Exercises 1.1–1.12 – Solved
The first twelve end-of-chapter exercises, each worked in the board-exam layout: Given → Formula → Substitution → Answer with unit. They cover Coulomb's law, quantisation and conservation of charge, forces at the centre of a square, field lines, the field and force at the midpoint of two charges, the dipole moment, the torque on a dipole, and electrons moved by rubbing. Try each one in your notebook first. Then open the hint, step through the solution and check the final answer.
- Electric charge (വൈദ്യുത ചാർജ്) is additive, conserved and quantised: q = ne, e = 1.6 × 10−19 C (page 6).
- Coulomb's law (കൂളോം നിയമം): F = 14πε0 q1q2r2, with k = 1/4πε0 = 9 × 109 N m2 C−2 (page 7). Forces from several charges add as vectors by the superposition principle (അധ്യാരോപണ തത്വം) (page 8).
- Electric field (വൈദ്യുത മണ്ഡലം) of a point charge: E = kQ/r², away from + and towards −; force on a charge q is F = qE (page 9). Field lines: page 10.
- Dipole moment (ദ്വിധ്രുവ മൊമെന്റ്) p = q × 2a, from −q to +q; torque in a uniform field τ = pE sin θ (page 11).
Write the data with units, convert to SI (cm → m, μC → C) before substituting, write the formula, substitute, and give the answer with its unit and direction. For a force or field, “6 × 10−3 N” is only half an answer; “6 × 10−3 N, repulsive” is a full one.
Exercise 1.1 · Force between two charged spheres
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?
Small spheres act like point charges (ബിന്ദു ചാർജ്). Use F = kq1q2/r² with r in metres. Both charges are positive. Will they attract or repel?
Halve the distance and the force becomes four times as large: Coulomb's force follows an inverse square law (വിപരീത വർഗ നിയമം).
Formula: F = kq1q2/r², k = 9 × 109 N m² C−2.
Substitution: F = 9 × 109 × 2 × 10−7 × 3 × 10−7 / (0.30)² = 5.4 × 10−4 / 0.09.
Answer: F = 6 × 10−3 N, repulsive (both charges are positive).
6 × 10−3 N is about the weight of a 0.6 g paper clip, from only 0.2–0.3 μC of charge. That is why a comb charged by dry hair can lift bits of paper against the pull of the whole Earth.
Exercise 1.2 · Finding the distance from the force
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?
(a) Rearrange F = k|q1q2|/r² to r = √(k|q1q2|/F). (b) Think of Newton's third law. Does the bigger charge feel a bigger force?
Formula: F = k|q1q2|/r², so r = √(k|q1q2|/F).
Substitution: r² = 9 × 109 × 4 × 10−7 × 8 × 10−7 / 0.2 = 2.88 × 10−3/0.2 = 1.44 × 10−2 m².
Answer (a): r = 0.12 m (12 cm).
(b) By Newton's third law the force on the second sphere is equal and opposite: 0.2 N, attractive, on the second sphere and directed towards the first sphere.
“The −0.8 μC sphere has twice the charge, so it feels twice the force.” No. Both charges appear in the same product q1q2, so the two forces are always equal in size and opposite in direction.
Working backwards from a measured force to a distance (or a charge) is exactly what Coulomb did with his torsion balance. Board papers often ask the “reverse” question, so practise rearranging the formula before substituting.
Exercise 1.3 · Electric force versus gravity
1.3 Check that the ratio ke²/(G me mp) is dimensionless. Look up a table of physical constants and determine the value of this ratio. What does the ratio signify?
Write the units of k (N m² C−2) and G (N m² kg−2). Multiply by e² (C²) and by memp (kg²). What is left in each? Then ke²/r² is the electric force between an electron and a proton, and Gmemp/r² is the gravitational one.
Value: with 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 × 2.56 × 10−38) / (6.67 × 10−11 × 9.1 × 10−31 × 1.67 × 10−27) = 2.30 × 10−28/1.01 × 10−67 ≈ 2.3 × 1039.
Meaning: it is the ratio of the electric force to the gravitational force between an electron and a proton at any distance (r² cancels). The electric force is about 1039 times stronger.
This is why chemistry, friction, the strength of a rope and the stiffness of your bones are all electrical. Gravity wins only for huge, almost exactly neutral bodies such as planets and stars, where the + and − charges cancel and the masses simply add up.
Exercise 1.4 · Quantisation of charge
1.4 (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?
(a) Stairs or a ramp? Which values of charge are allowed? (b) How many electrons make up even 1 μC? Would one electron more or less be noticed?
(b) Everyday charges are huge compared with e. For example 1 μC = 10−6 C contains n = 10−6/1.6 × 10−19 = 6.25 × 1012 electronic charges. Adding or removing one electron changes it by about one part in 1013, far too little to measure. So at the macroscopic level the steps are invisible and charge behaves as if it were continuous. (Similarly, a line of dots looks continuous from far away.)
Millikan proved quantisation in 1912 by balancing tiny charged oil drops (Exercise 1.25 on page 16). Today single-electron transistors and quantum dots count electrons one at a time, so at the nanoscale the steps matter again.
Exercise 1.5 · Rubbing and conservation of charge
1.5 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.
Before rubbing, are the rod and the cloth charged? Rubbing does not make electrons. It only moves them. Add the two charges after rubbing.
Rubbing does not create charge. It only transfers electrons from one body to the other: some electrons leave the glass and go to the silk. The glass is left with a positive charge +q, and the silk gains an equal negative charge −q.
Total charge after rubbing = (+q) + (−q) = 0, the same as before. The pair is an isolated system, so its total charge is unchanged. This is exactly the law of conservation of charge (ചാർജ് സംരക്ഷണം). The same happens with every pair rubbed together (plastic and fur, polythene and wool): one gains exactly what the other loses.
The spark when you touch a door handle after walking on a nylon carpet is the same story: electrons moved from carpet to you (or the other way), and the spark is them finding their way back.
Exercise 1.6 · Charges at the corners of a square
1.6 Four point charges qA = 2 μC, qB = −5 μC, qC = 2 μC and qD = −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?
A and C are opposite corners, and so are B and D. All four corners are the same distance from the centre. Compare the forces from A and C. Do the same for B and D.
Distance: each corner is r = a/√2 = 0.0707 m from O, so r² = 5 × 10−3 m².
Forces (F = kq1q2/r²): FA = FC = 9 × 109 × 2 × 10−6 × 10−6/(5 × 10−3) = 3.6 N, both repulsive, along the diagonal AC and in opposite directions. FB = FD = 9 N, both attractive, along BD and in opposite directions.
Answer: by superposition, the net force on the 1 μC charge is zero.
Symmetry saves time: in the exam you need not calculate 3.6 N and 9 N at all. Equal charges at equal distances on opposite sides always cancel. The same idea will make Gauss's law easy on page 13.
Exercise 1.7 · Why field lines are continuous and never cross
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.
The tangent to a field line at a point gives the direction of the electric field there. (a) Would a test charge feel a force in the gap? (b) How many directions can the field have at one point?
(b) The tangent to a field line at a point gives the direction of E there. If two lines crossed at a point P, there would be two tangents, so E at P would point in two different directions at once. But the field at a point is a single vector (the vector sum of the fields of all the charges), so it has only one direction. Hence two field lines can never cross.
These two rules are the quickest way to judge field-line sketches. You will use them to spot the impossible pictures of Fig 1.35 in Exercise 1.26 (page 16).
Exercise 1.8 · Field and force at the midpoint
1.8 Two point charges qA = 3 μC and qB = −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?
The field of +qA at O points away from A. The field of −qB at O points towards B. Do these add or cancel? For (b), F = qE; a negative q reverses the direction.
Formula: E = k|q|/r² for each charge; then add the vectors.
Substitution: EA = 9 × 109 × 3 × 10−6/(0.1)² = 2.7 × 106 N/C along OB (away from +qA); EB = 2.7 × 106 N/C, also along OB (towards −qB).
Answer (a): E = 5.4 × 106 N/C along OB (from A towards B).
(b) F = qE = 1.5 × 10−9 × 5.4 × 106 = 8.1 × 10−3 N, directed along OA: the test charge is negative, so the force is opposite to E.
Writing E = 2.7 × 106 − 2.7 × 106 = 0 “because the charges are opposite”. Signs of charges are not directions of fields. Draw the two arrows at O first: here they point the same way.
The space between opposite charges is where the field is strongest, as between the plates of a capacitor or between a thundercloud base and the ground just before a lightning stroke.
Exercise 1.9 · Total charge and dipole moment
1.9 A system has two charges qA = 2.5 × 10−7 C and qB = −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?
Total charge is an ordinary (signed) sum. The dipole moment (ദ്വിധ്രുവ മൊമെന്റ്) has size q × (separation) and points from − to +. Which of A and B is lower on the z-axis?
Dipole moment: p = q × 2a, with q = 2.5 × 10−7 C and 2a = AB = 30 cm = 0.30 m.
p = 2.5 × 10−7 × 0.30 = 7.5 × 10−8 C m, directed from B (−q) to A (+q), i.e. along the negative z-direction.
Total charge zero does not mean “no electric effect”. This system is an electric dipole (വൈദ്യുത ദ്വിധ്രുവം), like a water molecule, and it still produces a field and feels a torque. That is why water dissolves salt and heats up in a microwave oven.
Exercise 1.10 · Torque on a dipole
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 N C−1. Calculate the magnitude of the torque acting on the dipole.
In a uniform electric field (ഏകീകൃത വൈദ്യുത മണ്ഡലം) the two forces ±qE are equal and opposite (no net force) but they form a couple. τ = pE sin θ, with θ the angle between p and E.
Formula: τ = pE sin θ.
Substitution: τ = 4 × 10−9 × 5 × 104 × sin 30° = 2 × 10−4 × 0.5.
Answer: τ = 10−4 N m. It tends to turn p into line with E.
This torque (ടോർക്ക് (ബലാഘൂർണം)) is what a microwave oven uses: its field reverses about 2.45 × 109 times a second, the water dipoles in food keep twisting to follow it, and the jostling heats the food.
Exercise 1.11 · Electrons moved by rubbing
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?
Negative means the polythene gained electrons. n = |q|/e. Each electron has a mass of 9.1 × 10−31 kg.
(a) Formula: q = ne, so n = |q|/e = 3 × 10−7/1.6 × 10−19 = 1.875 × 1012 ≈ 2 × 1012 electrons, transferred from the wool to the polythene.
(b) Yes, because electrons have mass. Mass transferred = n me = 1.875 × 1012 × 9.1 × 10−31 ≈ 1.7 × 10−18 kg. This is negligibly small, far below anything a balance can detect.
A few trillion electrons sound like a lot, but they weigh less than a single bacterium, which is around 10−15 kg. In practice charging changes charge, not weight.
Exercise 1.12 · Copper spheres: charge doubled, distance halved
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? 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?
(a) Plain Coulomb's law. (b) Use ratios: doubling each charge multiplies q1q2 by 4; halving r divides r² by 4. What is the total factor?
(a) Formula: F = kqAqB/r². Substitution: F = 9 × 109 × (6.5 × 10−7)²/(0.5)² = 9 × 109 × 4.225 × 10−13/0.25 = 1.52 × 10−2 N.
Answer (a): F ≈ 1.5 × 10−2 N (repulsive).
(b) F′ = k(2q)(2q)/(r/2)² = 16 kq²/r² = 16F = 16 × 1.52 × 10−2 ≈ 0.24 N (repulsive).
Ratio reasoning (“×4 on top, ÷4 below → ×16”) is the fastest route in MCQs and entrance tests. Exercise 1.13 on the next page continues with these same spheres.
Check yourself
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