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mimidoudou93
@mimidoudou93
January 2021
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BONJOUR AIDER MOI SVP POUR L’EXERCICE 3 et 4
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scoladan
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Bonjour,
suite..
Ex 4)
1) ci-joint
2) diagonale du cube :
. diagonale de la face inférieure : l = √(a² + a²) = a√2
. diagonale du cube : D² = l² + a²
donc D = √[(a√2)² + a²] = a√3
Donc distance entre Cs⁺ et Cl⁻ : d = D/2 = a√3/2
3)a)
F = k x |Q₁Q₂|/r²
k = 1/4πε₀ = 9.10⁹SI
Q₁ = Q₂ = |e| = 1,6.10⁻¹⁹ C
et r = a = 0,205.10⁻⁹ m
soit F = 9.10⁹ x (1,6.10⁻¹⁹)²/(0,205.10⁻⁹)² ≈ 5,5.10⁻⁹ N
b) idem F' = k x e²/d² = k x e²/(3a²/4) = 4ke²/3a² = ...
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Thanks 1
mimidoudou93
Bonjour, merciii beaucoup
mimidoudou93
Mais il y a juste quelque chose que je n’est pas compris c’est pour la 3)a) c’est le 3 à l’envers accompagné d’un petit zéro sa signifie quoi ?
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Verified answer
Bonjour,suite..
Ex 4)
1) ci-joint
2) diagonale du cube :
. diagonale de la face inférieure : l = √(a² + a²) = a√2
. diagonale du cube : D² = l² + a²
donc D = √[(a√2)² + a²] = a√3
Donc distance entre Cs⁺ et Cl⁻ : d = D/2 = a√3/2
3)a)
F = k x |Q₁Q₂|/r²
k = 1/4πε₀ = 9.10⁹SI
Q₁ = Q₂ = |e| = 1,6.10⁻¹⁹ C
et r = a = 0,205.10⁻⁹ m
soit F = 9.10⁹ x (1,6.10⁻¹⁹)²/(0,205.10⁻⁹)² ≈ 5,5.10⁻⁹ N
b) idem F' = k x e²/d² = k x e²/(3a²/4) = 4ke²/3a² = ...