Chapter 1: Electric Charges and Fields
Key Formulas:
• Coulomb's Law: F = kq₁q₂/r², k = 9 × 109 Nm²/C²
• Electric Field: E = F/q = kQ/r²
• Gauss's Law: ∮ E · dA = qenc/ε₀
• Force on charge in E-field: F = qE
Q1. Two point charges q₁ = 2 × 10−6 C and q₂ = −2 × 10−6 C are placed 15 cm apart. Find the electric force between them. Is it attractive or repulsive?
Solution
F = k|q₁||q₂|/r²
= (9 × 109)(2 × 10−6)(2 × 10−6)/(0.15)²
= (9 × 109)(4 × 10−12)/(2.25 × 10−2)
= 36 × 10−3/(2.25 × 10−2)
= 1.6 N
Since charges are opposite, force is attractive.
Q2. An electric dipole of dipole moment p is placed in a uniform electric field E. Write the expression for the torque experienced by the dipole. Identify two orientations for which torque is (i) maximum (ii) zero.
Solution
Torque: τ = p × E = pE sinθ
(i) Maximum when sinθ = 1, i.e., θ = 90° (dipole ⊥ to E)
(ii) Zero when sinθ = 0, i.e., θ = 0° or 180° (dipole || to E)
Q3. Using Gauss's law, find the electric field due to an infinitely long straight uniformly charged wire with linear charge density λ.
Solution
Consider a Gaussian cylinder of radius r and length l.
By symmetry, E is radial and constant on the curved surface.
∮ E · dA = E × 2πrl = qenc/ε₀ = λl/ε₀
E = λ/(2πε₀r)
Q4. Calculate the electric field at a point on the equatorial line of an electric dipole of length 2a with dipole moment p, at a distance r from the centre (r >> a).
Solution
For equatorial line (r >> a):
E = −kp/r³ or E = kp/(4πε₀r³)
Direction is opposite to the dipole moment (antiparallel to p).
Q5. A point charge +q is placed at the centre of a cube of side a. What is the electric flux through one face of the cube?
Solution
Total flux through cube = q/ε₀ (by Gauss's law)
By symmetry, flux through each face is equal.
Flux through one face = q/(6ε₀)
Q6. Two point charges −q and +q are placed at distance d apart. Find the electric field at a point on the perpendicular bisector at distance d/2 from the midpoint.
Solution
Distance from each charge to the point:
r = √((d/2)² + (d/2)²) = d/√2
E₁ = E₂ = kq/r² = 2kq/d²
Horizontal components cancel; vertical components add:
E = 2E₁ cosθ = 2(2kq/d²)(d/2)/(d/√2)
= 2(2kq/d²)(1/√2)
E = 2√2 kq/d² (along the perpendicular bisector)
Q7. A uniformly charged thin spherical shell of radius R carries charge Q. Find the electric field at a point (i) outside the shell (ii) on the surface (iii) inside the shell. Draw the variation of E with r.
Solution
By Gauss's law:
(i) For r > R: E = kQ/r² (behaves as point charge)
(ii) For r = R: E = kQ/R²
(iii) For r < R: E = 0 (no charge enclosed)
The field is zero inside, increases linearly from 0 at r=0... no, the field inside a shell is zero. Outside, it decreases as 1/r².
Q8. Two charges +q and +4q are placed at a distance d apart. Find the point on the line joining them where the electric field is zero.
Solution
Let the point be at distance x from +q (between the charges, since both are positive).
kq/x² = k(4q)/(d−x)²
1/x² = 4/(d−x)²
(d−x)/x = 2
d−x = 2x ⇒ x = d/3
The point is at distance d/3 from +q (between the charges).
Q9. Calculate the electric flux through a square of side 10 cm placed in a uniform electric field of 100 N/C, if the electric field makes an angle of 60° with the normal to the square.
Solution
Φ = E · A cosθ
A = (0.1)² = 0.01 m²
Φ = 100 × 0.01 × cos 60°
= 1 × 0.5 = 0.5 N·m²/C
Q10. An electric dipole of dipole moment p is placed along the x-axis with its centre at the origin. Find the electric field at a point on the y-axis at distance d from the origin (d >> a).
Solution
On the equatorial line (y-axis) of the dipole:
E = −kp/d³ = −p/(4πε₀d³)
The field is antiparallel to the dipole moment (along −x direction).
Q11. A charge Q is distributed uniformly over a thin ring of radius R. Find the electric field at a point on the axis of the ring at distance x from its centre.
Solution
Consider a ring element of radius r and width dr.
dq = (Q/2πR) · 2πr dr = (Q/R) r dr
By symmetry, only axial component survives:
dE = (kx dq)/(r² + x²)3/2
E = kQx/(R² + x²)3/2
= Qx/(4πε₀(R² + x²)3/2)
Q12. What is the force between two small charged spheres of charges 2 × 10−7 C and 3 × 10−7 C placed 30 cm apart in air? If the force is reduced by a factor of 10 by a medium, what is the dielectric constant of the medium?
Solution
In air:
F = kq₁q₂/r² = (9 × 109)(2 × 10−7)(3 × 10−7)/(0.3)²
= (9 × 109)(6 × 10−14)/0.09
= 6 × 10−3 N
In medium: F' = F/K = F/10
K = 10 (dielectric constant of the medium)
Q13. A point charge q is placed at a distance d from the centre of an uncharged conducting sphere of radius R (d > R). Find the potential at the centre of the sphere.
Solution
The potential at the centre is the sum of the potential due to the point charge and the induced charges.
Due to point charge: V₁ = kq/d
Due to induced charges: The total induced charge is zero, and by properties of conductors, the potential due to induced charges at the centre cancels the tangential component...
For a conductor, the potential is the same everywhere inside and equals the surface potential.
V = kq/d (potential at centre = potential at surface = constant throughout the conductor)
Q14. Derive the expression for electric field due to a uniformly charged thin disc of radius R and surface charge density σ at a point on its axis at distance x from its centre.
Solution
Consider a ring element of radius r and width dr.
dq = σ · 2πr dr
dE = kx dq/(r² + x²)3/2
E = ∫ kx(σ · 2πr dr)/(r² + x²)3/2
Let u = r² + x², du = 2r dr:
E = πσx ∫ du/u3/2
E = (σ/2ε₀)[1 − x/√(R² + x²)]
For R → ∞ (infinite sheet): E = σ/(2ε₀)
Q15. A charge +Q is placed at each corner of a square of side a. Find the electric field at the centre of the square.
Solution
By symmetry, the fields due to charges at opposite corners cancel each other.
Distance from each corner to centre = a/√2
Each charge produces E = kQ/(a/√2)² = 2kQ/a²
Charges at opposite corners produce equal and opposite fields.
E = 0 at the centre of the square.
Q16. An oil drop of mass 3.2 × 10−15 kg is in equilibrium between two horizontal parallel plates 1 cm apart. If the drop carries a charge of 8 × 10−19 C, find the potential difference between the plates. (Take g = 10 m/s²)
Solution
At equilibrium: qE = mg
E = mg/q = (3.2 × 10−15)(10)/(8 × 10−19)
= 3.2 × 10−14/(8 × 10−19) = 4 × 104 N/C
V = Ed = (4 × 104)(0.01) = 400 V