Model Question Paper – Solved

Electric flux (വൈദ്യുത ഫ്ലക്സ്) Δφ = E·ΔS = (N C⁻¹) × (m²) = N m² C⁻¹. (b) N C⁻¹ is the unit of electric field, (c) C m of dipole moment, (d) C m⁻² of surface charge density.

Rubbing only transfers electrons. Glass loses some electrons to silk, so the rod is left with a deficit of electrons (positive) and the silk gains them (negative). Protons are bound in the nuclei and do not move (a); (c) would make the rod negative; (d) is wrong because charge is never created – conservation of charge (ചാർജ് സംരക്ഷണം).

Gauss’s law gives E = λ/(2πε₀r), so E ∝ 1/r. A point charge gives 1/r², a dipole 1/r³, and an infinite plane sheet a field independent of r.

F ∝ q₁q₂/r². New force = F × 2 ÷ (½)² = F × 2 × 4 = 8F.

Axial: E = 2p/(4πε₀r³); equatorial: E = p/(4πε₀r³) for r ≫ a. Both fall as 1/r³ – faster than the 1/r² of a single charge, because the fields of +q and −q nearly cancel far away.

Dipole moment (ദ്വിധ്രുവ മൊമെന്റ്) p = q × 2a, directed along the axis from the negative charge to the positive charge. (Field lines go the other way outside – from + to −.)

τ = pE sin θ is largest when sin θ = 1, i.e. θ = 90°; then τ_max = pE. At 0° and 180° the torque is zero; at 45° it is pE/√2.

The forces qE and −qE are equal and opposite, so the net force is zero; but they act at different points, so they form a couple: torque τ = p × E. (In a non-uniform field there is a net force too.)

Gauss’s law (ഗോസ് നിയമം): φ = q_enclosed/ε₀ = (2q − q)/ε₀ = q/ε₀. The outside charge +3q sends in as many field lines as it takes out – no net flux. (a) wrongly adds the outside charge; (c) uses only the outside charge.

A Gaussian sphere inside the shell encloses no charge, so E × 4πr² = 0 ⇒ E = 0 everywhere inside. Outside, E = q/(4πε₀r²) as if all the charge were at the centre.

The Gaussian pillbox has flux 2EA through its two flat faces and encloses σA, so 2EA = σA/ε₀ ⇒ E = σ/(2ε₀), the same at every distance. σ/ε₀ is the field just outside a charged conductor.

Electrostatic field lines (വൈദ്യുത മണ്ഡലരേഖകൾ) begin on + charges and end on − charges (or go to infinity), so they cannot form closed loops. (a), (b) and (d) are true properties.

k = 1/(4πε₀) ≈ 9 × 10⁹ N m² C⁻², with permittivity of free space (ശൂന്യസ്ഥലത്തിന്റെ പെർമിറ്റിവിറ്റി) ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻² – option (b) is ε₀’s number with the wrong unit. (d) is the gravitational constant G.

Quantisation of charge (ചാർജിന്റെ ക്വാണ്ടീകരണം): q = ne. Divide by e = 1.6 × 10⁻¹⁹ C: (a) n = 2, (b) n = 3, (c) n = 1.5 – not a whole number, impossible; (d) n = 10.

The metal body acts as a conducting shell: the charge stays on its outer surface and the field inside is zero – electrostatic shielding (സ്ഥിതവൈദ്യുത കവചനം). Tall trees, open ground and hill tops are where lightning strikes.

E = F/q₀: newton per coulomb, N C⁻¹ (equivalently V m⁻¹).

Charging by induction (പ്രേരണം വഴി ചാർജ് ചെയ്യൽ): the nearby charge redistributes the free electrons; earthing or separating the parts leaves a net charge. The inducing body loses nothing.

Charge can be transferred, but it is never created or destroyed in an isolated system – conservation of charge (e.g. a neutron → a proton + an electron: 0 = +e − e).

In a polar molecule (ധ്രുവീയ തന്മാത്ര) the centres of positive and negative charge do not coincide, so it has a permanent dipole moment. CO₂, CH₄ are non-polar.

φ = q_enclosed/ε₀ = 0/ε₀ = zero – as many field lines leave as enter.

Statement: Coulomb’s law (കൂളോം നിയമം): the electrostatic force between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them; it acts along the line joining them (repulsive for like, attractive for unlike charges).

F = 14πε₀ |q₁q₂|r², with 1/(4πε₀) = 9 × 10⁹ N m² C⁻².

Vector form: F₂₁ (force on q₂ due to q₁) = 14πε₀ q₁q₂r₂₁² r̂₂₁, where r̂₂₁ is the unit vector from q₁ to q₂; and F₁₂ = −F₂₁.

Quantisation of charge: the charge on any body is always an integral multiple of a basic unit of charge e (the charge of an electron or a proton); charge exists only in discrete packets.

q = ne, where n = 0, ±1, ±2, ±3, … and e = 1.6 × 10⁻¹⁹ C. (For macroscopic charges n is huge, so quantisation can be ignored.)

The dipole moment of an electric dipole (വൈദ്യുത ദ്വിധ്രുവം) is the product of the magnitude of either charge and the distance between the two charges: p = q × 2a. It is a vector directed along the axis from −q to +q.

SI unit: coulomb metre (C m).

Statement: Gauss’s law: the total electric flux through any closed surface is 1/ε₀ times the net charge enclosed by the surface.

φ = ∮_S E · dS = q_enclosed/ε₀. Charges outside S give zero net flux; the closed surface is called a Gaussian surface.

(i) The tangent to a field line gives the direction of E at that point. If two lines crossed, there would be two directions of the field at the crossing point – impossible, since the field has only one direction at a point.

(ii) Electrostatic field lines start on positive charges and end on negative charges (or at infinity); they are not continuous around a closed path, so they cannot form closed loops. (Go further: the electrostatic field is conservative.)

Electric flux through an area is the number of field lines crossing it normally; through a small area ΔS, Δφ = E · ΔS = E ΔS cos θ, θ being the angle between E and the area vector (ഏരിയ സദിശം) along the normal. It is a scalar.

SI unit: N m² C⁻¹ (also V m).

Electrostatic shielding: the electric field inside a cavity of a conductor is always zero, whatever the outside field. So a region enclosed by a conductor is protected (shielded) from external electric fields.

Application (any one): sensitive instruments are kept in metal boxes; a car or bus with a metal body is safe during lightning; coaxial cables have an outer metal braid.

Given: q = +3.2 μC = 3.2 × 10⁻⁶ C; e = 1.6 × 10⁻¹⁹ C

Formula: q = ne ⇒ n = q/e

Substitution: n = 3.2 × 10⁻⁶1.6 × 10⁻¹⁹

Answer: n = 2 × 10¹³ electrons (removed, so the sphere is positive)

(1) Place the two metal spheres A and B on insulating stands, touching each other (one conductor).

(2) Bring a negatively charged rod near A, without touching. Free electrons are repelled to the far side of B: A gets a positive induced charge, B a negative one.

(3) Keeping the rod in place, separate the spheres slightly.

(4) Remove the rod. The charges spread uniformly over each sphere: A is positive and B negative, with equal magnitudes. The rod loses no charge.

(1) Additivity (ചാർജിന്റെ സങ്കലനീയത): the total charge of a system is the algebraic sum of the individual charges, e.g. +2 + 5 − 3 = +4 units. Charge is a scalar.

(2) Conservation: the total charge of an isolated system remains constant. In rubbing, electrons are only transferred; in a neutron decay n → p + e⁻ the total stays zero.

(3) Quantisation: every charge is an integral multiple of e: q = ne, e = 1.6 × 10⁻¹⁹ C.

A dipole (charges ±q, separation 2a) makes an angle θ with a uniform field E. Force on +q = qE (along E), on −q = −qE – equal and opposite, so the net force is zero.

They act at different points and form a couple. Torque = magnitude of one force × perpendicular distance between their lines of action = qE × 2a sin θ.

Since p = q × 2a: τ = pE sin θ; in vector form τ = p × E. It tends to turn p towards E (maximum pE at θ = 90°, zero at 0° and 180°).

An electric field line is a curve drawn so that the tangent at each point gives the direction of the electric field at that point; their closeness shows the strength of the field.

(a) Radial straight lines going outwards. (b) Lines start on +q, curve round and end on −q. (c) Lines repel each other; there is a neutral point N midway where E = 0.

The electric field (വൈദ്യുത മണ്ഡലം) at a point is the force experienced by a unit positive test charge (പരീക്ഷണ ചാർജ്) placed at that point: E = F/q₀ (q₀ → 0, so it does not disturb the source).

A test charge q₀ at distance r from a point charge Q feels (Coulomb) F = 14πε₀ Qq₀r² r̂. Dividing by q₀: E = 14πε₀ Qr² r̂, unit N C⁻¹. It points radially outwards for Q > 0 and inwards for Q < 0.

Given: q₁ = 4 × 10⁻⁶ C, q₂ = −6 × 10⁻⁶ C, r = 20 cm = 0.20 m

Formula: F = 14πε₀ |q₁q₂|r²

Substitution: F = 9 × 10⁹ × 4 × 10⁻⁶ × 6 × 10⁻⁶(0.20)² = 0.2160.04

Answer: F = 5.4 N, attractive (unlike charges)

Given: R = 0.20 m, q = 4 × 10⁻⁶ C; r₁ = 0.10 m (inside), r₂ = 0.30 m (outside)

Formula: σ = q/(4πR²); inside E = 0; outside E = 14πε₀ qr²

Substitution: σ = 4 × 10⁻⁶4 × 3.14 × 0.04;   E₂ = 9 × 10⁹ × 4 × 10⁻⁶(0.30)²

Answer: (a) σ = 7.96 × 10⁻⁶ C m⁻² (b) E = 0 (c) E = 4 × 10⁵ N C⁻¹, radially outward

Given: q = 17.7 × 10⁻⁹ C, ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻²

Formula: φ = q/ε₀; one face φ/6 (by symmetry)

Substitution: φ = 17.7 × 10⁻⁹8.854 × 10⁻¹²

Answer: (a) φ ≈ 2 × 10³ N m² C⁻¹ (b) φ/6 ≈ 3.33 × 10² N m² C⁻¹ (c) No – the flux depends only on the charge enclosed, not on the size of the surface.

Dipole: −q at A, +q at B, AB = 2a, centre O; P on the axis at OP = r (on the side of +q).

Field at P due to −q (distance r + a): E_−q = 14πε₀ q(r + a)², towards A (along −p̂).

Field at P due to +q (distance r − a): E_+q = 14πε₀ q(r − a)², away from B (along +p̂).

E_+q > E_−q, so E = 14πε₀ q [1(r − a)² − 1(r + a)²] = 14πε₀ 4qar(r² − a²)².

With p = 2qa: E = 14πε₀ 2pr(r² − a²)² p̂. For r ≫ a: E = 14πε₀ 2pr³, directed along p (from −q to +q).

Dipole: −q at A, +q at B, AB = 2a, centre O; P on the perpendicular bisector with OP = r. Each charge is √(r² + a²) from P.

Magnitudes: E_+q = E_−q = 14πε₀ qr² + a²; E_+q along BP (away from B), E_−q along PA (towards A).

Resolve: the components perpendicular to the axis are equal and opposite and cancel; the components parallel to the axis add, both pointing opposite to p.

E = 2E_+q cos θ with cos θ = a√(r² + a²) ⇒ E = 14πε₀ 2qa(r² + a²)^3/2.

E = −14πε₀ p(r² + a²)^3/2 p̂; for r ≫ a, E = −14πε₀ pr³ – opposite to p. (Half the axial value at the same distance.)

Gauss’s law: the total flux through a closed surface = q_enclosed/ε₀.

Wire (അനന്തമായ നീളമുള്ള നേർകമ്പി) with linear charge density (രേഖീയ ചാർജ് സാന്ദ്രത) λ. By symmetry E is radial and has the same magnitude at all points at distance r. Take a Gaussian cylinder of radius r and length l, coaxial with the wire.

Flat ends: E is parallel to the end faces (perpendicular to dS) – no flux. Curved surface: E ∥ dS, so flux = E × 2πrl.

Charge enclosed = λl. Gauss: E × 2πrl = λl/ε₀.

E = λ2πε₀r n̂, radially outward for λ > 0 (E ∝ 1/r).

Infinite plane sheet (അനന്തമായ സമതല ഷീറ്റ്) with surface charge density (ഉപരിതല ചാർജ് സാന്ദ്രത) σ. By symmetry E is normal to the sheet, pointing away on both sides (σ > 0) and equal in magnitude at equal distances.

Gaussian surface: a cylinder (pillbox) of cross-section A crossing the sheet, its axis normal to the sheet.

Curved surface: E ⊥ dS – no flux. Each flat face: flux EA. Total φ = 2EA.

Charge enclosed = σA. Gauss: 2EA = σA/ε₀.

E = σ2ε₀ n̂ – independent of the distance from the sheet.

Thin spherical shell (നേർത്ത ഗോളീയ കവചം) of radius R, total charge q = 4πR²σ. By symmetry E is radial and the same at every point at a given distance r from the centre.

(i) r > R: Gaussian sphere of radius r: flux = E × 4πr²; charge enclosed = q.

E × 4πr² = q/ε₀ ⇒ E = 14πε₀ qr² r̂ – as if all the charge were at the centre.

(ii) r < R: a Gaussian sphere inside encloses no charge: E × 4πr² = 0 ⇒ E = 0 everywhere inside.

Given: Q = 2 × 10⁻⁶ C; q_A = 3 × 10⁻⁶ C, OA = 0.30 m; q_B = −4 × 10⁻⁶ C, OB = 0.40 m; OA ⊥ OB.

Formula: F = 14πε₀ |q₁q₂|r²; superposition: F = F₁ + F₂.

Step 1: F₁ (due to A) = 9 × 10⁹ × 2 × 10⁻⁶ × 3 × 10⁻⁶(0.30)² = 0.6 N, repulsive – along AO produced (away from A).

Step 2: F₂ (due to B) = 9 × 10⁹ × 2 × 10⁻⁶ × 4 × 10⁻⁶(0.40)² = 0.45 N, attractive – along OB (towards B).

Step 3: F₁ ⊥ F₂: F = √(0.6² + 0.45²) = √0.5625 = 0.75 N.

Step 4: tan θ = F₂/F₁ = 0.45/0.6 = 0.75 ⇒ θ = 36.9° with AO produced, on the side of B.

Given: q₁ = 4 × 10⁻⁶ C at A, q₂ = 9 × 10⁻⁶ C at B, AB = 0.50 m.

Formula: E = 14πε₀ qr²; for like charges the null point lies between them, nearer the smaller charge.

(a) At N, AN = x: q₁x² = q₂(0.5 − x)² ⇒ 0.5 − xx = √(9/4) = 1.5 ⇒ 0.5 = 2.5x ⇒ x = 0.2 m = 20 cm from the +4 μC charge (there each field is 9 × 10⁵ N C⁻¹, opposite).

(b) At the midpoint M (r = 0.25 m): E₁ = 9 × 10⁹ × 4 × 10⁻⁶(0.25)² = 5.76 × 10⁵ N C⁻¹ towards B; E₂ = 9 × 10⁹ × 9 × 10⁻⁶(0.25)² = 1.296 × 10⁶ N C⁻¹ towards A.

Resultant E = E₂ − E₁ = 7.2 × 10⁵ N C⁻¹, towards A (the +4 μC charge).

Given: q = 8 × 10⁻⁹ C, 2a = 5 × 10⁻³ m, r = 0.20 m (r ≫ a), E = 5 × 10⁴ N C⁻¹, θ = 30°.

(a) p = q × 2a = 8 × 10⁻⁹ × 5 × 10⁻³ = 4 × 10⁻¹¹ C m, from −q to +q.

(b) E_axial = 14πε₀ 2pr³ = 9 × 10⁹ × 2 × 4 × 10⁻¹¹(0.20)³ = 90 N C⁻¹, along p.

(c) E_equatorial = 14πε₀ pr³ = 45 N C⁻¹, opposite to p.

(d) τ = pE sin θ = 4 × 10⁻¹¹ × 5 × 10⁴ × 0.5 = 1 × 10⁻⁶ N m.

(e) τ_max = pE (at θ = 90°) = 2 × 10⁻⁶ N m.