaussian surface in q_enclosed
  • Confusing σ/(2ε₀) with σ/ε₀ (two faces vs one)
  • Non-integer N in Q = Ne

Memory Aids & Mnemonics

Coulomb: "Inverse square like gravity" — F ∝ 1/r²
Dipole torque: τ = pE sinθ — max when perpendicular (θ=90°)
Gauss five: Point · Line · Shell-out · Shell-in-zero · Sheet
Like/unlike: Like repel, unlike attract — Franklin: glass +, rubber −

Which Formula When?

  • Discrete charge on body? → Q = Ne
  • Two point charges? → Coulomb F = kq₁q₂/r²
  • Many charges? → Superposition
  • Force on test charge? → F = qE
  • Polar molecule / dipole? → p = 2ql, τ = pE sinθ
  • Symmetric distribution? → Gauss Φ = q/ε₀
  • Hollow sphere inside? → E = 0
  • Large charged plate? → E = σ/(2ε₀)

QUICK REFERENCE — Ch 15 Electrostatics

Q = N eF = k q₁q₂/r²F = q₁q₂/(4πε₀r²)F_net = F₁₂ + F₁₃ + …E = F/q₀ = kq/r² r̂F = qEp = q(2l) = 2qlE_axial = 2p/(4πε₀r³)E_eq = p/(4πε₀r³)τ = pE sin θΦ_E = q_enclosed/ε₀E = q/(4πε₀r²)E = λ/(2πε₀r)Shell: E = Q/(4πε₀r²) (r > R)Shell: E = 0 (r < R)E = σ/(2ε₀)

Constants: e = 1.6×10⁻¹⁹ C · k = 9×10⁹ N·m²·C⁻² · ε₀ = 8.85×10⁻¹² C²·N⁻¹·m⁻²

Gauss fields: point q/(4πε₀r²) · line λ/(2πε₀r) · shell out Q/(4πε₀r²) · shell in 0 · sheet σ/(2ε₀)

Tip: Draw the Gaussian surface before picking the E formula — symmetry decides everything.

PYQ — Previous Year Questions

Extracted from NIOS Physics (312) board exam papers in your PDF. Chapter L15 — Electric Charge and Electric Field only. Use Model Answer for marking points; Explanation for concept clarity.

Chapter 15 — Electric Charge and Electric Field (L15)

11 questions · Section A (MCQ & objective) + Section B (short/long) · Sources: 312/TUS/104A, 312/MAY/204A–C, 68/ESS/1-312-A, Marking Scheme

Section A — Multiple Choice (1 mark)

PYQ1. Two point charges are each halved in magnitude and the distance between them is also halved. The electrostatic force between them — (A) remains unchanged   (B) becomes half   (C) becomes one-fourth   (D) becomes four times

1 mark · Section A Q5 · 312/MAY/204A

Model Answer

Answer: (A) remains unchanged

F′ = k(q/2)(q/2)/(r/2)² = kq²/r² = F

Explanation

Halving q twice gives factor ¼ in numerator; halving r gives factor 4 in denominator — they cancel. Coulomb's law F ∝ q₁q₂/r² (L15 §15.2).

PYQ2. If both charges are doubled and separation is doubled, the Coulomb force — (A) doubles   (B) halves   (C) becomes one-fourth   (D) remains unchanged

1 mark · Section A Q3 · 312/MAY/204B

Model Answer

Answer: (D) remains unchanged

F′ = k(2q)(2q)/(2r)² = 4kq²/4r² = F

Explanation

Factor 4 from charges cancels factor 4 from r². Same scaling trick as PYQ1 — useful board pattern.

PYQ3. Charges become 2q each and separation becomes r/2. Force compared to original F is — (A) 2F   (B) 4F   (C) 8F   (D) 16F

1 mark · Section A Q2 · 312/MAY/204C

Model Answer

Answer: (D) 16F

F′ = k(2q)(2q)/(r/2)² = 4kq² × 4/r² = 16F

Explanation

Charge factor 4 × distance factor 4 = 16. Inverse-square law amplifies small distance changes sharply.

PYQ4. The SI unit of electric field intensity is — (A) N·C   (B) C·N⁻¹   (C) V·m⁻¹   (D) V·m

1 mark · Section A Q8 · 312/TUS/104A

Model Answer

Answer: (C) V·m⁻¹

Also written N·C⁻¹ since E = F/q₀.

Explanation

E = F/q₀ → N/C. Also E = −dV/dr → volt per metre. Both are equivalent SI units (L15 §15.3).

PYQ5. Dimensional formula of electric field E is — (A) MLT⁻²A⁻¹   (B) MLT⁻²A⁻²   (C) MLT⁻³A⁻¹   (D) MLT⁻³A⁻²

1 mark · Section A Q6 · 312/MAY/204A

Model Answer

Answer: (C) MLT⁻³A⁻¹

E = F/q → [M L T⁻²]/[A T] = M L T⁻³ A⁻¹

Explanation

Force [MLT⁻²], charge [AT]. Divide to get field dimensions. Cross-check: kq/r² also gives same result.

PYQ6. Ratio of electrostatic force in air to that in a medium of dielectric constant K (same charges and distance) is — (A) 1 : K   (B) K : 1   (C) K² : 1   (D) 1 : K²

1 mark · Section A Q6 · 68/ESS/1-312-A

Model Answer

Answer: (B) K : 1

F_air / F_medium = K (force in medium = F_air / K)

Explanation

Medium reduces field and force by factor εᵣ = K. F_medium = F_vacuum × (ε₀/ε) = F_air/K (L15 §15.2).

PYQ7. An oil drop of mass m carrying n electronic charges is held stationary in a uniform vertical electric field E (gravity downward). The field magnitude is — (A) mg·e/n   (B) mg/(n·e)   (C) n·e/mg   (D) m·g·n/e

1 mark · Section A Q7 · 312/TUS/104A

Model Answer

Answer: (B) mg/(n·e)

Balance: qE = mg → (n e) E = mg → E = mg/(ne)

Explanation

Millikan-type balance — electric upward force equals weight. Links quantisation Q = ne with E = F/q₀ (L15 §15.1.2, §15.3).

PYQ8. Two concentric spherical Gaussian surfaces enclose the same point charge +q at the centre. Ratio of electric flux through the larger surface to the smaller is — (A) 1 : 1   (B) 2 : 1   (C) 1 : 2   (D) 1 : 4

1 mark · Section A (alternate) · 312/TUS/104A

Model Answer

Answer: (A) 1 : 1

Both surfaces enclose the same q → Φ = q/ε₀ for each → ratio 1 : 1

Explanation

Gauss's law: net flux depends only on enclosed charge, not on surface size or shape (L15 §15.4).

PYQ9. Match the SI unit (any two): (a) Electric flux → ?   (b) Electric field → ?   Options: (i) N·C⁻¹   (ii) N·m²·C⁻¹   (iii) C·m   (iv) V·m

2 marks (1×2) · Section A Q21 · 68/ESS/1-312-A (Marking Scheme)

Model Answer

(a) Electric flux ↔ (ii) N·m²·C⁻¹

(b) Electric field ↔ (i) N·C⁻¹ (same as V·m⁻¹)

Explanation

Flux Φ = E·Δs → (N/C)(m²). Field is force per unit charge. Dipole moment p has unit C·m (option iii).

PYQ10. The number of electrons that constitute −1 C of charge is — (A) 6.25×10¹⁸   (B) 6.4×10²⁷   (C) 9.0×10¹⁸   (D) 1.6×10¹⁹

1 mark · Section A Q11 · 68/ESS/1-312-A

Model Answer

Answer: (A) 6.25×10¹⁸

N = 1/(1.6×10⁻¹⁹) = 6.25×10¹⁸ electrons

Explanation

Quantisation Q = Ne. One coulomb = 1/(e) electrons.

PYQ11. An electric dipole of dipole moment p is held in a uniform electric field E. Obtain expressions for (i) torque and (ii) potential energy. Also obtain the work done to turn the dipole from stable to unstable equilibrium.

3 marks · Section B Q39 (OR) · 312/TUS/104A

Model Answer

(i) τ = pE sin θ = p × E

(ii) U = −p·E = −pE cos θ

Stable → unstable: θ: 0° → 180°, W = pE(cos0° − cos180°) = 2pE

Explanation

Direct from TUS 104A Section B Q39 OR. Marking scheme integrates τ = pE sin θ for U.

Problem Solving — L15 Electric Charge and Electric Field

Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). Explanations open by default.

Question 1 of 6Coulomb

Find force between charges +2 μC and +3 μC separated by 0.30 m in vacuum (k = 9×10⁹).

F = k |q₁q₂|/r²

Solution — step by step with formulas

  1. F = 9e9 × (2e-6)(3e-6)/(0.09) = 0.60 N (repulsive).

Final answer: F = 0.60 N repulsive

Formulas used in this problem

F = k |q₁q₂|/r²

Textbook formal language

Coulomb’s law gives electrostatic force between point charges along the joining line.

Working formula set for this problem: F = k |q₁q₂|/r². In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Like charges push apart; plug into kq₁q₂/r².

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Coulomb’s law

Force is inverse-square and obeys Newton’s third law pairwise.

Link to chapter notes (L15 — Coulomb’s law): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: F = k |q₁q₂|/r². In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write F = k |q₁q₂|/r² before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 2 of 6Field

Define electric field. Field 200 N·C⁻¹ acts on +2 μC. Find force.

E = F/q₀
E = kQ/r²

Solution — step by step with formulas

  1. F = qE = 2e-6 × 200 = 4.0×10⁻⁴ N along E.

Final answer: F = 4.0×10⁻⁴ N along field

Formulas used in this problem

E = F/q₀
E = kQ/r²

Textbook formal language

E is force per unit positive test charge at a point.

Working formula set for this problem: E = F/q₀; E = kQ/r². In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Field tells how hard a +1 C charge would be pushed; multiply by your charge.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Electric field

Field lines start on + and end on −; density indicates |E|.

Link to chapter notes (L15 — Electric field): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: E = F/q₀; E = kQ/r². In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write E = F/q₀; E = kQ/r² before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 3 of 6Superposition

Two equal positive charges produce fields at midpoint. What is E_net and why?

E_net = Σ E_i

Solution — step by step with formulas

  1. Equal opposite fields cancel; E_net = 0 at midpoint between equal like charges.

Final answer: E_net = 0

Formulas used in this problem

E_net = Σ E_i

Textbook formal language

Electric field is a vector; net field is the vector sum of individual fields.

Working formula set for this problem: E_net = Σ E_i. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Each charge pushes a test + charge away; at the middle the two pushes cancel.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Field superposition

Use components carefully for non-collinear arrangements.

Link to chapter notes (L15 — Field superposition): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: E_net = Σ E_i. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write E_net = Σ E_i before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 4 of 6Dipole

Define electric dipole moment. When is torque on a dipole in uniform E maximum?

p = q × 2a
τ = pE sinθ

Solution — step by step with formulas

  1. p = q·(2a) from − to +.
  2. τ_max when θ = 90° (p ⟂ E).

Final answer: τ_max = pE at θ = 90°

Formulas used in this problem

p = q × 2a
τ = pE sinθ

Textbook formal language

Dipole moment is a vector from negative to positive charge.

Working formula set for this problem: p = q × 2a; τ = pE sinθ. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Torque tries to align the dipole with the field; strongest when sideways.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Electric dipole

Net force on dipole is zero in uniform field; non-uniform fields can translate it.

Link to chapter notes (L15 — Electric dipole): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: p = q × 2a; τ = pE sinθ. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write p = q × 2a; τ = pE sinθ before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 5 of 6Flux/Gauss

State Gauss’s law. Why is E outside a uniformly charged spherical shell like a point charge?

Φ = q_encl/ε₀

Solution — step by step with formulas

  1. Flux through closed surface = q_encl/ε₀.
  2. Spherical symmetry ⇒ E·4πr² = Q/ε₀ ⇒ E = kQ/r².

Final answer: E = kQ/r² outside shell

Formulas used in this problem

Φ = q_encl/ε₀

Textbook formal language

Gauss’s law relates flux of E to enclosed charge; symmetry fixes E.

Working formula set for this problem: Φ = q_encl/ε₀. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Outside, shell’s charge looks concentrated at the centre for the field strength.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Gauss’s law idea

Inside a charged conducting shell in electrostatics, E = 0 in the metal.

Link to chapter notes (L15 — Gauss’s law idea): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: Φ = q_encl/ε₀. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write Φ = q_encl/ε₀ before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 6 of 6Conductor

Why is electric field zero inside the material of a conductor in electrostatic equilibrium?

E = 0 inside conductor

Solution — step by step with formulas

  1. Free charges rearrange until E = 0 inside; else currents would flow.

Final answer: E = 0 inside conductor (electrostatics)

Formulas used in this problem

E = 0 inside conductor

Textbook formal language

Electrostatic equilibrium forbids steady currents; hence net E vanishes in the bulk.

Working formula set for this problem: E = 0 inside conductor. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

If field remained inside, free electrons would keep moving—so they move until field dies.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Conductor in electrostatics

Excess charge resides on the outer surface of an isolated conductor.

Link to chapter notes (L15 — Conductor in electrostatics): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: E = 0 inside conductor. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write E = 0 inside conductor before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).