Chapter 1: Electric Charges and Fields

1. Introduction to Electric Charge

Historically, it was discovered that amber rubbed with wool or silk cloth attracts light objects. There are two kinds of electrification, and we find that like charges repel and unlike charges attract each other. The property which differentiates the two kinds of charges is called the polarity of charge .

Conductors and Insulators

Substances that readily allow the passage of electricity through them are called conductors (e.g., metals, human bodies) [4]. Those which offer high resistance to the passage of electricity are called insulators (e.g., glass, plastic, nylon) .

Basic Properties of Electric Charge

• Additivity of Charges: If a system contains n charges, the total charge is the algebraic sum of all individual charges: q = q₁ + q₂ + … + q_n [4].
• Conservation of Charge: Within an isolated system, the total charge is always conserved. Charges can neither be created nor destroyed [5].
• Quantisation of Charge: All free charges are integral multiples of a basic unit of charge denoted by e. q = ne, where n is an integer. The value of e is 1.602 × 10^-19 C .

2. Coulomb’s Law

Coulomb’s law states that the electrical force between two point charges varies inversely as the square of the distance between them and is directly proportional to the product of the magnitude of the two charges .

Formula: F = k · (|q₁ q₂|) / r²

Where constant k = 1 / (4πε₀) = 9 × 10⁹ N m² C^-2. ε₀ is the permittivity of free space.

Forces between Multiple Charges (Superposition Principle)

The force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time.

3. Electric Field and Field Lines

The electric field produced by a charge Q at a point r is defined as the force exerted on a unit positive test charge placed at that point [14].

Formula: E = 1 / (4πε₀) · (Q / r²)

Electric Field Lines Properties:

• Field lines start from positive charges and end at negative charges.
• In a charge-free region, electric field lines can be taken to be continuous curves without any breaks.
• Two field lines can never cross each other. If they did, the field at the point of intersection will not have a unique direction.
• Electrostatic field lines do not form any closed loops.

4. Electric Flux & Gauss’s Law

Electric Flux (Δφ): It is proportional to the number of field lines cutting the area element. Δφ = E · ΔS = E ΔS cosθ.

Gauss’s Law: The electric flux through a closed surface S is equal to q / ε₀, where q is the total charge enclosed by S.

Formula: φ = q / ε₀

5. Electric Dipole

An electric dipole is a pair of equal and opposite point charges q and -q, separated by a distance 2a. The dipole moment vector p has magnitude 2aq and is directed from -q to q .

• Field on Axis: E = 2p / (4πε₀ r³) for r >> a .
• Field on Equatorial Plane: E = p / (4πε₀ r³) for r >> a .
• Torque in uniform external field: τ = p × E .

6. Applications of Gauss’s Law

• Field due to an infinitely long straight uniformly charged wire: E = λ / (2πε₀ r) .
• Field due to a uniformly charged infinite plane sheet: E = σ / (2ε₀) .
• Field due to a uniformly charged thin spherical shell: Outside the shell (r ≥ R): E = q / (4πε₀ r²). Inside the shell (r < R): E = 0 .

Textbook Exercise Solutions (Complete)

Q 1.1: 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?
Solution:
Given, q₁ = 2 × 10^-7 C, q₂ = 3 × 10^-7 C, r = 30 cm = 0.3 m.
Force F = (1 / 4πε₀) · (q₁q₂ / r²)
F = (9 × 10⁹ × 2 × 10^-7 × 3 × 10^-7) / (0.3)² = 6 × 10^-3 N.
Force is repulsive as both charges are positive.

Q 1.2: The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge -0.8 μC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first? Solution:
(a) F = 0.2 N, q₁ = 0.4 × 10^-6 C, q₂ = 0.8 × 10^-6 C.
r² = (k · q₁q₂) / F = (9 × 10⁹ × 0.4 × 10^-6 × 0.8 × 10^-6) / 0.2 = 0.0144 m².
r = 0.12 m = 12 cm.
(b) By Newton’s third law, the force on the second sphere is equal and opposite. So, force is 0.2 N (attractive).

Q 1.3: Check that the ratio ke² / G m_e m_p is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Solution:
Units: k (N m² / C²), e² (C²), G (N m² / kg²), m_em_p (kg²).
Ratio units = (N m² C^-2 · C²) / (N m² kg^-2 · kg²) = 1 (Dimensionless).
Value = (9×10⁹ × (1.6×10^-19)²) / (6.67×10^-11 × 9.1×10^-31 × 1.67×10^-27) ≈ 2.3 × 10³⁹.
It signifies that electrostatic force is enormously stronger than gravitational force.

Q 1.4: (a) Explain the meaning of ‘electric charge of a body is quantised’. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic charges?
Solution:
(a) It means charge on a body is always an integral multiple of elementary charge e (1.6 × 10^-19 C). q = ne.
(b) At the macroscopic level, charges are very large. Because ‘e’ is extremely small, the discrete nature of charge is lost and it appears continuous.

Q 1.5: When a glass rod is rubbed with a silk cloth, charges appear on both. Explain how this observation is consistent with the law of conservation of charge.
Solution:
Before rubbing, both are neutral (net charge zero). When rubbed, electrons transfer from glass to silk. Glass acquires a positive charge (+q) and silk acquires an equal negative charge (-q). Net charge is still (+q – q = 0). Hence, charge is conserved.

Q 1.6: Four point charges q_A = 2 μC, q_B = -5 μC, q_C = 2 μC, and q_D = -5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?
Solution:
The charge at centre O is 1 μC. Charges at opposite corners A and C are equal (2 μC), so they exert equal and opposite repulsive forces on the 1 μC charge, cancelling each other out. Similarly, charges at B and D (-5 μC) exert equal and opposite attractive forces, cancelling each other. Thus, the net force is Zero.

Q 1.7: (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not? (b) Explain why two field lines never cross each other at any point.
Solution:
(a) A field line represents the path of a unit positive test charge. Since the electrostatic force is continuous, the test charge moves continuously. It cannot jump from one point to another, so there are no breaks.
(b) If two field lines crossed, it would mean that at the point of intersection, there are two different directions of the electric field (two tangents), which is physically impossible.

Q 1.8: Two point charges q_A = 3 μC and q_B = -3 μC are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude 1.5 × 10^-9 C is placed at this point, what is the force experienced by the test charge?
Solution:
(a) Midpoint O is 10 cm (0.1 m) from both charges.
Field due to q_A: E_A = (9×10⁹ × 3×10^-6) / (0.1)² = 2.7 × 10⁶ N/C (towards O).
Field due to q_B: E_B = (9×10⁹ × 3×10^-6) / (0.1)² = 2.7 × 10⁶ N/C (towards B).
Net field E = E_A + E_B = 5.4 × 10⁶ N/C directed along OB.
(b) Force F = qE = (1.5 × 10^-9) × (5.4 × 10⁶) = 8.1 × 10^-3 N. Since the test charge is negative, the force is opposite to the field, i.e., directed along OA.

Q 1.9: A system has two charges q_A = 2.5 × 10^-7 C and q_B = -2.5 × 10^-7 C located at points A: (0, 0, -15 cm) and B: (0, 0, +15 cm), respectively. What are the total charge and electric dipole moment of the system? [42]
Solution:
Total charge = q_A + q_B = 2.5 × 10^-7 – 2.5 × 10^-7 = 0.
Distance 2a = 15 cm + 15 cm = 30 cm = 0.3 m.
Dipole moment p = q × 2a = (2.5 × 10^-7) × 0.3 = 7.5 × 10^-8 C m. The direction is from negative to positive charge, i.e., from B to A (along negative z-axis).

Q 1.10: An electric dipole with dipole moment 4 × 10^-9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 10⁴ NC^-1. Calculate the magnitude of the torque acting on the dipole. [42]
Solution:
Torque τ = pE sinθ = (4 × 10^-9) × (5 × 10⁴) × sin(30°)
τ = 20 × 10^-5 × 0.5 = 10^-4 N m.

Q 1.11: A polythene piece rubbed with wool is found to have a negative charge of 3 × 10^-7 C. (a) Estimate the number of electrons transferred (from which to which?) (b) Is there a transfer of mass from wool to polythene?
Solution:
(a) q = ne ⇒ n = q / e = (3 × 10^-7) / (1.6 × 10^-19) = 1.875 × 10¹² electrons. Since polythene is negative, electrons are transferred from wool to polythene.
(b) Yes, there is a transfer of mass because electrons have mass. Mass transferred = n × m_e = 1.875 × 10¹² × 9.1 × 10^-31 = 1.7 × 10^-18 kg.

Q 1.12: (a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10^-7 C? (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved? [42]
Solution:
(a) F = k · (q₁q₂ / r²) = (9 × 10⁹ × (6.5 × 10^-7)²) / (0.5)² = 1.52 × 10^-2 N.
(b) New charges: q₁‘ = 2q₁, q₂‘ = 2q₂. New distance r’ = r/2.
New Force F’ = k (2q₁)(2q₂) / (r/2)² = 16 × (k q₁q₂ / r²) = 16 × 1.52 × 10^-2 = 0.243 N.

Q 1.13: Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?
Solution:
Particles 1 and 2 bend towards the positive plate, so they must be negatively charged. Particle 3 bends towards the negative plate, so it is positively charged. Deflection in an electric field is directly proportional to the charge-to-mass ratio (e/m). Since particle 3 shows the maximum deflection, it has the highest charge-to-mass ratio.

Q 1.14: Consider a uniform electric field E = 3 × 10³ î N/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?
Solution:
Area A = 10 cm × 10 cm = 0.01 m².
(a) Plane is parallel to yz plane, so area vector is along x-axis (A = 0.01 î). Flux φ = E · A = (3 × 10³) × 0.01 = 30 N m² C^-1.
(b) Angle θ = 60°. φ = EA cos(60°) = 30 × 0.5 = 15 N m² C^-1.

Q 1.15: What is the net flux of the uniform electric field of Exercise 1.14 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?
Solution:
For any closed surface in a uniform electric field, the number of field lines entering equals the number of field lines leaving. Thus, the net flux through the cube is Zero.

Q 1.16: Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 10³ Nm²/C. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?
Solution:
(a) From Gauss’s Law, φ = q / ε₀ ⇒ q = φ × ε₀ = 8.0 × 10³ × 8.854 × 10^-12 = 0.07 μC.
(b) No, we cannot conclude that there are no charges inside. We can only conclude that the net charge inside is zero. There could be equal amounts of positive and negative charges inside.

Q 1.17: A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm. What is the magnitude of the electric flux through the square?
Solution:
We can imagine the square as one face of a cube of edge 10 cm, with the charge q at its center. Total flux through the cube is q / ε₀. Since the cube has 6 identical faces, the flux through one square face is φ = (1/6) × (q / ε₀).
φ = (1/6) × (10 × 10^-6) / (8.854 × 10^-12) = 1.88 × 10⁵ N m² C^-1.

Q 1.18: A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
Solution:
By Gauss’s Law, the net flux through a closed surface depends only on the enclosed charge. φ = q / ε₀.
φ = (2.0 × 10^-6) / (8.854 × 10^-12) = 2.26 × 10⁵ N m² C^-1.

Q 1.19: A point charge causes an electric flux of -1.0 × 10³ Nm²/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?
Solution:
(a) Flux depends only on the charge enclosed, not on the radius of the surface. So, the flux would remain -1.0 × 10³ N m² C^-1.
(b) q = φ × ε₀ = (-1.0 × 10³) × (8.854 × 10^-12) = -8.854 × 10^-9 C.

Q 1.20: A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 10³ N/C and points radially inward, what is the net charge on the sphere?
Solution:
E = k · q / r². The field points inward, so charge is negative.
q = (E × r²) / k = (1.5 × 10³ × (0.2)²) / (9 × 10⁹) = 6.67 × 10^-9 C.
Therefore, net charge is -6.67 nC.

Q 1.21: A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m². (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?
Solution:
Radius r = 2.4 / 2 = 1.2 m. σ = 80 × 10^-6 C/m².
(a) Charge q = σ × Area = σ × 4πr² = 80 × 10^-6 × 4 × 3.14 × (1.2)² = 1.45 × 10^-3 C.
(b) Total flux φ = q / ε₀ = (1.45 × 10^-3) / (8.854 × 10^-12) = 1.6 × 10⁸ N m² C^-1.

Q 1.22: An infinite line charge produces a field of 9 × 10⁴ N/C at a distance of 2 cm. Calculate the linear charge density.
Solution:
Electric field E = λ / (2πε₀r). We know 1 / (4πε₀) = 9 × 10⁹, so 1 / (2πε₀) = 18 × 10⁹.
9 × 10⁴ = λ × 18 × 10⁹ / 0.02
λ = (9 × 10⁴ × 0.02) / (18 × 10⁹) = 10^-7 C/m.

Q 1.23: Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10^-22 C/m². What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?
Solution:
(a) In the outer region of the first plate, the fields due to the positive and negative plates are equal and opposite, cancelling each other out. So, E = 0.
(b) Similarly, in the outer region of the second plate, E = 0.
(c) Between the plates, the fields due to both plates add up: E = σ/2ε₀ + σ/2ε₀ = σ/ε₀.
E = (17.0 × 10^-22) / (8.854 × 10^-12) = 1.92 × 10^-10 N/C.