Additional Exercises 1.25–1.34 – Solved
The ten additional exercises: Millikan's oil drop, judging the five field-line pictures of Fig 1.35, a dipole in a non-uniform field, a conductor with a cavity (Fig 1.36), the field in a hole, the field of a long wire from Coulomb's law (with the integral), quarks, why a null point is never stable, and an electron deflected between two plates, with an animation of its parabolic path. Numericals are in exam layout. Derivations are step by step, as you would write them in the board exam.
- Motion under a constant force: s = ½at², projectile path y = gx²/(2u²) (page 3). Force on a charge F = qE (page 9).
- Rules of electric field lines (വൈദ്യുത മണ്ഡലരേഖകൾ): start on +, end on −, never cross, no closed loops, and they meet a conductor's surface at right angles (page 10).
- Dipole moment (ദ്വിധ്രുവ മൊമെന്റ്) p from −q to +q; torque τ = p × E (page 11).
- Gauss's law (ഗോസ് നിയമം) φ = qenc/ε0. Inside a conductor (ചാലകം) E = 0; just outside it E = σ/ε0, normal to the surface; plane sheet σ/(2ε0) (pages 12–13).
Exercise 1.25 · Millikan's oil drop
1.25 An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 N C−1 (Millikan's oil drop experiment). The density of the oil is 1.26 g cm−3. Estimate the radius of the drop. (g = 9.81 m s−2; e = 1.60 × 10−19 C.)
“Held stationary”: the electric force balances the weight, qE = mg. Write the mass as density × volume, m = (4/3)πr³ρ, and convert ρ to kg m−3 (× 1000).
Formula: stationary, so qE = mg = (4/3)πr³ρg, giving r³ = 3qE/(4πρg).
Substitution: r³ = 3 × 1.92 × 10−18 × 2.55 × 104/(4π × 1.26 × 103 × 9.81) = 1.469 × 10−13/1.553 × 105 = 9.46 × 10−19 m³.
Answer: r ≈ 9.82 × 10−7 m, about 10−3 mm (1 μm).
Millikan balanced hundreds of such drops and found that their charges were always whole-number multiples of 1.6 × 10−19 C. That was the experimental proof of the quantisation of charge (ചാർജിന്റെ ക്വാണ്ടീകരണം) and the measurement of e (Nobel Prize 1923).
Exercise 1.26 · Which pictures cannot be field lines? (Fig 1.35)
1.26 Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?
Test each picture against four rules: lines start on + and end on − (or go to infinity); they meet a conductor at right angles; they never cross; they never form closed loops.
Score: 0 / 0
(b) Cannot. Some lines start on the negative charge and end on a positive charge below it, and some join two positive charges. Field lines start on + charges and end on − charges.
(c) Can. This is the correct pattern for two equal positive charges: lines start on the charges, go off to infinity, never cross, and leave a neutral point midway.
(d) Cannot. The field lines cross each other. At a crossing the field would have two directions. The lines also run from one positive charge to another.
(e) Cannot. The two small curves in the gap are closed loops. Electrostatic field lines never form closed loops (the electrostatic field is conservative).
So (a), (b), (d) and (e) cannot represent electrostatic field lines; only (c) can.
Board papers often show one sketch and ask “What is wrong with this?”. Run the four rules in order: start/end, right angles at conductors, no crossing, no loops. (Closed loops do occur for the magnetic field and for electric fields induced by changing magnetic fields, but never for charges at rest.)
Exercise 1.27 · A dipole in a non-uniform field
1.27 In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 105 N C−1 per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10−7 C m in the negative z-direction?
The two charges of the dipole sit where the field is slightly different, so the forces no longer cancel. Net force F = pz(dE/dz). Keep the sign of pz. For the torque, what is the angle between p and E?
Force: take −q at z + 2a and +q at z (since p points down, + is the lower one). Net force = qE(z) − qE(z + 2a) = −q(2a)(dE/dz) = pz(dE/dz) along z.
F = (−10−7)(105) = −10−2 N, so F = 10−2 N in the negative z-direction.
Torque: τ = p × E = pE sin 180° = 0, because p is anti-parallel to E. (This is an unstable equilibrium position: a small tilt would grow.)
A non-uniform field gives a dipole a net force. This explains why a charged comb attracts neutral bits of paper: it induces dipoles in the paper and its field is stronger nearer the comb. In this exercise the dipole points against the field, so it is pushed towards the weaker field instead.
Exercise 1.28 · Conductor with a cavity (Fig 1.36)
1.28 (a) A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor. (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Fig. 1.36(b)]. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.
Draw a Gaussian surface (ഗോസിയൻ ഉപരിതലം) that lies inside the metal of A, all round the cavity. What is E at every point of it? What does Gauss's law then say about the charge it encloses?
(b) With B (charge q) in the cavity, the same surface S still has E = 0 on it, so the total charge enclosed is zero: q + (charge on the cavity wall) = 0. The cavity wall therefore carries −q (induced). A as a whole still has charge Q (conservation of charge), so its outer surface carries Q − (−q) = Q + q.
(c) Enclose the instrument completely in a closed metal box (a Faraday cage), preferably earthed. The field inside a charge-free cavity of a conductor is zero, whatever the field outside. This is electrostatic shielding (സ്ഥിതവൈദ്യുത കവചനം).
A car is a rough Faraday cage, so its passengers are safe in a lightning storm. MRI rooms, sensitive lab amplifiers and the screened cables of microphones and TV aerials are all shielded the same way.
Exercise 1.29 · The field in a tiny hole
1.29 A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ/2ε0) n̂, where n̂ is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.
Imagine the hole filled with its tiny patch of charge. Split the field at that spot into two parts: from the patch (which acts like a small plane sheet) and from the rest of the conductor. You know the total just outside (σ/ε0) and just inside (0).
The same split explains the outward pressure on a charged surface: each patch is pushed by the field of the rest, σ/2ε0, not by the full σ/ε0. This is why a charged soap bubble expands slightly.
Exercise 1.30 · Field of a long wire without Gauss's law
1.30 Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss's law. [Hint: Use Coulomb's law directly and evaluate the necessary integral.] (The textbook prints “E” for the charge density; it means λ.)
Take an element dx of the wire at distance x from the foot O of the perpendicular from P. Elements at +x and −x pair up, so the components along the wire cancel. Substitute x = r tan θ; the integral becomes ∫cos θ dθ from −π/2 to +π/2.
Compare the effort: Gauss's law gave this result in three lines. Symmetry plus Gauss's law beats direct integration whenever the symmetry is high (a line, a plane or a sphere). For less symmetric shapes, integration like this, done by computer, is how engineers design high-voltage insulators.
Exercise 1.31 · Quarks in protons and neutrons
1.31 It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so-called ‘up’ quark (denoted by u) of charge +(2/3)e, and the ‘down’ quark (denoted by d) of charge (−1/3)e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.
Three quarks, each u or d. List all four possibilities (uuu, uud, udd, ddd) and add the charges. Which gives +e, which gives 0?
Neutron = udd: (+2/3 − 1/3 − 1/3)e = 0 ✔.
(The other combinations give uuu = +2e and ddd = −e. These are real but short-lived particles, the Δ++ and Δ−, not the proton and neutron.)
Do quarks break the rule q = ne? Quarks are never found free. They are always bound in groups whose total charge is a whole multiple of e, so every charge we can isolate is still quantised in units of e.
Exercise 1.32 · A null point is never a stable equilibrium
1.32 (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable. (b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.
Stable equilibrium (സന്തുലനം (സ്ഥിര/അസ്ഥിര)) needs a restoring force for a small displacement in every direction. What would the field around the null point look like then? Draw a tiny sphere around it and apply Gauss's law.
(b) Two equal charges +Q at A and B: at the midpoint O the two fields cancel, so O is a null point. Displace a + test charge along AB: it comes closer to one charge, whose push becomes stronger, so it is pushed back to O (stable along the line). Displace it perpendicular to AB: both charges now push it partly outward along the perpendicular, so it is driven further away (unstable). Hence the equilibrium at O is unstable, as (a) predicts.
This is Earnshaw's theorem: no arrangement of fixed charges can hold a charged particle in stable equilibrium. Ion traps used in atomic clocks and quantum computers get round it with rapidly changing fields (the Paul trap) or by adding a magnetic field (the Penning trap).
Exercise 1.33 · Deflection between charged plates
1.33 A particle of mass m and charge (−q) enters the region between the two charged plates initially moving along x-axis with speed vx (like particle 1 in Fig. 1.33). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL²/(2m vx²). Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.
The force qE is constant and vertical, so treat x and y separately, exactly like a horizontally thrown ball. Along x: no force, so constant speed. Along y: constant acceleration a = qE/m from rest.
This formula is the basis of the old cathode-ray-tube TV and the oscilloscope: deflecting plates steered the electron beam across the screen. It is also the principle of Fig 1.33 (Exercise 1.14 on page 15): for the same speed, y ∝ q/m.
Exercise 1.34 · Where does the electron strike the plate?
1.34 Suppose that the particle in Exercise 1.33 is an electron projected with velocity vx = 2.0 × 106 m s−1. If E between the plates separated by 0.5 cm is 9.1 × 102 N/C, where will the electron strike the upper plate? (|e| = 1.6 × 10−19 C, me = 9.1 × 10−31 kg.)
Find a = eE/m. The electron enters along the lower plate (as in the textbook's solution), so it must rise the full gap, y = 0.5 cm, to hit the upper plate. Use y = ½at² for t, then x = vxt.
Plates drawn to scale: 4 cm long, 0.5 cm apart. The electron enters at the lower plate. Halve E or double vx and see how far it gets.
Acceleration: a = eE/m = 1.6 × 10−19 × 9.1 × 102/9.1 × 10−31 = 1.6 × 1014 m s−2 (towards the upper, + plate).
Time to rise 0.5 cm: y = ½at², so t = √(2y/a) = √(2 × 5 × 10−3/1.6 × 1014) = √(6.25 × 10−17) ≈ 7.9 × 10−9 s.
Distance along the plate: x = vxt = 2.0 × 106 × 7.9 × 10−9 = 1.6 × 10−2 m.
Answer: the electron strikes the upper plate about 1.6 cm from the edge where it entered.
The question does not say where the electron enters. The standard answer assumes it enters right next to the lower plate, so it must rise the full 0.5 cm. If it entered midway between the plates, it would rise only 0.25 cm and strike at x = 2.0 × 106 × √(2 × 2.5 × 10−3/1.6 × 1014) ≈ 1.1 cm. State your assumption in the exam.
The electron crosses 1.6 cm in under 10 nanoseconds, with an acceleration 1013 times g. That is why electron beams can be steered so quickly in oscilloscopes, electron microscopes and the electron guns of X-ray tubes.