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State Board · Class 11 · Physics

Lesson 10: Electrostatics

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

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Electric charge and its properties

From daily life

A comb rubbed on dry hair picks up bits of paper. Link to earlier classes : static electricity and charge of classes 9-10 now extend to Coulomb's law, electric field and Gauss's law.

Simple definition

Electric charge is the property of matter that gives rise to electric force.

Ways of charging−friction+contact−+inductionq = n e ; e = 1.6 × 10⁻¹⁹ C

In the figure : Three ways of charging.

Rule 1: Properties of charge — Quantisation : q = n e (e = 1.6 × 10⁻¹⁹ C); conservation : total charge constant; additivity : algebraic sum.

  • Easy : 1 μC = n × 1.6 × 10⁻¹⁹ ⇒ n = 6.25 × 10¹² electrons.
  • Medium : Gain of 5 × 10¹² electrons ⇒ q = −0.8 μC.
  • Exam (CET) : 3.2 × 10⁻¹⁹ C = ? e (1) 2e (2) 3e (3) 0.5e (4) 1e → answer 1)

Rule 2: Methods — Friction : equal and opposite charges; conduction : same kind of charge; induction : without touching, opposite charge.

  • Easy : Glass rod + silk ⇒ glass +, silk −.
  • Medium : Bring a negative rod near a neutral sphere and earth it ⇒ the sphere becomes +.
  • Exam (CET) : Charge obtained by induction? (1) opposite (2) same (3) zero (4) double → answer 1)

Rule 3: Conductors and insulators — Conductors have free electrons; insulators do not. Charge stays on a conductor's surface.

  • Easy : Copper, salt water = conductors; rubber, glass = insulators.
  • Medium : Earthing lets the charge flow away.
  • Exam (CET) : Where is charge on a conductor? (1) on the surface (2) inside (3) at the centre (4) everywhere → answer 1)

Derivation : Charging by induction

  • Step 1 : Bring a negative rod near a neutral sphere : near side +, far side −.
  • Step 2 : Earth the sphere : − electrons flow to earth.
  • Step 3 : Remove the earth connection.
  • Step 4 : Remove the rod ⇒ the sphere keeps a + charge, spread over it.
Remember

q = n e; total q is conserved; like repel, unlike attract.

Common mistake / CET trap

Charge is never half an e; friction moves charge, it does not create it.

Shortcut

Memory clue : "q = n e".

Spot the difference

MethodTouchCharge
Frictionyesopposite
Conductionyessame
Inductionnoopposite

Practice questions

Easy
  1. 1What is quantisation of charge?
  2. 2Value of e?
  3. 3What is induction?
Medium
  1. 4How many electrons in 2 μC?
  2. 58 × 10⁻¹⁹ C = ? e
  3. 6+3 μC and −5 μC touch; charge on each?
Challenging
  1. 7Mass change when a body gains 10¹² electrons?
  2. 8Explain conservation of charge.
Show answers (try them yourself first)
  1. 1q = n e (integer multiples)
  2. 21.6 × 10⁻¹⁹ C
  3. 3Charging without contact
  4. 41.25 × 10¹³
  5. 55e
  6. 6−1 μC each
  7. 79.1 × 10⁻¹⁹ kg increase
  8. 8Total q does not change

Coulomb's law

From daily life

Two rubbed balloons pushed together repel each other.

Simple definition

Coulomb's law gives the force between two point charges.

Coulomb's lawq₁q₂rF = k q₁ q₂ ÷ r²k = 1/4πε₀ = 9 × 10⁹ N m² C⁻²

In the figure : F ∝ q₁q₂ and F ∝ 1/r².

Rule 1: The law — F = k q₁ q₂ ÷ r²; k = 1/(4π ε₀) = 9 × 10⁹ N m² C⁻²; ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻². Vector form F₁₂ = k q₁q₂ r̂ ÷ r².

  • Easy : 1 μC and 1 μC, 1 m apart ⇒ F = 9 × 10⁻³ N.
  • Medium : 2 μC and 3 μC, 0.3 m apart ⇒ F = 0.60 N.
  • Exam (CET) : r halved ⇒ F? (1) four times (2) double (3) half (4) ¼ → answer 1) Trick : inverse square.

Rule 2: In a medium — F_medium = F_vacuum ÷ K (K = relative permittivity; water ≈ 80). F = q₁ q₂ ÷ 4π ε r², ε = K ε₀.

  • Easy : The 0.6 N above in water ⇒ 0.0075 N.
  • Medium : K = 4 ⇒ force ¼.
  • Exam (CET) : Force in water? (1) less (2) more (3) same (4) zero → answer 1)

Rule 3: Compare with gravity — Electron-proton : F_e / F_g ≈ 10³⁹. Gravity depends on mass, electric force on charge; both inverse-square.

  • Easy : F_e ÷ F_g = 2.27 × 10³⁹ (same r).
  • Medium : Gravity is only attractive; electric force can be either.
  • Exam (CET) : F_e ÷ F_g approx? (1) 10³⁹ (2) 10⁹ (3) 1 (4) 10⁻³⁹ → answer 1)

Derivation : Vector form of Coulomb's law

  • Step 1 : The force on q₂ due to q₁ acts along r̂₂₁ (from q₁ to q₂).
  • Step 2 : Magnitude k |q₁ q₂| / r².
  • Step 3 : Sign : q₁ q₂ > 0 ⇒ repulsion, < 0 ⇒ attraction.
  • Step 4 : F₂₁ = k q₁ q₂ r̂₂₁ ÷ r²; F₁₂ = −F₂₁.
Remember

F = kq₁q₂/r²; F ÷ K in a medium.

Common mistake / CET trap

Signs fix the direction, not the magnitude; r in metres.

Shortcut

Memory clue : "k q q over r squared".

Spot the difference

ForceDepends onDirection
Electricq₁q₂/r²attractive/repulsive
Gravitym₁m₂/r²attractive

Practice questions

Easy
  1. 1State Coulomb's law.
  2. 2Value of k?
  3. 3Both charges doubled ⇒ F?
Medium
  1. 45 μC, 4 μC, 0.5 m : F?
  2. 5F = 8 N; r tripled : F?
  3. 620 N in a medium with K = 5 ⇒ F in vacuum?
Challenging
  1. 7Two 1 C charges 1 km apart : F?
  2. 8Why is F_e/F_g so large?
Show answers (try them yourself first)
  1. 1F = kq₁q₂/r²
  2. 29 × 10⁹ N m² C⁻²
  3. 3Four times
  4. 40.72 N
  5. 58/9 N
  6. 6100 N
  7. 79 × 10³ N
  8. 8Masses are tiny compared with charges

Superposition of forces and equilibrium

From daily life

Three children push one ball from different sides — the net force is their sum.

Simple definition

Superposition: each charge exerts its force independent of others; add the forces.

Superposition and equilibriumqqqF = √3 k q² ÷ a²balance : x/(d − x) = √(q₁/q₂)

In the figure : Green : the resultant force.

Rule 1: Superposition — Net force = vector sum of the pair forces : F = F₁ + F₂ + … .

  • Easy : On a line : 2 μC (x = 0), −2 μC (x = 0.2), 1 μC (x = 0.1) ⇒ both act along +x ⇒ 1.8 + 1.8 = 3.6 N.
  • Medium : Equilateral triangle a = 0.1 m, q = 1 μC ⇒ F = √3 × 0.9 = 1.56 N.
  • Exam (CET) : Equal charges at the corners (side a) ⇒ F on each? (1) √3kq²/a² (2) kq²/a² (3) 2kq²/a² (4) 0 → answer 1)

Rule 2: Equilibrium point — Like charges q₁, q₂ : the balance point lies between; x ÷ (d − x) = √(q₁ ÷ q₂) (x from q₁).

  • Easy : 4 μC and 1 μC, d = 0.3 m ⇒ x = 0.2 m (from the 4 μC).
  • Medium : 9 μC and 1 μC, d = 0.4 ⇒ x = 0.3 m.
  • Exam (CET) : Equal q, q ⇒ balance point? (1) midpoint (2) outside (3) none (4) at q → answer 1)

Rule 3: Equilibrium of a system — For equilibrium ΣF = 0. Three equal q at corners and Q at the centre ⇒ Q = − q ÷ √3.

  • Easy : q = 3 μC ⇒ Q = −1.73 μC.
  • Medium : This equilibrium is unstable (Earnshaw).
  • Exam (CET) : Q at the centre? (1) −q/√3 (2) +q/√3 (3) −q (4) 0 → answer 1)

Derivation : Equilibrium point

  • Step 1 : At distance x from q₁ and d − x from q₂ the forces on q₀ are equal.
  • Step 2 : k q₁ q₀ / x² = k q₂ q₀ / (d − x)².
  • Step 3 : q₁/x² = q₂/(d − x)².
  • Step 4 : x ÷ (d − x) = √(q₁ ÷ q₂).
Remember

F = ΣFᵢ (vector); x/(d − x) = √(q₁/q₂).

Common mistake / CET trap

Do not add angled forces algebraically; the balance point is between only for like charges.

Shortcut

Memory clue : "Add as vectors".

Spot the difference

CaseForce
Along a linealgebraic sum
At an anglevector sum
EquilibriumΣF = 0

Practice questions

Easy
  1. 1State the superposition principle.
  2. 2Condition for equilibrium?
  3. 3Balance point of two equal like charges?
Medium
  1. 4Equilateral, 1 μC, a = 0.3 m : F?
  2. 516 μC and 4 μC, d = 0.6 : x?
  3. 6q = 6 μC : Q at the centre?
Challenging
  1. 7Why is this equilibrium unstable?
  2. 8Derive F for the equilateral triangle.
Show answers (try them yourself first)
  1. 1F = ΣFᵢ (vector)
  2. 2ΣF = 0
  3. 3Midpoint
  4. 40.173 N
  5. 50.4 m
  6. 6−3.46 μC
  7. 7It does not return after a small displacement
  8. 8Two forces at 60° : 2F cos30°

Electric field and lines of force

From daily life

Iron filings show lines around a magnet; a charge also has a field around it.

Simple definition

Electric field is the force per unit positive charge.

Field lines+−E = F/q = kQ/r²

In the figure : Out of +, into −.

Live figure

Electric field and lines of force

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

Open in the app

Rule 1: Electric field — E = F ÷ q₀ (N/C = V/m); point charge : E = k Q ÷ r²; away from +, toward −. Force F = q E.

  • Easy : Q = 2 μC, r = 0.3 ⇒ E = 2.0 × 10⁵ N/C.
  • Medium : Force on 5 nC = 5 × 10⁻⁹ × 2 × 10⁵ = 1 × 10⁻³ N.
  • Exam (CET) : r doubled ⇒ E? (1) ¼ (2) ½ (3) 2 (4) 4 → answer 1)

Rule 2: Uniform field — Between parallel plates E = V ÷ d; electron acceleration a = e E ÷ m.

  • Easy : V = 500, d = 0.01 ⇒ E = 5 × 10⁴ V/m; a = 8.8 × 10¹⁵ m/s².
  • Medium : A proton's a is ≈ 1/1836 of an electron's.
  • Exam (CET) : Unit of E? (1) N/C (2) N (3) C (4) J → answer 1)

Rule 3: Field lines — From + to −; never cross; density ∝ E; the tangent gives E's direction; no closed loops.

  • Easy : Dense lines ⇒ larger E.
  • Medium : If lines crossed, a point would have two E — impossible.
  • Exam (CET) : Lines run? (1) + to − (2) − to + (3) closed loops (4) crossing → answer 1)

Derivation : E = kQ/r²

  • Step 1 : Place a test charge q₀ : Coulomb force F = k Q q₀ / r².
  • Step 2 : E = F / q₀.
  • Step 3 : q₀ cancels.
  • Step 4 : E = k Q ÷ r² (along r̂).
Remember

E = F/q; E = kQ/r²; F = qE; E = V/d.

Common mistake / CET trap

E is a vector; the force on − is opposite to E.

Shortcut

Memory clue : "E = force per charge".

Spot the difference

FieldE
Point chargekQ/r²
PlatesV/d
Sheetσ/2ε₀

Practice questions

Easy
  1. 1Define E.
  2. 2Properties of field lines?
  3. 3F = ?
Medium
  1. 4Q = 5 μC, r = 0.5 : E?
  2. 5E = 3 × 10⁴ : F on 2 μC?
  3. 6V = 200, d = 0.004 : E?
Challenging
  1. 7Derive E = kQ/r².
  2. 8Electron in 2 × 10⁴ V/m : a?
Show answers (try them yourself first)
  1. 1E = F/q₀
  2. 2+ to −, never cross
  3. 3qE
  4. 41.8 × 10⁵ N/C
  5. 50.06 N
  6. 65 × 10⁴ V/m
  7. 7Use a test charge
  8. 83.5 × 10¹⁵ m/s²

Electric flux and Gauss's law

From daily life

How much rain passes through an umbrella depends on its tilt — that is flux.

Simple definition

Electric flux measures the field lines passing through a surface.

Gauss's lawqGaussian surfaceΦ = q ÷ ε₀sphere : E = kQ/r²line : λ/(2πε₀r)sheet : σ/2ε₀

In the figure : The outward flux is q/ε₀.

Rule 1: Flux — Φ = E A cosθ (N m² C⁻¹); closed surface : Φ = ∮ E·dA. Outward is +.

  • Easy : E = 100 N/C, A = 0.5 m², normal ⇒ Φ = 50 N m²/C.
  • Medium : θ = 60° ⇒ Φ = 25.
  • Exam (CET) : Surface parallel to E ⇒ Φ? (1) 0 (2) EA (3) −EA (4) infinite → answer 1)

Rule 2: Gauss's law — Φ = ∮ E·dA = q_in ÷ ε₀; only the charge inside counts. Cube with the charge at its centre ⇒ each face q/(6ε₀).

  • Easy : q = 1 μC ⇒ Φ = 1.13 × 10⁵ N m²/C.
  • Medium : One face of the cube ⇒ 1.88 × 10⁴.
  • Exam (CET) : Charge outside the surface contributes to Φ? (1) no (2) yes (3) half (4) double → answer 1)

Rule 3: Symmetric applications — Sphere/shell : E = kQ/r² (outside), 0 (inside). Line : E = λ/(2π ε₀ r). Sheet : E = σ/(2ε₀). Conductor surface : σ/ε₀.

  • Easy : λ = 1 μC/m, r = 0.5 m ⇒ E = 3.60 × 10⁴ N/C.
  • Medium : σ = 1 μC/m² ⇒ E = 5.65 × 10⁴ N/C.
  • Exam (CET) : E inside a hollow shell? (1) 0 (2) kQ/r² (3) σ/ε₀ (4) infinite → answer 1) Trick : zero inside.

Derivation : E of an infinite line

  • Step 1 : Gaussian surface : a cylinder of radius r and length l.
  • Step 2 : E is normal to the curved face and constant; Φ = E (2π r l).
  • Step 3 : Enclosed charge q = λ l.
  • Step 4 : E 2π r l = λ l / ε₀ ⇒ E = λ ÷ (2π ε₀ r).
Remember

Φ = EA cosθ; Φ = q/ε₀; E = λ/2πε₀r; σ/2ε₀.

Common mistake / CET trap

Only enclosed charge enters Φ, but E is due to all charges.

Shortcut

Memory clue : "Flux = enclosed charge ÷ ε₀".

Spot the difference

DistributionE (outside)Inside
Spherical shellkQ/r²0
Lineλ/2πε₀r—
Sheetσ/2ε₀—

Practice questions

Easy
  1. 1State Gauss's law.
  2. 2Formula for Φ?
  3. 3Unit of Φ?
Medium
  1. 4q = 5 μC : Φ?
  2. 5σ = 2 μC/m² : E?
  3. 6λ = 2 μC/m, r = 1 m : E?
Challenging
  1. 7Prove E of a sphere using Gauss's law.
  2. 8Why E = σ/ε₀ at a conductor surface?
Show answers (try them yourself first)
  1. 1Φ = q_in/ε₀
  2. 2E A cosθ
  3. 3N m² C⁻¹
  4. 45.65 × 10⁵
  5. 51.13 × 10⁵ N/C
  6. 63.60 × 10⁴ N/C
  7. 7Spherical surface : E 4πr² = Q/ε₀
  8. 8E = 0 inside; the field exists only outside

Electric dipole

From daily life

A water molecule is slightly + on one side and − on the other.

Simple definition

A dipole is a pair of equal and opposite charges a small distance apart.

Electric dipole−+p2ap = q (2a)axial E = 2kp/r³equatorial E = kp/r³τ = p E sinθ ; U = − p E cosθ

In the figure : p points from − to +.

Rule 1: Dipole moment — p = q (2a) (C m); from − to +. Equal and opposite charges; net q = 0.

  • Easy : q = 1 nC, 2a = 2 cm ⇒ p = 2.0 × 10⁻¹¹ C m.
  • Medium : p of water ≈ 6.2 × 10⁻³⁰ C m.
  • Exam (CET) : Unit of p? (1) C m (2) C (3) N (4) V → answer 1)

Rule 2: Field of a dipole — r ≫ a : axial E = 2 k p ÷ r³; equatorial E = k p ÷ r³ (opposite to p). E ∝ 1/r³.

  • Easy : p = 2 × 10⁻¹¹, r = 0.5 ⇒ axial 2.88 N/C; equatorial 1.44.
  • Medium : Axial : equatorial = 2 : 1.
  • Exam (CET) : r doubled ⇒ E? (1) 1/8 (2) ¼ (3) ½ (4) 8 → answer 1)

Rule 3: In a uniform field — Net force 0; τ = p E sinθ; U = − p E cosθ. θ = 0 stable, 180° unstable equilibrium.

  • Easy : p = 4 × 10⁻⁹, E = 10⁴, θ = 30° ⇒ τ = 2 × 10⁻⁵ N m.
  • Medium : Rotating 0° → 90° needs W = pE = 4 × 10⁻⁵ J.
  • Exam (CET) : θ = 0 ⇒ τ? (1) 0 (2) pE (3) maximum (4) infinite → answer 1)

Derivation : Axial field

  • Step 1 : On the axis at r : E = kq [1/(r − a)² − 1/(r + a)²].
  • Step 2 : = kq × 4 a r ÷ (r² − a²)².
  • Step 3 : r ≫ a ⇒ (r² − a²)² ≈ r⁴.
  • Step 4 : E = 4kqa/r³ = 2 k p ÷ r³ (p = 2qa).
Remember

p = 2qa; axial 2kp/r³; τ = pE sinθ.

Common mistake / CET trap

Equatorial E is opposite to p; net force 0 but τ ≠ 0 in a uniform field.

Shortcut

Memory clue : "Axial = double".

Spot the difference

PositionEDirection
Axial2kp/r³along p
Equatorialkp/r³opposite to p

Practice questions

Easy
  1. 1Formula for p?
  2. 2Formula for τ?
  3. 3Net charge on a dipole?
Medium
  1. 4q = 2 nC, 2a = 1 cm : p?
  2. 5p = 10⁻¹⁰, r = 1 m : axial E?
  3. 6p = 3 × 10⁻⁹, E = 2 × 10⁴, 90° : τ?
Challenging
  1. 7Derive the axial formula.
  2. 8When is U maximum/minimum?
Show answers (try them yourself first)
  1. 1q × 2a
  2. 2pE sinθ
  3. 30
  4. 42 × 10⁻¹¹ C m
  5. 51.8 N/C
  6. 66 × 10⁻⁵ N m
  7. 7Difference of two charges' fields, r ≫ a
  8. 8180° maximum (+pE); 0° minimum (−pE)

Continuous charge distributions

From daily life

Charge spread on a glass disc can be treated as countless point charges.

Simple definition

Continuous distribution: charge with density λ, σ, ρ; E = ∫ k dq r̂ / r².

Charged ringPx√(x²+R²)EE = kQx / (x²+R²)^(3/2)

In the figure : E on the axis of a ring.

Rule 1: Densities — dq = λ dl (line), σ dA (surface), ρ dV (volume). E = ∫ k dq / r² (vector).

  • Easy : Rod 0.5 m, λ = 2 μC/m ⇒ Q = 1 μC.
  • Medium : Disc r = 0.1, σ = 5 μC/m² ⇒ Q = π r² σ = 0.157 μC.
  • Exam (CET) : Unit of λ? (1) C/m (2) C (3) C/m² (4) C/m³ → answer 1)

Rule 2: Ring on its axis — E = k Q x ÷ (x² + R²)^{3/2}; 0 at x = 0; x ≫ R : kQ/x²; maximum at x = R/√2.

  • Easy : Q = 3 μC, R = x = 0.1 ⇒ E = 9.55 × 10⁵ N/C.
  • Medium : Maximum : x = R/√2 = 0.071 m.
  • Exam (CET) : E at the centre of a ring? (1) 0 (2) kQ/R² (3) infinite (4) maximum → answer 1)

Rule 3: Disc and sheet — E = (σ ÷ 2ε₀)[1 − x ÷ √(x² + R²)]; R → ∞ : σ/2ε₀ (infinite sheet).

  • Easy : σ = 1 μC/m², x = R ⇒ E = 1.65 × 10⁴ N/C.
  • Medium : x ≪ R ⇒ E ≈ σ/2ε₀.
  • Exam (CET) : R → ∞ ⇒ E? (1) σ/2ε₀ (2) 0 (3) σ/ε₀ (4) infinite → answer 1)

Derivation : E on the axis of a ring

  • Step 1 : Small dq : dE = k dq / (x² + R²); components : axial and perpendicular.
  • Step 2 : Perpendicular components cancel on opposite sides.
  • Step 3 : Axial : dE cosθ, cosθ = x/√(x² + R²).
  • Step 4 : ∫dq = Q ⇒ E = kQx ÷ (x² + R²)^{3/2}.
Remember

E_ring = kQx/(x² + R²)^{3/2}; E_sheet = σ/2ε₀.

Common mistake / CET trap

Perpendicular parts cancel; E = 0 at x = 0.

Shortcut

Memory clue : "Ring : on the axis, zero at the centre".

Spot the difference

DistributionAxial ELimit
RingkQx/(x²+R²)^{3/2}x ≫ R : kQ/x²
Disc(σ/2ε₀)(1−x/√..)R → ∞ : σ/2ε₀
Infinite sheetσ/2ε₀constant

Practice questions

Easy
  1. 1Kinds of dq?
  2. 2E at a ring's centre?
  3. 3E of an infinite sheet?
Medium
  1. 4Q = 2 μC, R = 0.2, x = 0.2 : E?
  2. 5R = 0.3 : where is E maximum?
  3. 6σ = 4 μC/m² : sheet E?
Challenging
  1. 7Prove the x for maximum E.
  2. 8Disc E for x ≫ R?
Show answers (try them yourself first)
  1. 1λ dl, σ dA, ρ dV
  2. 20
  3. 3σ/2ε₀
  4. 41.59 × 10⁵ N/C
  5. 50.212 m
  6. 62.26 × 10⁵ N/C
  7. 7dE/dx = 0 ⇒ x = R/√2
  8. 8kQ/x² (like a point charge)

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
Chargeq = n eC; A T
CoulombF = k q₁q₂/r²N; M L T⁻²
FieldE = F/q = kQ/r²N/C; M L T⁻³ A⁻¹
FluxΦ = EA cosθN m²/C
GaussΦ = q/ε₀N m²/C
LineE = λ/(2πε₀r)N/C
Dipolep = 2qa; τ = pE sinθC m; N m
RingkQx/(x²+R²)^{3/2}N/C

🎯 Most important points for the exam

  • q = n e; total q is conserved.
  • F = kq₁q₂/r²; ÷ K in a medium.
  • Add forces as vectors; equilibrium ΣF = 0.
  • E = F/q; lines go + to −, never cross.
  • Φ = q_in/ε₀; E = 0 inside a shell.
  • Axial E = 2kp/r³, equatorial kp/r³.
  • Ring : E = kQx/(x² + R²)^{3/2}.
Study mantra

q → F → E → Φ → dipole → continuous distributions.