Q1.1 What is the force between two small charged spheres having charges of 2 × 10⁻⁷ C and 3 × 10⁻⁷ C placed 30 cm apart in air?
Given:
q₁ = 2 × 10⁻⁷ C
q₂ = 3 × 10⁻⁷ C
r = 30 cm = 0.30 m
k = 9 × 10⁹ Nm²/C²
Formula:
F = k q₁ q₂ / r²
Substituting values:
F = (9 × 10⁹ × 2 × 10⁻⁷ × 3 × 10⁻⁷) / (0.30)²
F = (9 × 6 × 10⁻⁵) / 0.09
F = 6 × 10⁻³ N
Since both charges are positive, force is repulsive.
Final Answer: 6 × 10⁻³ N (Repulsive)
Q1.2 The electrostatic force on a small sphere of charge 0.4 μC due to another sphere of charge −0.8 μC in air is 0.2 N. Find (a) distance between spheres (b) force on second sphere.
Given:
F = 0.2 N
q₁ = 0.4 × 10⁻⁶ C
q₂ = 0.8 × 10⁻⁶ C
k = 9 × 10⁹ Nm²/C²
Formula:
F = k q₁ q₂ / r²
Rearranging:
r = √(k q₁ q₂ / F)
Substituting:
r = √[(9 × 10⁹ × 0.4 × 10⁻⁶ × 0.8 × 10⁻⁶) / 0.2]
r = √(1.44 × 10⁻³ / 0.2)
r = √(7.2 × 10⁻³)
r ≈ 0.085 m (≈ 8.5 cm)
(b) By Newton’s Third Law, force on second sphere = 0.2 N (Attractive)
Distance ≈ 0.085 m
Force on second sphere = 0.2 N (Attractive)
Q1.3 Show that the ratio ke² / Gmₑmₚ is dimensionless. What does it represent?
The ratio compares electric force and gravitational force between electron and proton.
Units of:
ke² → Nm²/C² × C² = Nm²
Gmₑmₚ → Nm²/kg² × kg² = Nm²
Thus units cancel.
Numerical value ≈ 10³⁹.
It is dimensionless and shows electric force is about 10³⁹ times stronger than gravitational force.
Q1.4 (a) Explain quantisation of charge. (b) Why can it be ignored for macroscopic bodies?
(a) Electric charge exists in discrete units.
q = ne
Where e = 1.6 × 10⁻¹⁹ C and n is integer.
(b) For large bodies, number of electrons transferred is extremely large, so charge appears continuous.
Charge is quantised in multiples of e but negligible for large bodies.
Q1.5 Define electric field and derive its expression for a point charge.
Electric field at a point is defined as force experienced by unit positive test charge.
E = F / q₀
From Coulomb’s law:
F = kQq₀ / r²
Substituting:
E = kQ / r²
Direction: Radially outward for positive charge.
Electric field due to point charge: E = kQ / r²
Q1.6 Define electric dipole and dipole moment.
Electric dipole consists of two equal and opposite charges separated by small distance.
Dipole moment:
p = q × 2a
Unit: C·m
Direction: From negative to positive charge.
Dipole moment p = q × separation
Q1.7 Derive electric field on axial line of dipole.
Using superposition principle and assuming r >> a:
E = (1 / 4πε₀) × (2p / r³)
Field is along dipole axis.
E_axial = (1 / 4πε₀) × (2p / r³)
Q1.8 Derive electric field on equatorial line of dipole.
On equatorial line, horizontal components cancel.
E = (1 / 4πε₀) × (p / r³)
Direction opposite to dipole moment.
E_equatorial = (1 / 4πε₀) × (p / r³)
Q1.9 Define electric flux. Give its mathematical expression and SI unit.
Electric flux through a surface is defined as the total number of electric field lines passing through that surface.
If a surface of area A is placed in an electric field E making angle θ with the normal to the surface, then
Φ = E A cosθ
Where
Φ = electric flux
E = electric field
A = area
θ = angle between electric field and normal to surface
SI Unit: Nm²/C
Electric flux Φ = EA cosθ
SI unit = Nm²/C
Q1.10 State Gauss’s Law in electrostatics.
Gauss’s Law states that the total electric flux through any closed surface is equal to the net charge enclosed divided by permittivity of free space.
Φ = Q_enclosed / ε₀
Where
ε₀ = 8.854 × 10⁻¹² C²/Nm²
Important point: It depends only on enclosed charge, not on shape of surface.
Φ = Q_enclosed / ε₀
Q1.11 Using Gauss’s law, derive electric field due to an infinitely long straight charged wire.
Consider a cylindrical Gaussian surface of radius r and length L.
Charge enclosed = λL
From Gauss's law:
Φ = E (2πrL)
E (2πrL) = λL / ε₀
Cancel L:
E = λ / (2πε₀ r)
Thus electric field decreases inversely with distance.
E = λ / (2πε₀ r)
Q1.12 Using Gauss’s law, derive electric field due to an infinite plane sheet of charge.
Consider a cylindrical pillbox Gaussian surface.
Charge enclosed = σA
Total flux = 2EA
From Gauss’s law:
2EA = σA / ε₀
Cancel A:
E = σ / (2ε₀)
Field is uniform and independent of distance.
E = σ / (2ε₀)
Q1.13 Using Gauss’s law, find electric field due to uniformly charged spherical shell.
Case 1: Outside shell (r > R)
Charge enclosed = Q
E (4πr²) = Q / ε₀
E = kQ / r²
Case 2: Inside shell (r < R)
Charge enclosed = 0
E = 0
Thus field inside shell is zero.
Outside: E = kQ/r²
Inside: E = 0
Q1.14 What is electrostatic shielding?
Electrostatic shielding is the phenomenon of protecting a region from external electric field by enclosing it inside a conducting shell.
Since electric field inside conductor is zero, any object inside remains unaffected.
Example: Faraday cage.
Inside conductor, E = 0 → shielding occurs.
Q1.15 Derive expression for torque on electric dipole in uniform electric field.
In uniform field E, forces on charges form a couple.
Torque = Force × perpendicular distance
Force on each charge = qE
Perpendicular separation = 2a sinθ
Thus,
τ = qE (2a sinθ)
Since p = q(2a),
τ = pE sinθ
Maximum torque when θ = 90°.
τ = pE sinθ
Q1.16 Derive expression for potential energy of dipole in uniform electric field.
Work done in rotating dipole from 90° to angle θ gives potential energy.
U = -pE cosθ
When θ = 0° → U minimum
When θ = 180° → U maximum
U = -pE cosθ
Q1.17 Why do electric field lines never intersect each other?
Electric field at any point has a unique direction.
If two field lines intersect, it would mean that at the point of intersection the electric field has two different directions simultaneously, which is physically impossible.
Therefore, electric field lines can never intersect.
Electric field at a point has only one direction → hence field lines never intersect.
Q1.18 What are the properties of electric field lines?
Important properties of electric field lines:
1. They originate from positive charge and terminate on negative charge.
2. The tangent at any point gives direction of electric field.
3. No two field lines intersect.
4. Closer the lines, stronger the electric field.
5. They are perpendicular to surface of conductor.
Field lines show direction and strength of electric field.
Q1.19 Explain why electric field inside a conductor is zero in electrostatic equilibrium.
In a conductor, free electrons are present.
If electric field exists inside conductor, electrons will experience force and start moving.
They continue to move until they rearrange themselves in such a way that internal electric field cancels the applied field.
Thus, in electrostatic equilibrium:
E_inside = 0
Electric field inside conductor in electrostatic equilibrium is zero.
Q1.20 Why is electric field at the surface of conductor always perpendicular to the surface?
If electric field had a component parallel to the surface, free charges would move along the surface.
But in electrostatic equilibrium, charges are at rest.
Hence, tangential component must be zero and electric field is perpendicular (normal) to surface.
Electric field at conductor surface is always normal to the surface.
Q1.21 What is electrostatic shielding? Give an example.
Electrostatic shielding is the phenomenon in which a region is protected from external electric field by enclosing it inside a conductor.
Since electric field inside conductor is zero, the inside region remains unaffected.
Example: Sensitive electronic devices kept inside metallic enclosure (Faraday cage).
Electrostatic shielding protects interior from external electric field.
Q1.22 Why does charge reside only on the outer surface of a conductor?
If charge were present inside conductor, it would create electric field inside.
But we know that electric field inside conductor is zero in electrostatic equilibrium.
Therefore, excess charge must reside on outer surface.
Excess charge on conductor resides only on outer surface.
Q1.23 What is surface charge density?
Surface charge density is defined as charge per unit area.
σ = Q / A
SI unit: C/m²
Surface charge density σ = Q/A
Q1.24 What is linear charge density?
Linear charge density is defined as charge per unit length.
λ = Q / L
SI unit: C/m
Linear charge density λ = Q/L
Q1.25 What is volume charge density?
Volume charge density is defined as charge per unit volume.
ρ = Q / V
Where
ρ = volume charge density
Q = total charge
V = volume
SI unit: C/m³
Volume charge density ρ = Q/V (Unit: C/m³)
Q1.26 What is dielectric constant? Write its effect on Coulomb’s force.
Dielectric constant (K) of a medium is defined as the ratio of electrostatic force between two charges in vacuum to the force between them in that medium.
K = F_vacuum / F_medium
Effect on Coulomb’s force:
F = (1 / 4πε₀K) × (q₁q₂ / r²)
Thus, force decreases by a factor K when medium is introduced.
Force reduces by factor K in dielectric medium.
Q1.27 What is polarization of dielectric?
When a dielectric is placed in an external electric field, the positive and negative charges inside its molecules slightly separate.
This separation produces induced dipole moments.
This phenomenon is called polarization.
Polarization is the alignment of induced dipoles inside dielectric.
Q1.28 Define electric dipole moment and its direction.
Electric dipole moment is defined as product of magnitude of one charge and separation between the two charges.
p = q × d
Direction: From negative charge to positive charge.
SI unit: C·m
Dipole moment p = qd (Direction: − to +)
Q1.29 Why do sharp points on a charged conductor leak charge?
At sharp points, surface curvature is very small.
Electric field is inversely proportional to radius of curvature.
Thus electric field becomes very strong at sharp points.
Strong field ionizes surrounding air and charge escapes.
Strong electric field at sharp points causes charge leakage.
Q1.30 Why is electrostatic force conservative in nature?
A force is conservative if work done depends only on initial and final positions, not on path.
In electrostatics, work done in moving a charge between two points is independent of path.
Hence electrostatic force is conservative.
Electrostatic force is conservative because work done is path independent.
Q1.31 What happens to electric field inside a cavity within a conductor?
If conductor is in electrostatic equilibrium and cavity contains no charge:
Electric field inside cavity = 0.
If charge is placed inside cavity:
Field exists inside cavity but remains zero inside conducting material.
E inside cavity = 0 (if no internal charge)
Q1.32 State superposition principle in electrostatics.
The net electric field at a point due to several charges is equal to the vector sum of electric fields due to individual charges.
E_total = E₁ + E₂ + E₃ + ...
Net field is vector sum of individual fields.
Q1.33 Why is electric field zero inside hollow charged spherical conductor?
Using Gauss’s Law:
For Gaussian surface inside hollow conductor:
Charge enclosed = 0
Φ = Q_enclosed / ε₀ = 0
Therefore,
E = 0
Electric field inside hollow conductor is zero.
Q1.34 Why does a charged conductor have maximum charge density at sharp edges?
Surface charge density is higher where radius of curvature is smaller.
Sharp edges have very small radius of curvature.
Thus charge accumulates more at sharp edges.
This leads to strong electric field at those points.
Maximum charge density occurs at sharp edges.