Chap.1 Ex.++ en


Electrostatics – Problems

Whenever needed: the elementary charge e = 1.6×10−19 C; the constant k = 9×109 SI.

I – Electrons & Charges

Two metallic spheres (A) and (B) are electrified as follows:

  • (A) carries an excess of 2.5×1013 electrons;
  • (B) carries an excess of 1.5×1013 electrons.
(B)(A)excess of 1.5×1013 e⁻excess of 2.5×1013 e⁻
  1. Determine, in µC, QA & QB the respective charges of (A) and (B).
  2. Represent the forces of interaction exerted between (A) and (B).
  3. (A) loses 8×1012 electrons.
    1. Determine the charge of the electrons lost by (A).
    2. Show that the charge of (A) becomes Q′A = −2.72 µC.
    3. The charges lost by (A) are gained by (B). Determine Q′B the final charge of (B).
  4. Calculate QA + QB the initial sum of charges carried by the system [(A) & (B)].
  5. Compare the sum previously calculated to the sum of final charge. Conclude.
Show the detailed solution

1) Charges QA and QB

An excess of electrons means a negative charge: Q = −N·e.

QA = −2.5×1013 × 1.6×10−19 = −4×10−6 C
QB = −1.5×1013 × 1.6×10−19 = −2.4×10−6 C
QA = −4 µC    QB = −2.4 µC

2) Forces of interaction

Both charges are negative (same sign), so the spheres repel each other. The two forces act along the line (AB), have the same magnitude (action–reaction) and opposite directions:

  • FB/A: applied on A, directed away from B;
  • FA/B: applied on B, directed away from A.
−−−−−−−−(B)−−−−−−−−(A)FA/BFB/A

3) (A) loses 8×1012 electrons

a) Charge of the lost electrons (each electron carries −e):

qlost = −8×1012 × 1.6×10−19 = −1.28×10−6 C = −1.28 µC

b) The new charge of A equals its old charge minus the charge that left it:

Q′A = QA − qlost = −4 − (−1.28) = −2.72 µC ✓

Check with electrons: A had 2.5×1013 in excess, it loses 0.8×1013, so 1.7×1013 remain in excess: −1.7×1013 × 1.6×10−19 = −2.72×10−6 C ✓

c) B gains exactly the charge that A lost:

Q′B = QB + qlost = −2.4 + (−1.28) = −3.68 µC
Q′B = −3.68 µC

4) Initial sum of charges

QA + QB = −4 + (−2.4) = −6.4 µC

5) Comparison and conclusion

Q′A + Q′B = −2.72 + (−3.68) = −6.4 µC

The final sum equals the initial sum.

Conclusion: the total charge of an isolated system is conserved (conservation of electric charge).

II – Exchange of Electrons

Two identical metallic spheres (A) & (B) carry respectively the charges QA = −4 µC & QB = +8 µC. (A) & (B) are made to touch each other until the electrostatic equilibrium takes place and then separated again.

−−−−−−−−(A)++++++++(B)QA = −4 µCQB = +8 µC
  1. Charging by contact is one mode of electrification. Give the names of the other modes.
  2. Determine the charges Q′A & Q′B carried by (A) & (B) after contact.
  3. Do the spheres attract or repel each other after contact? Justify.
  4. Specify which sphere gains electrons.
  5. Determine the number of electrons gained.
Show the detailed solution

1) Other modes of electrification

Electrification by friction (rubbing) and electrification by induction (influence).

2) Charges after contact

The system is isolated, so the total charge is conserved. The spheres are identical, so at equilibrium the total charge is shared equally:

QA + QB = Q′A + Q′B and Q′A = Q′B
Q′A = Q′B = (−4 + 8) / 2 = +2 µC
Q′A = Q′B = +2 µC

3) Attraction or repulsion?

Both spheres now carry charges of the same sign (+2 µC each), so they repel each other.

4) Which sphere gains electrons?

  • Sphere A: charge goes from −4 µC to +2 µC. It became more positive, so it lost electrons.
  • Sphere B: charge goes from +8 µC to +2 µC. It became less positive, so it gained electrons.
Sphere (B) gains electrons (they move from A to B).

5) Number of electrons gained

The charge variation of B is:

ΔQB = Q′B − QB = 2 − 8 = −6 µC
n = |ΔQB| / e = 6×10−6 / 1.6×10−19 = 3.75×1013
n = 3.75×1013 electrons
A first loses its 4 µC excess of electrons (to become neutral) and then loses 6 µC worth of electrons in total to end at +2 µC. The same number of electrons that A loses is gained by B.

III – Sign of a Charged Metallic Sphere

A metallic sphere (A), carrying a charge QA = −3.8 µC, is rubbed against a negatively charged rod.

−−−−−−−−(A)−−−−−−rubbingnegatively charged rod
Electrification of (A)
−−−−−−−−(A)(B)QA = −3.8 µCQB = ?
Before the contact of question 3
  1. What do we call this mode of electrification?
  2. Determine the number of electrons in excess carried by this sphere.
  3. The sphere (A) is put in contact with an identical sphere (B), carrying a charge QB, and then separated. We notice that the charge of (A) after contact becomes Q′A = −2 µC.
    1. Does (A) gain or lose electrons? Justify.
    2. Specify the charge of (B) after contact.
    3. Determine QB.
Show the detailed solution

1) Mode of electrification

Electrification by friction (rubbing).

2) Number of excess electrons

QA is negative, so A has an excess of electrons:

n = |QA| / e = 3.8×10−6 / 1.6×10−19 = 2.375×1013
n ≈ 2.4×1013 electrons in excess

3) Contact with the identical sphere (B)

a) The charge of A goes from −3.8 µC to −2 µC. It became less negative, which means it now has fewer electrons in excess.

(A) loses electrons.

b) The spheres are identical, so after contact they carry the same charge:

Q′B = Q′A = −2 µC

c) Conservation of charge (isolated system):

QA + QB = Q′A + Q′B
−3.8 + QB = −2 + (−2) = −4
QB = −4 + 3.8 = −0.2 µC
QB = −0.2 µC
Check: (−3.8 + (−0.2)) / 2 = −2 µC ✓. Also, B goes from −0.2 to −2 µC, so it gains 1.8 µC worth of electrons, exactly what A loses. ✓

IV – Electrostatic Equilibrium & Interaction

Two identical point metallic spheres (A) and (B) are separated by a distance of 20 cm and electrified as follows:

(A): excess of 108 electrons     (B): deficit of 109 electrons.

−−−−−−−−(A)++++++++(B)excess of 108 e⁻deficit of 109 e⁻d = 20 cm
  1. Determine the charges qA and qB carried by (A) and (B) respectively.
  2. (A) and (B) are placed near each other.
    1. Specify whether the spheres attract or repel each other.
    2. Represent the electrostatic force acting between the spheres and then show that its magnitude is equal to F = 5.76×10−10 N.
  3. (A) & (B), being electrically isolated, are made to touch each other and then separated again.
    1. Determine the new charges q′A & q′B of (A) and (B).
    2. Do the two spheres (A) and (B) attract or repel each other? Justify.
    3. Determine the direction of displacement of electrons and then verify that the number of electrons exchanged is 5.5×108.
Show the detailed solution

1) Charges qA and qB

Excess of electrons → negative; deficit of electrons → positive.

qA = −108 × 1.6×10−19 = −1.6×10−11 C
qB = +109 × 1.6×10−19 = +1.6×10−10 C

2) Interaction between the spheres

a) qA < 0 and qB > 0 (opposite signs) → the spheres attract each other.

b) Representation (see the first row of the figure below): FB/A acts on A towards B, FA/B acts on B towards A, along the line (AB), with equal magnitudes.

Magnitude (Coulomb's law), with d = 20 cm = 0.2 m:

F = k·|qA|·|qB| / d²
F = 9×109 × 1.6×10−11 × 1.6×10−10 / (0.2)²
F = 9×109 × 2.56×10−21 / 0.04 = 2.304×10−11 / 0.04
F = 5.76×10−10 N ✓

3) After contact and separation

a) Identical spheres, isolated system: the total charge is conserved and shared equally.

q′A = q′B = (qA + qB) / 2 = (−1.6×10−11 + 16×10−11) / 2 = 7.2×10−11 C
q′A = q′B = +7.2×10−11 C

b) Both charges are positive (same sign) → the spheres repel each other (second row of the figure).

2) Before contact: opposite charges → attraction−−−−−−−−(A)++++++++(B)FB/AFA/B3) After contact: same sign → repulsion++++++++(A)++++++++(B)FB/AFA/B

c) Direction of the electrons:

  • A: from −1.6×10−11 C to +7.2×10−11 C → its charge increased → A lost electrons.
  • B: from +16×10−11 C to +7.2×10−11 C → its charge decreased → B gained electrons.
Electrons move from (A) to (B).

Number of electrons exchanged (use the charge variation of B):

|ΔqB| = 16×10−11 − 7.2×10−11 = 8.8×10−11 C
n = |ΔqB| / e = 8.8×10−11 / 1.6×10−19 = 5.5×108
n = 5.5×108 electrons ✓

V – Charge Carried by a Ball

Given a small and light conducting ball (S), carrying a charge q and held by an insulating string, and a neutral plastic rod (R).

(S)(R)insulating stringneutral plastic rod

Part A – Electrification of the rod (R)

The rod is electrified when it is rubbed against a piece of wool. It thus acquires a negative charge.

−−−−−−(R)woolrubbing
  1. What is meant by «a body is electrified»?
  2. a) Explain how the rod has acquired its charge.
    b) What do we call this charging process of the rod?
  3. Is the piece of wool charged? Why?

Part B – Sign of q

We approach the rod (R) near the ball (S). We then observe that (S) is repelled. What is the sign of the charge on (S)? Justify.

q ?(S)−−−(R)(S) is repelled

Part C – Value of q

(R) is removed and then (S) is placed in contact with a neutral ball (S′), identical to (S). When electrostatic equilibrium is attained, we place (S) and (S′) at two fixed points A and B, 6 cm apart, respectively. We observe that (S′) repels (S).

(S)(S')AB6 cm
  1. a) Explain what happens when (S) and (S′) are put in contact.
    b) What is meant by «electrostatic equilibrium»?
    c) Determine, in terms of q, the charges q1 and q2 of (S) and (S′) at the electrostatic equilibrium.
  2. Given that (S′) repels (S) with an electrostatic force F of magnitude 2.5 N. Determine q.
Show the detailed solution

Part A – Electrification of the rod

1) A body is electrified when it acquires a net electric charge: it has an excess or a deficit of electrons, so its number of electrons is no longer equal to its number of protons.

2) a) When the rod is rubbed against the wool, electrons are transferred from the wool to the rod. The rod then has more electrons than protons, which gives it a negative charge.

b) This process is called electrification by friction (rubbing).

3) Yes, the wool is charged. It lost the electrons that the rod gained, so it has a deficit of electrons and carries a positive charge, equal in magnitude to the charge of the rod (conservation of charge).

Part B – Sign of q

The ball (S) is repelled by the negatively charged rod. Only charges of the same sign repel each other (opposite charges attract, and a neutral conducting ball would be attracted by induction).

(S)−−−(R)ball is pushed away
The charge q of (S) is negative (q < 0).

Part C – Value of q

1) a) (S) has an excess of electrons and (S′) is neutral. When they touch, some electrons move from (S) to (S′), until equilibrium is reached. Both balls then carry negative charges.

b) Electrostatic equilibrium means that there is no more net displacement of electrons between the two balls: the charge distribution no longer changes (the two conductors are at the same potential).

c) The system {S, S′} is isolated, so the total charge is conserved. The balls are identical, so the charge is shared equally:

q + 0 = q1 + q2 and q1 = q2
q1 = q2 = q / 2

2) Coulomb's law with d = AB = 6 cm = 0.06 m and F = 2.5 N:

F = k·|q1|·|q2| / d² = k·(|q|/2)² / d² = k·q² / (4·d²)
q² = 4·F·d² / k = 4 × 2.5 × (0.06)² / (9×109) = 4 × 2.5 × 3.6×10−3 / (9×109)
q² = 3.6×10−2 / (9×109) = 4×10−12 C²
|q| = 2×10−6 C = 2 µC

From Part B, q is negative:

q = −2 µC   (so q1 = q2 = −1 µC)
Check: F = 9×109 × (10−6)² / (0.06)² = 9×10−3 / 3.6×10−3 = 2.5 N ✓
Exercices d'Électrostatique - Avec Solutions

Problèmes d'Électrostatique

Chaque fois que nécessaire : la charge élémentaire e = 1,6 × 10⁻¹⁹ C ; la constante k = 9 × 10⁹ SI.

I- Électrons & Charges

Deux sphères métalliques (A) et (B) sont électrisées comme suit :
— (A) porte un excès de 2,5 × 10¹³ électrons ;
— (B) porte un excès de 1,5 × 10¹³ électrons ;
1. Déterminer, en μC, Q_A et Q_B les charges respectives de (A) et (B).

Pour (A) : Excès d'électrons (charge négative).

Q_A = -n × e = -(2.5 × 10¹³) × (1.6 × 10⁻¹⁹) = -4.0 × 10⁻⁶ C = -4 μC

Pour (B) : Excès d'électrons (charge négative).

Q_B = -(1.5 × 10¹³) × (1.6 × 10⁻¹⁹) = -2.4 × 10⁻⁶ C = -2.4 μC

2. Représenter les forces d'interaction exercées entre (A) et (B).

Les deux sphères portent des charges de même signe (négatives), elles exercent donc des forces de répulsion. Les vecteurs forces pointent à l'opposé l'un de l'autre le long de la ligne reliant leurs centres.

3. (A) perd 8 × 10¹² électrons.
a) Déterminer la charge des électrons perdus par (A).
b) Montrer que la charge de (A) devient Q'_A = -2.72 μC.
c) Les charges perdues par (A) sont gagnées par (B). Déterminer Q'_B.

a) La charge perdue correspond à celle des électrons (négative) :

q_perdue = n × e = (8 × 10¹²) × (1.6 × 10⁻¹⁹) = 1.28 × 10⁻⁶ C = +1.28 μC (en valeur absolue)

b) Puisque (A) perd des électrons (charge négative), sa charge augmente (devient moins négative) :

Q'_A = Q_A - q_perdue = -4 - (-1.28) = -4 + 1.28 = -2.72 μC

c) (B) gagne les électrons perdus par (A) :

Q'_B = Q_B + q_perdue = -2.4 + (-1.28) = -3.68 μC

4. Calculer Q_A + Q_B la somme initiale des charges portées par le système {(A) & (B)}.

Q_initiale = Q_A + Q_B = -4 + (-2.4) = -6.4 μC

5. Comparer la somme calculée précédemment à la somme des charges finales. Conclure.

Q_finale = Q'_A + Q'_B = -2.72 + (-3.68) = -6.4 μC

Conclusion : La somme initiale est égale à la somme finale. Cela démontre la conservation de la charge électrique dans un système isolé.

II- Échange d'Électrons

Deux sphères métalliques identiques (A) et (B) portent respectivement les charges Q_A = -4 μC et Q_B = +8 μC. On met (A) et (B) en contact jusqu'à ce que l'équilibre électrostatique soit atteint, puis on les sépare à nouveau.
1. L'électrisation par contact est un mode d'électrisation. Donner les noms des autres modes.

Les autres modes sont : Frottement (Friction) et Influence (Induction).

2. Déterminer les charges Q'_A et Q'_B portées par (A) et (B) après le contact.

Pour des sphères identiques en contact, la charge totale est partagée équitablement :

Q_totale = Q_A + Q_B = -4 + 8 = +4 μC

Q'_A = Q'_B = Q_totale / 2 = +4 / 2 = +2 μC

3. Les sphères s'attirent-elles ou se repoussent-elles après le contact ? Justifier.

Elles se repoussent. Après le contact, elles ont la même charge (+2 μC), et des charges de même signe se repoussent.

4. Préciser quelle sphère gagne des électrons.

La sphère (B) gagne des électrons. (Sa charge passe de +8 μC à +2 μC, elle devient donc moins positive, ce qui signifie qu'elle a gagné des charges négatives).

5. Déterminer le nombre d'électrons gagnés.

Charge gagnée par (B) = Q'_B - Q_B = +2 - (+8) = -6 μC = -6 × 10⁻⁶ C

n = |Q| / e = (6 × 10⁻⁶) / (1.6 × 10⁻¹⁹) = 3.75 × 10¹³ électrons

III- Signe d'une Sphère Métallique Chargée

Une sphère métallique (A), portant une charge Q_A = -3,8 μC, est frottée contre une tige chargée négativement.
1. Comment appelle-t-on ce mode d'électrisation ?

C'est l'électrisation par frottement.

2. Déterminer le nombre d'électrons en excès portés par cette sphère.

n = |Q_A| / e = (3.8 × 10⁻⁶) / (1.6 × 10⁻¹⁹) = 2.375 × 10¹³ électrons

3. La sphère (A) est mise en contact avec une sphère identique (B), portant une charge Q_B, puis elles sont séparées. On constate que la charge de (A) après contact devient Q'_A = -2 μC.
a) Est-ce que (A) gagne ou perd des électrons ? Justifier.
b) Préciser la charge de (B) après contact.
c) Déterminer Q_B.

a) (A) perd des électrons. Sa charge passe de -3.8 μC à -2 μC, elle devient donc moins négative (elle a perdu des charges négatives).

b) Les deux sphères étant identiques et métalliques, la charge se répartit équitablement après contact. La charge finale de (A) est -2 μC, donc la charge de (B) après contact est aussi Q'_B = -2 μC.

c) En appliquant la conservation de la charge : Q_A + Q_B = Q'_A + Q'_B

-3.8 + Q_B = -2 + (-2)

Q_B = -4 + 3.8 = -0.2 μC



IV- Équilibre & Interaction Électrostatiques

Deux sphères métalliques ponctuelles identiques (A) et (B) sont séparées par une distance de 20 cm et électrisées comme suit :
(A) : excès de 10⁸ électrons     (B) : déficit de 10¹⁰ électrons
1. Déterminer les charges q_A et q_B portées respectivement par (A) et (B).

(A) : Excès de 10⁸ électrons (charge négative).

q_A = -10⁸ × 1.6 × 10⁻¹⁹ = -1.6 × 10⁻¹¹ C

(B) : Déficit de 10¹⁰ électrons (charge positive).

q_B = +10¹⁰ × 1.6 × 10⁻¹⁹ = +1.6 × 10⁻⁹ C

2. (A) et (B) sont placées l'une près de l'autre.
a) Préciser si les sphères s'attirent ou se repoussent.
b) Représenter la force électrostatique agissant entre les sphères, puis montrer que sa valeur est égale à F = 5,76 × 10⁻¹⁰ N.

a) Elles ont des charges de signes opposés (A négative, B positive), donc elles s'attirent.

b) d = 20 cm = 0.2 m

F = k × |q_A × q_B| / d² = (9 × 10⁹) × (1.6 × 10⁻¹¹ × 1.6 × 10⁻⁹) / (0.2)²

F = (9 × 10⁹) × (2.56 × 10⁻²⁰) / 0.04 = 5.76 × 10⁻⁹ N

Note : La valeur calculée est 5.76 × 10⁻⁹ N et non 10⁻¹⁰ N. Il semble y avoir une faute de frappe dans l'énoncé original.

3. (A) et (B), électriquement isolées, sont mises en contact puis séparées à nouveau.
a) Déterminer les nouvelles charges q'_A et q'_B de (A) et (B).
b) Les deux sphères (A) et (B) s'attirent-elles ou se repoussent-elles ? Justifier.
c) Déterminer le sens de déplacement des électrons, puis vérifier que le nombre d'électrons échangés est 5,5 × 10⁸.

a) Q_totale = q_A + q_B = -1.6 × 10⁻¹¹ + 1.6 × 10⁻⁹ = 1.584 × 10⁻⁹ C

q'_A = q'_B = Q_totale / 2 = 1.584 × 10⁻⁹ / 2 = 7.92 × 10⁻¹⁰ C

b) Elles ont maintenant la même charge (positive), donc elles se repoussent.

c) Les électrons se déplacent de (A) vers (B) (car A était négative et B positive).

Charge perdue par A = q'_A - q_A = 7.92 × 10⁻¹⁰ - (-1.6 × 10⁻¹¹) = 8.08 × 10⁻¹⁰ C

n = 8.08 × 10⁻¹⁰ / 1.6 × 10⁻¹⁹ = 5.05 × 10⁹ électrons

Note : Le résultat est 5.05 × 10⁹ et non 5.5 × 10⁸. Il semble y avoir une erreur dans les données de l'exercice original.

V- Charge Portée par une Bille

On considère une petite bille conductrice légère (S), portant une charge q et suspendue par un fil isolant, ainsi qu'une tige en plastique neutre (R).

Partie A : Électrisation de la tige (R)

La tige est électrisée lorsqu'elle est frottée contre un morceau de laine. Elle acquiert ainsi une charge négative.
1. Que signifie « un corps est électrisé » ?
2. a) Expliquer comment la tige a acquis sa charge. b) Comment appelle-t-on ce procédé ?
3. Le morceau de laine est-il chargé ? Pourquoi ?

1. Cela signifie que le corps a acquis une charge électrique (un déséquilibre entre les électrons et les protons).

2. a) Par frottement, des électrons sont transférés de la laine vers la tige, qui acquiert donc une charge négative (elle a gagné des électrons).
b) Cela s'appelle l'électrisation par frottement.

3. Oui, le morceau de laine est chargé. D'après la conservation de la charge, puisque la tige a gagné des électrons (elle est devenue négative), la laine a perdu des électrons (elle est devenue positive).

Partie B : Signe de q

On approche la tige (R) de la bille (S). On observe ensuite que (S) est repoussée.
Quel est le signe de la charge sur (S) ? Justifier.

La tige (R) est chargée négativement. Puisque (S) est repoussée, elle doit avoir une charge de même signe. Donc la charge q sur (S) est négative.

Partie C : Valeur de q

(R) est retirée, puis (S) est mise en contact avec une bille neutre (S'), identique à (S). Lorsque l'équilibre électrostatique est atteint, on place (S) et (S') en deux points fixes A et B séparés de 6 cm. On observe que (S') repousse (S).
1. a) Expliquer ce qui se passe lorsque (S) et (S') sont mises en contact.
b) Que signifie « équilibre électrostatique » ?
c) Déterminer, en fonction de q, les charges q_1 et q_2 de (S) et (S') à l'équilibre électrostatique.
2. Sachant que (S') repousse (S) avec une force électrostatique F de valeur 2,5 N, déterminer q.

1. a) Lors du contact, la charge totale se répartit équitablement entre les deux sphères identiques pour atteindre l'équilibre.

b) L'équilibre électrostatique signifie que les charges cessent de se déplacer et se répartissent uniformément sur la surface des conducteurs.

c) Charge totale = q + 0 = q. Comme elles sont identiques : q_1 = q_2 = q / 2

2. d = 6 cm = 0.06 m

F = k × |q_1 × q_2| / d² = k × (q/2)² / d² = k × q² / (4 × d²)

2.5 = (9 × 10⁹ × q²) / (4 × 0.06²) = (9 × 10⁹ × q²) / 0.0144

q² = (2.5 × 0.0144) / (9 × 10⁹) = 0.036 / (9 × 10⁹) = 4 × 10⁻¹² C²

q = √(4 × 10⁻¹²) = 2 × 10⁻⁶ C = 2 μC

Puisque le signe est négatif (d'après la partie B) : q = -2 μC


Electrostatics Problems - With Solutions

Electrostatics Problems

Whenever necessary: elementary charge e = 1.6 × 10⁻¹⁹ C ; constant k = 9 × 10⁹ SI.

Exercise 9- Electrons & Charges

Two metallic spheres (A) and (B) are electrified as follows:
— (A) carries an excess of 2.5 × 10¹³ electrons;
— (B) carries an excess of 1.5 × 10¹³ electrons.
1. Determine, in μC, Q_A and Q_B, the respective charges of (A) and (B).

For (A): Excess of electrons (negative charge).

Q_A = -n × e = -(2.5 × 10¹³) × (1.6 × 10⁻¹⁹) = -4.0 × 10⁻⁶ C = -4 μC

For (B): Excess of electrons (negative charge).

Q_B = -(1.5 × 10¹³) × (1.6 × 10⁻¹⁹) = -2.4 × 10⁻⁶ C = -2.4 μC

2. Draw the interaction forces exerted between (A) and (B).

Both spheres carry charges of the same sign (negative), so they exert repulsive forces on each other. The force vectors point away from each other along the line connecting their centers.

3. (A) loses 8 × 10¹² electrons.
a) Determine the charge of the electrons lost by (A).
b) Show that the charge of (A) becomes Q'_A = -2.72 μC.
c) The charges lost by (A) are gained by (B). Determine Q'_B.

a) The lost charge corresponds to that of the electrons (negative):

q_lost = n × e = (8 × 10¹²) × (1.6 × 10⁻¹⁹) = 1.28 × 10⁻⁶ C = +1.28 μC (in absolute value)

b) Since (A) loses electrons (negative charge), its charge increases (becomes less negative):

Q'_A = Q_A - q_lost = -4 - (-1.28) = -4 + 1.28 = -2.72 μC

c) (B) gains the electrons lost by (A):

Q'_B = Q_B + q_lost = -2.4 + (-1.28) = -3.68 μC

4. Calculate Q_A + Q_B, the initial sum of charges carried by the system {(A) & (B)}.

Q_initial = Q_A + Q_B = -4 + (-2.4) = -6.4 μC

5. Compare the sum calculated above to the sum of the final charges. Conclude.

Q_final = Q'_A + Q'_B = -2.72 + (-3.68) = -6.4 μC

Conclusion: The initial sum equals the final sum. This demonstrates the conservation of electric charge in an isolated system.

Exercise 10- Electron Exchange

Two identical metallic spheres (A) and (B) carry charges Q_A = -4 μC and Q_B = +8 μC respectively. (A) and (B) are put in contact until electrostatic equilibrium is reached, then separated again.
1. Electrification by contact is a mode of electrification. Give the names of the other modes.

The other modes are: Friction and Induction (Influence).

2. Determine the charges Q'_A and Q'_B carried by (A) and (B) after contact.

For identical spheres in contact, the total charge is shared equally:

Q_total = Q_A + Q_B = -4 + 8 = +4 μC

Q'_A = Q'_B = Q_total / 2 = +4 / 2 = +2 μC

3. Do the spheres attract or repel each other after contact? Justify.

They repel. After contact, they have the same charge (+2 μC), and like charges repel.

4. Specify which sphere gains electrons.

Sphere (B) gains electrons. (Its charge goes from +8 μC to +2 μC, so it becomes less positive, meaning it gained negative charges).

5. Determine the number of electrons gained.

Charge gained by (B) = Q'_B - Q_B = +2 - (+8) = -6 μC = -6 × 10⁻⁶ C

n = |Q| / e = (6 × 10⁻⁶) / (1.6 × 10⁻¹⁹) = 3.75 × 10¹³ electrons

Exercise 11- Sign of a Charged Metallic Sphere

A metallic sphere (A), carrying a charge Q_A = -3.8 μC, is rubbed against a negatively charged rod.
1. What is this mode of electrification called?

It is electrification by friction.

2. Determine the number of excess electrons carried by this sphere.

n = |Q_A| / e = (3.8 × 10⁻⁶) / (1.6 × 10⁻¹⁹) = 2.375 × 10¹³ electrons

3. Sphere (A) is put in contact with an identical sphere (B), carrying a charge Q_B, then they are separated. It is observed that the charge of (A) after contact becomes Q'_A = -2 μC.
a) Does (A) gain or lose electrons? Justify.
b) Specify the charge of (B) after contact.
c) Determine Q_B.

a) (A) loses electrons. Its charge goes from -3.8 μC to -2 μC, so it becomes less negative (it lost negative charges).

b) Since both spheres are identical and metallic, the charge is distributed equally after contact. The final charge of (A) is -2 μC, so the charge of (B) after contact is also Q'_B = -2 μC.

c) Applying conservation of charge: Q_A + Q_B = Q'_A + Q'_B

-3.8 + Q_B = -2 + (-2)

Q_B = -4 + 3.8 = -0.2 μC



Exercise 12- Electrostatic Equilibrium & Interaction

Two identical point-like metallic spheres (A) and (B) are separated by a distance of 20 cm and electrified as follows:
(A): excess of 10⁸ electrons     (B): deficit of 10¹⁰ electrons
1. Determine the charges q_A and q_B carried respectively by (A) and (B).

(A): Excess of 10⁸ electrons (negative charge).

q_A = -10⁸ × 1.6 × 10⁻¹⁹ = -1.6 × 10⁻¹¹ C

(B): Deficit of 10¹⁰ electrons (positive charge).

q_B = +10¹⁰ × 1.6 × 10⁻¹⁹ = +1.6 × 10⁻⁹ C

2. (A) and (B) are placed near each other.
a) Specify whether the spheres attract or repel each other.
b) Draw the electrostatic force acting between the spheres, then show that its value is equal to F = 5.76 × 10⁻¹⁰ N.

a) They have charges of opposite signs (A negative, B positive), so they attract each other.

b) d = 20 cm = 0.2 m

F = k × |q_A × q_B| / d² = (9 × 10⁹) × (1.6 × 10⁻¹¹ × 1.6 × 10⁻⁹) / (0.2)²

F = (9 × 10⁹) × (2.56 × 10⁻²⁰) / 0.04 = 5.76 × 10⁻⁹ N

Note: The calculated value is 5.76 × 10⁻⁹ N, not 10⁻¹⁰ N. There seems to be a typo in the original problem statement.

3. (A) and (B), electrically isolated, are put in contact then separated again.
a) Determine the new charges q'_A and q'_B of (A) and (B).
b) Do the two spheres (A) and (B) attract or repel each other? Justify.
c) Determine the direction of electron movement, then verify that the number of exchanged electrons is 5.5 × 10⁸.

a) Q_total = q_A + q_B = -1.6 × 10⁻¹¹ + 1.6 × 10⁻⁹ = 1.584 × 10⁻⁹ C

q'_A = q'_B = Q_total / 2 = 1.584 × 10⁻⁹ / 2 = 7.92 × 10⁻¹⁰ C

b) They now have the same charge (positive), so they repel each other.

c) Electrons move from (A) to (B) (since A was negative and B was positive).

Charge lost by A = q'_A - q_A = 7.92 × 10⁻¹⁰ - (-1.6 × 10⁻¹¹) = 8.08 × 10⁻¹⁰ C

n = 8.08 × 10⁻¹⁰ / 1.6 × 10⁻¹⁹ = 5.05 × 10⁹ electrons

Note: The result is 5.05 × 10⁹, not 5.5 × 10⁸. There seems to be an error in the original problem data.

Exercise 13- Charge Carried by a Small Sphere

Consider a small light conductive sphere (S), carrying a charge q and suspended by an insulating thread, as well as a neutral plastic rod (R).

Part A: Electrification of the Rod (R)

The rod is electrified when rubbed against a piece of wool. It thus acquires a negative charge.
1. What does "a body is electrified" mean?
2. a) Explain how the rod acquired its charge. b) What is this process called?
3. Is the piece of wool charged? Why?

1. It means the body has acquired an electric charge (an imbalance between electrons and protons).

2. a) Through friction, electrons are transferred from the wool to the rod, which thus acquires a negative charge (it gained electrons).
b) This is called electrification by friction.

3. Yes, the piece of wool is charged. According to conservation of charge, since the rod gained electrons (became negative), the wool lost electrons (became positive).

Part B: Sign of q

The rod (R) is brought near the sphere (S). It is then observed that (S) is repelled.
What is the sign of the charge on (S)? Justify.

The rod (R) is negatively charged. Since (S) is repelled, it must have a charge of the same sign. Therefore, the charge q on (S) is negative.

Part C: Value of q

(R) is removed, then (S) is put in contact with a neutral sphere (S'), identical to (S). When electrostatic equilibrium is reached, (S) and (S') are placed at two fixed points A and B separated by 6 cm. It is observed that (S') repels (S).
1. a) Explain what happens when (S) and (S') are put in contact.
b) What does "electrostatic equilibrium" mean?
c) Determine, as a function of q, the charges q_1 and q_2 of (S) and (S') at electrostatic equilibrium.
2. Knowing that (S') repels (S) with an electrostatic force F of value 2.5 N, determine q.

1. a) During contact, the total charge is distributed equally between the two identical spheres to reach equilibrium.

b) Electrostatic equilibrium means that the charges stop moving and are distributed uniformly over the surface of the conductors.

c) Total charge = q + 0 = q. Since they are identical: q_1 = q_2 = q / 2

2. d = 6 cm = 0.06 m

F = k × |q_1 × q_2| / d² = k × (q/2)² / d² = k × q² / (4 × d²)

2.5 = (9 × 10⁹ × q²) / (4 × 0.06²) = (9 × 10⁹ × q²) / 0.0144

q² = (2.5 × 0.0144) / (9 × 10⁹) = 0.036 / (9 × 10⁹) = 4 × 10⁻¹² C²

q = √(4 × 10⁻¹²) = 2 × 10⁻⁶ C = 2 μC

Since the sign is negative (from Part B): q = -2 μC

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