Exercises 1.13–1.24 – Solved
Exercises 1.13–1.24, worked in the exam layout Given → Formula → Substitution → Answer with unit. They cover sharing charge by touching, the three tracks of Fig 1.33, flux through a square and a cube, a black box, the point charge above a square (Fig 1.34), Gaussian cubes and spheres, a charged conducting sphere, a line charge and two parallel plates. Most of them are Gauss's law in action: flux depends only on the charge enclosed.
- Coulomb's law (കൂളോം നിയമം) F = kq1q2/r². Identical conducting spheres in contact share their total charge equally (page 7).
- Electric flux (വൈദ്യുത ഫ്ലക്സ്) through a small flat area: Δφ = E·ΔS = EΔS cos θ, where the area vector (ഏരിയ സദിശം) is along the normal; unit N m² C−1 (page 10).
- Gauss's law (ഗോസ് നിയമം): flux through any closed surface φ = qenclosed/ε0, with ε0 = 8.854 × 10−12 C² N−1 m−2 (page 12).
- Results from Gauss's law (page 13): infinite line E = λ/(2πε0r); infinite plane sheet E = σ/(2ε0); outside a charged sphere or shell E = q/(4πε0r²), as if all the charge were at the centre.
Exercise 1.13 · Sharing charge with a third sphere
1.13 Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?
Two identical conductors (ചാലകം) in contact end up with equal charges: each gets half the total. Do it twice, then use F ∝ qAqB. From Exercise 1.12(a), F = 1.52 × 10−2 N.
C touches A: total q is shared equally, so qA = q/2 and qC = q/2.
C touches B: total q + q/2 = 3q/2 is shared, so qB = 3q/4 (and C keeps 3q/4 as it is taken away).
Formula: F′ = k(q/2)(3q/4)/r² = (3/8) kq²/r² = (3/8)F.
Answer: F′ = 3/8 × 1.52 × 10−2 N ≈ 5.7 × 10−3 N (still repulsive).
This “halving by touching” trick is how Coulomb got charges of q/2, q/4, … without being able to measure charge directly. That let him check that F ∝ q1q2.
Exercise 1.14 · Three tracks in a uniform field (Fig 1.33)
1.14 Figure 1.33 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?
Which plate is +? The field points from + to −. A negative particle is pulled towards the + plate. For particles entering with the same speed, the sideways deflection is y = (q/m)EL²/(2v²): the bigger q/m, the bigger the bend.
Signs: particles 1 and 2 bend upward, towards the + plate and against E, so they are negative. Particle 3 bends downward, along E, so it is positive.
Highest q/m: the deflection on leaving the plates is y = (q/m)·EL²/(2vx²). For the same E, L and entry speed, y ∝ q/m. Particle 3 is deflected the most, so it has the highest charge-to-mass ratio.
The comparison assumes the three particles enter with the same speed, as the textbook answer does. A slower particle spends longer between the plates and bends more even with the same q/m. Exercise 1.33 on page 16 derives the formula for y.
Bending charged particles in fields to sort them by q/m is the idea behind J.J. Thomson's discovery of the electron and today's mass spectrometers, used for drug testing, carbon dating and finding pollutants.
Exercise 1.15 · Flux through a square
1.15 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?
Flux is φ = EΔS cos θ, where θ is the angle between E and the normal to the surface, not the surface itself. A plane parallel to the yz plane has its normal along x.
Formula: φ = E·ΔS = EΔS cos θ.
(a) Plane parallel to yz, so the normal is along x and θ = 0°: φ = 3 × 103 × 10−2 × 1 = 30 N m²/C.
(b) θ = 60°: φ = 3 × 103 × 10−2 × 0.5 = 15 N m²/C.
A solar panel works the same way: it catches the most sunlight when its face is square-on to the rays (θ = 0) and less as it tilts (cos θ). Flux measures how much field “passes through” a surface.
Exercise 1.16 · Net flux through a cube
1.16 What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?
Take the six faces one at a time. Area vectors of a closed surface point outward. For which faces is the outward normal along +x or −x? For the rest, θ = 90°.
Left face (outward normal −x, θ = 180°): φ = −EA = −120 N m²/C. Right face (normal +x): φ = +120 N m²/C. Other four faces: normals ⊥ E, so φ = 0.
Answer: net flux = −120 + 120 + 0 = zero. Every field line that enters also leaves. This agrees with Gauss's law: there is no charge inside the cube.
A uniform field always gives zero net flux through a closed surface. Net flux appears only when a surface encloses a source (charge) of the field. This is the heart of Gauss's law.
Exercise 1.17 · The black box
1.17 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?
The surface of the box is a closed Gaussian surface (ഗോസിയൻ ഉപരിതലം). Gauss's law gives the net charge enclosed: q = ε0φ. Could +q and −q together give a net of zero?
(a) Formula: φ = q/ε0, so q = ε0φ = 8.854 × 10−12 × 8.0 × 103 = 7.08 × 10−8 C ≈ 0.07 μC (positive, since the net flux is outward).
(b) No. Zero net flux only tells us that the net (total) charge inside is zero. The box could contain equal positive and negative charges, for example a dipole, or many charges adding to zero. The flux out of the + charges is cancelled by the flux into the − charges.
Gauss's law lets you “weigh” the charge in a closed region without opening it, just by measuring the field on its surface. Atmospheric scientists estimate the net charge in a thundercloud in a similar way, from field measurements around it.
Exercise 1.18 · Point charge above a square (Fig 1.34)
1.18 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. 1.34. 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.)
Build a 10 cm cube on the square. Where is the charge now? It sits exactly at the centre of the cube (5 cm from every face). The six faces are identical, so each receives the same share of the total flux q/ε0.
Idea: the square is one face of a cube of edge 10 cm whose centre is the charge. By Gauss's law the total flux through the cube is q/ε0, and by symmetry the six faces share it equally.
Formula: φsquare = q/(6ε0).
Substitution: φ = 1.0 × 10−5/(6 × 8.854 × 10−12).
Answer: φ ≈ 1.88 × 105 N m²/C.
Symmetry plus Gauss's law replaces a hard surface integral by one division. Exam variants: a charge at a cube's corner (each of the three far faces gets q/24ε0) or on a face; always first ask “which closed surface puts the charge at its centre?”.
Exercise 1.19 · Charge at the centre of a cube
1.19 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?
Does the net flux through a closed surface depend on its size or shape? Only the enclosed charge matters.
Formula: Gauss's law, φ = q/ε0.
Substitution: φ = 2.0 × 10−6/8.854 × 10−12.
Answer: φ ≈ 2.26 × 105 N m²/C. The edge length 9.0 cm is not needed.
Using the edge length to find an area and then E × area. The field is not uniform over the cube's faces, so that does not work. Gauss's law gives the total directly.
“Only the enclosed charge matters” is what makes Gaussian surfaces so useful. You may choose any convenient closed surface, and the answer for the total flux will not change.
Exercise 1.20 · Doubling the Gaussian sphere
1.20 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?
(a) The bigger sphere encloses the same charge. Count the field lines crossing each sphere. (b) q = ε0φ. Keep the minus sign!
(a) Flux depends only on the enclosed charge (φ = q/ε0), not on the radius. Doubling the radius makes E four times smaller but the area four times larger. Answer (a): −1.0 × 103 N m²/C (unchanged).
(b) Formula: q = ε0φ = 8.854 × 10−12 × (−1.0 × 103).
Answer (b): q ≈ −8.85 × 10−9 C (about −8.9 nC). Negative flux (field lines going in) means a negative charge.
E ∝ 1/r² and area ∝ r² cancel exactly. That cancellation is why Gauss's law works, and it is a direct test of the inverse square law (വിപരീത വർഗ നിയമം).
Exercise 1.21 · Field outside a charged sphere
1.21 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?
Outside a charged sphere the field is the same as that of a point charge at the centre: E = kq/r², with r measured from the centre (20 cm, not 10 cm). Radially inward tells you the sign.
Formula: outside the sphere E = kq/r² (Gauss's law with a concentric Gaussian sphere of radius r), so |q| = Er²/k.
Substitution: |q| = 1.5 × 103 × (0.2)²/(9 × 109) = 60/(9 × 109) = 6.67 × 10−9 C.
Answer: the field points inward, so the charge is negative: q = −6.67 × 10−9 C (−6.67 nC).
Seen from outside, a charged sphere behaves exactly like a point charge at its centre, just as the Earth's gravity acts as if all its mass were at the centre. Engineers use this to estimate the field around the dome of a Van de Graaff generator.
Exercise 1.22 · Charge and flux of a charged sphere
1.22 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?
Surface charge density (ഉപരിതല ചാർജ് സാന്ദ്രത) σ = charge per area, so Q = σ × (surface area) = σ × 4πR². Use the radius (1.2 m), not the diameter. Then Gauss: φ = Q/ε0.
(a) Formula: Q = σ × 4πR². Substitution: Q = 8.0 × 10−5 × 4π × (1.2)² = 8.0 × 10−5 × 18.1. Answer (a): Q ≈ 1.45 × 10−3 C.
(b) Formula: φ = Q/ε0. Substitution: φ = 1.45 × 10−3/8.854 × 10−12. Answer (b): φ ≈ 1.64 × 108 ≈ 1.6 × 108 N m²/C.
1.45 mC is a huge static charge. In air, a sphere this size would spark over long before it held that much. Large charged domes are therefore made smooth and big, because sharp points leak charge (that is how lightning conductors work, page 17).
Exercise 1.23 · Linear charge density of a line charge
1.23 An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density.
For an infinitely long straight wire (അനന്തമായ നീളമുള്ള നേർകമ്പി), Gauss's law with a cylinder gives E = λ/(2πε0r). Since 1/(2πε0) = 2k = 1.8 × 1010, you can also write E = 2kλ/r.
Formula: E = λ/(2πε0r), so λ = 2πε0rE = rE/(2k).
Substitution: λ = 0.02 × 9 × 104/(2 × 9 × 109) = 1800/(1.8 × 1010).
Answer: λ = 10−7 C/m (0.1 μC/m). (With ε0 = 8.854 × 10−12 you get 1.001 × 10−7 C/m.)
The 1/r field of a line charge (weaker than a point charge's 1/r² fall-off) is used in electrostatic precipitators. A thin charged wire runs down a chimney, and the strong field near it charges smoke particles, which then drift to the walls.
Exercise 1.24 · Two parallel plates
1.24 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?
Treat each plate as an infinite plane sheet (അനന്തമായ സമതല ഷീറ്റ്) giving σ/(2ε0) on both sides: away from the + sheet, towards the − sheet. Add the two fields region by region.
Formula: each sheet gives E = σ/(2ε0), directed away from the + sheet and towards the − sheet.
(a) and (b) Outer regions: the two fields are equal and opposite, so E = 0 outside both plates.
(c) Between the plates: both fields point from the + plate to the − plate and add: E = σ/(2ε0) + σ/(2ε0) = σ/ε0 = 17.0 × 10−22/8.854 × 10−12 ≈ 1.92 × 10−10 N/C, directed from the positive plate to the negative plate.
This is the parallel-plate capacitor of the next chapter: a uniform field trapped between the plates and none outside. The same arrangement steers electron beams (Exercises 1.33–1.34) and, in inkjet printers, steers charged ink drops.