Exercises
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?
Answer
Given: q₁ = 2 × 10⁻⁷ C, q₂ = 3 × 10⁻⁷ C, r = 0.30 m.
F = kq₁q₂/r²
⇒ F = (9 × 10⁹ × 2 × 10⁻⁷ × 3 × 10⁻⁷)/(0.30)²
⇒ F = (5.4 × 10⁻⁴)/0.09
⇒ F = 6 × 10⁻³ N
Both charges are positive, so the force is repulsive.
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?
Answer
(a) Given: q₁ = 0.4 × 10⁻⁶ C, q₂ = 0.8 × 10⁻⁶ C (magnitude), F = 0.2 N.
r² = kq₁q₂/F
⇒ r² = (9 × 10⁹ × 0.4 × 10⁻⁶ × 0.8 × 10⁻⁶)/0.2
⇒ r² = 1.44 × 10⁻² m²
⇒ r = 0.12 m = 12 cm
(b) By Newton's third law, the second sphere experiences a force of the same magnitude, 0.2 N, in the opposite direction. Since the charges are unlike, the force is attractive, i.e., directed towards the first sphere.
3. Check that the ratio ke²/G mₑmₚ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Answer
ke²/r² is a force and Gmₑmₚ/r² is also a force. Hence ke²/(Gmₑmₚ) = (ke²/r²)/(Gmₑmₚ/r²) is a ratio of two forces and has no dimensions.
Dimensional check: [k] = N m² C⁻², [e²] = C², [G] = N m² kg⁻², [mₑmₚ] = kg²
⇒ (N m² C⁻² × C²)/(N m² kg⁻² × kg²) = 1, i.e., dimensionless.
Value: k = 9 × 10⁹ N m² C⁻², e = 1.6 × 10⁻¹⁹ C, G = 6.67 × 10⁻¹¹ N m² kg⁻², mₑ = 9.1 × 10⁻³¹ kg, mₚ = 1.67 × 10⁻²⁷ kg.
Ratio = (9 × 10⁹ × (1.6 × 10⁻¹⁹)²)/(6.67 × 10⁻¹¹ × 9.1 × 10⁻³¹ × 1.67 × 10⁻²⁷)
⇒ (2.304 × 10⁻²⁸)/(1.014 × 10⁻⁶⁷)
⇒ ≈ 2.3 × 10³⁹
Significance: it is the ratio of the electrostatic force to the gravitational force between an electron and a proton at any separation. The electric force is about 10³⁹ times stronger, so gravity is negligible at atomic scales.
4. (a) Explain the meaning of the statement ‘electric charge of a body is quantised’. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?
Answer
(a) Charge on any body can only be an integral multiple of the basic charge e (1.6 × 10⁻¹⁹ C), i.e., q = ne, where n = 0, ±1, ±2, … A body cannot carry a charge such as 0.5e or 2.7e, because charge is exchanged only as whole electrons (or protons).
(b) Macroscopic charges are enormous compared with e (for example 1 μC contains about 6 × 10¹² units of e). A change of one unit e is far too small to detect, so the charge behaves as if it varies continuously, and quantisation can be ignored.
5. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.
Answer
Rubbing does not create charge; it only transfers electrons from one body to the other. The glass rod loses some electrons and becomes positive, while the silk gains exactly those electrons and becomes equally negative. Before rubbing the total charge of the isolated pair was zero, and after rubbing it is still (+q) + (–q) = 0. Since charges always appear in equal and opposite amounts on the two bodies, the total charge of the system is conserved.
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?
Answer
The centre O is equidistant from all four corners (distance = 10/√2 cm ≈ 7.07 cm).
A and C are opposite corners carrying equal charges (2 μC each). Their forces on the 1 μC charge at O are equal in magnitude and opposite in direction, so they cancel.
Similarly, B and D are opposite corners carrying equal charges (–5 μC each), so their forces on the charge at O also cancel.
Hence the net force on the 1 μC charge at the centre is zero.
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?
Answer
(a) A field line traces the path a free positive test charge would tend to follow. In a charge-free region the field exists at every point and varies continuously, so a test charge moves smoothly without jumping from one point to another. Hence a field line is a continuous curve with no breaks (it starts on a positive charge and ends on a negative charge or at infinity).
(b) The tangent to a field line at a point gives the direction of the field there. If two lines crossed, there would be two tangents, i.e., two directions of the field at the same point, which is impossible because the field at a point has a unique direction. So field lines never intersect.
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?
Answer
(a) Distance of O from each charge: r = 10 cm = 0.1 m.
Field due to A: E_A = kq/r²
⇒ (9 × 10⁹ × 3 × 10⁻⁶)/(0.1)²
⇒ 2.7 × 10⁶ N/C, directed away from A, i.e., towards B.
Field due to B: E_B = 2.7 × 10⁶ N/C, directed towards B (B is negative).
Both point the same way, so E = E_A + E_B = 5.4 × 10⁶ N/C, along OB (from A towards B).
(b) F = qE
⇒ F = 1.5 × 10⁻⁹ × 5.4 × 10⁶
⇒ F = 8.1 × 10⁻³ N
Since the test charge is negative, the force is opposite to the field, i.e., directed along OA (towards A).
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?
Answer
Total charge = 2.5 × 10⁻⁷ + (–2.5 × 10⁻⁷) = 0.
Separation 2a = 15 + 15 = 30 cm = 0.30 m.
p = q × 2a
⇒ p = 2.5 × 10⁻⁷ × 0.30
⇒ p = 7.5 × 10⁻⁸ C m
The dipole moment points from the negative charge to the positive charge, i.e., from B to A, along the negative z-axis.
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 N C⁻¹. Calculate the magnitude of the torque acting on the dipole.
Answer
τ = pE sin θ
⇒ τ = 4 × 10⁻⁹ × 5 × 10⁴ × sin 30°
⇒ τ = 2 × 10⁻⁴ × 0.5
⇒ τ = 10⁻⁴ N m
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?
Answer
(a) n = q/e
⇒ n = (3 × 10⁻⁷)/(1.6 × 10⁻¹⁹)
⇒ n ≈ 1.87 × 10¹² electrons. Since polythene becomes negative, electrons are transferred from wool to polythene.
(b) Yes, because electrons have mass. Mass transferred = n × mₑ
⇒ 1.87 × 10¹² × 9.1 × 10⁻³¹ kg
⇒ ≈ 1.7 × 10⁻¹⁸ kg. This is negligibly small.
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? The radii of A and B are negligible compared to the distance of separation. (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
Answer
(a) F = kq²/r²
⇒ F = (9 × 10⁹ × (6.5 × 10⁻⁷)²)/(0.5)²
⇒ F = (3.80 × 10⁻³)/0.25
⇒ F ≈ 1.5 × 10⁻² N
(b) New charge = 2q on each, new distance = r/2.
F′ = k(2q)(2q)/(r/2)² = 16 kq²/r²
⇒ F′ = 16 × 1.52 × 10⁻² N ≈ 0.24 N
13. Figure 1.30 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?
Answer
Unlike charges attract. Particles 1 and 2 bend towards the positively charged plate, so they are negatively charged. Particle 3 bends towards the negatively charged plate, so it is positively charged.
For the same speed entering the field, the sideways deflection is proportional to q/m. Particle 3 shows the largest deflection, so it has the highest charge to mass ratio.
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?
Answer
Area A = (0.1)² = 10⁻² m².
(a) Plane parallel to yz plane
⇒ normal along x-axis, θ = 0°. φ = EA cos 0°
⇒ 3 × 10³ × 10⁻²
⇒ φ = 30 N m²/C
(b) θ = 60°. φ = EA cos 60°
⇒ 30 × 0.5
⇒ φ = 15 N m²/C
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?
Answer
The field lines entering through one face (perpendicular to x) leave through the opposite face; the flux through the other four faces is zero because the field is parallel to them. Inward flux = outward flux, so the net flux through the cube is zero. (Equivalently, the cube encloses no charge, so by Gauss's law φ = 0.)
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 N m²/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?
Answer
(a) Gauss's law: φ = q/ε₀
⇒ q = ε₀φ
⇒ q = 8.854 × 10⁻¹² × 8.0 × 10³
⇒ q ≈ 7.08 × 10⁻⁸ C ≈ 0.07 μC
(b) No. Zero net flux only means the net (algebraic sum of) charge enclosed is zero. The box could contain equal amounts of positive and negative charge.
17. A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.31. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)
Answer
Imagine a cube of edge 10 cm with the charge at its centre; the square is one of its six faces. By symmetry each face receives one-sixth of the total flux.
φ = (1/6)(q/ε₀)
⇒ φ = 10 × 10⁻⁶/(6 × 8.854 × 10⁻¹²)
⇒ φ ≈ 1.88 × 10⁵ N m²/C
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?
Answer
φ = q/ε₀
⇒ φ = 2.0 × 10⁻⁶/8.854 × 10⁻¹²
⇒ φ ≈ 2.26 × 10⁵ N m²/C. (The size of the cube does not matter.)
19. A point charge causes an electric flux of –1.0 × 10³ N m²/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?
Answer
(a) Flux depends only on the enclosed charge, which is unchanged. So φ = –1.0 × 10³ N m²/C.
(b) q = ε₀φ
⇒ q = 8.854 × 10⁻¹² × (–1.0 × 10³)
⇒ q ≈ –8.85 × 10⁻⁹ C ≈ –8.8 nC
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?
Answer
Outside the sphere the field is that of a point charge at the centre: E = kq/r².
q = Er²/k
⇒ q = (1.5 × 10³ × (0.20)²)/(9 × 10⁹)
⇒ q = 60/(9 × 10⁹)
⇒ q ≈ 6.67 × 10⁻⁹ C
The field points inward, so the charge is negative: q ≈ –6.67 nC.
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?
Answer
r = 1.2 m, σ = 80.0 × 10⁻⁶ C/m².
(a) Q = σ × 4πr²
⇒ Q = 80 × 10⁻⁶ × 4 × 3.14 × (1.2)²
⇒ Q = 80 × 10⁻⁶ × 18.1
⇒ Q ≈ 1.45 × 10⁻³ C
(b) φ = Q/ε₀
⇒ φ = 1.45 × 10⁻³/8.854 × 10⁻¹²
⇒ φ ≈ 1.64 × 10⁸ N m²/C
22. An infinite line charge produces a field of 9 × 10⁴ N/C at a distance of 2 cm. Calculate the linear charge density.
Answer
E = λ/(2πε₀r) = 2kλ/r
⇒ λ = Er/(2k)
⇒ λ = (9 × 10⁴ × 0.02)/(2 × 9 × 10⁹)
⇒ λ = 10⁻⁷ C/m
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?
Answer
Each plate gives a field σ/(2ε₀). Outside the pair the two fields are equal and opposite; between the plates they add.
(a) Outer region of first plate: E = 0
(b) Outer region of second plate: E = 0
(c) Between the plates: E = σ/ε₀
⇒ 17.0 × 10⁻²²/8.854 × 10⁻¹²
⇒ E ≈ 1.92 × 10⁻¹⁰ N/C, directed from the positive plate to the negative plate.
Important Extra Questions
State Coulomb's law in vector form.
Answer
The force on charge q₂ due to q₁ is F₂₁ = (1/4πε₀)(q₁q₂/r₂₁²) r̂₂₁, where r̂₂₁ is the unit vector from q₁ to q₂. Like charges give a repulsive force, unlike charges an attractive one, and F₁₂ = –F₂₁.
What is the net force and torque on an electric dipole placed in a uniform electric field?
Answer
Net force is zero because +qE and –qE cancel. Torque is τ = p × E, magnitude pE sin θ, which tends to align the dipole with the field.
Write the electric field on the axial and equatorial lines of a short dipole.
Answer
Axial line: E = 2kp/r³, along p. Equatorial line: E = kp/r³, opposite to p. Both fall off as 1/r³.
State Gauss's law.
Answer
The total electric flux through any closed surface equals 1/ε₀ times the net charge enclosed by it: ∮E·dS = q_enclosed/ε₀.
What is the electric field due to a uniformly charged thin spherical shell at a point inside it and outside it?
Answer
Inside (r < R): E = 0. Outside (r ≥ R): E = q/(4πε₀r²), as if all the charge were at the centre, directed radially.