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).
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:
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.
q₁, q₂ = charges (C)
r = separation (m)
Vector form: F₁₂ = k (q₁q₂/r²) r̂₁₂
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:
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:
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).
Direction: from −q to +q along the dipole axis
|p| = 2ql
Field parallel to p; magnitude twice the broad-on value
∝ 1/r³
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:
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.
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.
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:
q = total enclosed charge
Gaussian surface is imaginary — need not be a real physical surface
15.4.1 Point Charge
Recovers Coulomb's law: F = q₀E
15.4.2 Long Line Charge
Uniform linear charge density λ (C·m⁻¹). Cylindrical Gaussian surface of radius r, length l:
Field ∝ 1/r (not 1/r²)
Flat end caps contribute zero flux (E ⊥ Δs)
15.4.3 Uniformly Charged Spherical 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:
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.
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:
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
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
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
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
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
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
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
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
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
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
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.
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.
Find force between charges +2 μC and +3 μC separated by 0.30 m in vacuum (k = 9×10⁹).
Solution — step by step with formulas
- F = 9e9 × (2e-6)(3e-6)/(0.09) = 0.60 N (repulsive).
Final answer: F = 0.60 N repulsive
Formulas used in this problem
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).
Define electric field. Field 200 N·C⁻¹ acts on +2 μC. Find force.
Solution — step by step with formulas
- 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
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).
Two equal positive charges produce fields at midpoint. What is E_net and why?
Solution — step by step with formulas
- Equal opposite fields cancel; E_net = 0 at midpoint between equal like charges.
Final answer: E_net = 0
Formulas used in this problem
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).
Define electric dipole moment. When is torque on a dipole in uniform E maximum?
Solution — step by step with formulas
- p = q·(2a) from − to +.
- τ_max when θ = 90° (p ⟂ E).
Final answer: τ_max = pE at θ = 90°
Formulas used in this problem
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).
State Gauss’s law. Why is E outside a uniformly charged spherical shell like a point charge?
Solution — step by step with formulas
- Flux through closed surface = q_encl/ε₀.
- Spherical symmetry ⇒ E·4πr² = Q/ε₀ ⇒ E = kQ/r².
Final answer: E = kQ/r² outside shell
Formulas used in this problem
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).
Why is electric field zero inside the material of a conductor in electrostatic equilibrium?
Solution — step by step with formulas
- Free charges rearrange until E = 0 inside; else currents would flow.
Final answer: E = 0 inside conductor (electrostatics)
Formulas used in this problem
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).