14:I[70017,["/_next/static/immutable/chunks/2gtpnylaz50zo.js","/_next/static/immutable/chunks/22kg_k_j3c_ao.js","/_next/static/immutable/chunks/3s-qhi50ildb6.js"],""] 15:I[16632,["/_next/static/immutable/chunks/2gtpnylaz50zo.js","/_next/static/immutable/chunks/22kg_k_j3c_ao.js","/_next/static/immutable/chunks/3s-qhi50ildb6.js"],"default"] 16:I[3261,["/_next/static/immutable/chunks/2gtpnylaz50zo.js","/_next/static/immutable/chunks/22kg_k_j3c_ao.js","/_next/static/immutable/chunks/3s-qhi50ildb6.js"],"default"] 17:I[44578,["/_next/static/immutable/chunks/2gtpnylaz50zo.js","/_next/static/immutable/chunks/22kg_k_j3c_ao.js","/_next/static/immutable/chunks/3s-qhi50ildb6.js"],"default"] 18:T1999,{ "consigne": "Attention aux conversions : écris les distances en mètres. Réponds avec 2 chiffres significatifs.", "donnees": [ { "cle": "k", "tex": "k = 9{,}0 \\times 10^{9}\\ \\mathrm{N.m^2.C^{-2}}" }, { "cle": "G", "tex": "G = 6{,}67 \\times 10^{-11}\\ \\mathrm{N.m^2.kg^{-2}}" }, { "cle": "conversions", "tex": "1\\ \\mathrm{nC} = 10^{-9}\\ \\mathrm{C} \\quad 1\\ \\mu\\mathrm{C} = 10^{-6}\\ \\mathrm{C}" } ], "questions": [ { "enonce": "Deux charges de 20 nC sont distantes de 4,0 cm. Quelle relation donne l'intensité de la force que chacune exerce sur l'autre ?", "choix": [ "F = k \\times \\dfrac{|q_1 \\times q_2|}{d}", "F = k \\times |q_1 \\times q_2| \\times d^2", "F = k \\times \\dfrac{|q_1 \\times q_2|}{d^2}" ], "bonne": 2, "explication": "La force diminue en 1/d² quand la distance augmente. La distance est donc au carré et au dénominateur.", "donnees": [ "conversions" ] }, { "enonce": "Quelle est l'intensité de cette force ?", "aide": "20 nC = 2,0×10⁻⁸ C et 4,0 cm = 0,040 m.", "reponse": 0.00225, "tolerancePct": 4, "unite": "N", "explication": "F = 9,0×10⁹ × (2,0×10⁻⁸)² / (0,040)² = 2,3×10⁻³ N.", "donnees": [ "k", "conversions" ] }, { "enonce": "Deux charges valent 1,5 µC et 2,0 µC. Elles sont distantes de 30 cm. Quelle est l'intensité de la force que chacune exerce sur l'autre ?", "aide": "1,5 µC = 1,5×10⁻⁶ C et 30 cm = 0,30 m.", "reponse": 0.3, "tolerancePct": 4, "unite": "N", "explication": "F = 9,0×10⁹ × (1,5×10⁻⁶ × 2,0×10⁻⁶) / (0,30)² = 0,30 N.", "donnees": [ "k", "conversions" ] }, { "enonce": "Deux charges de 40 nC se repoussent avec une force F = 3,6×10⁻² N. Quelle distance les sépare ? Réponds en cm.", "aide": "Isole d² dans la loi de Coulomb, puis n'oublie pas la racine carrée.", "reponse": 2.0, "tolerancePct": 5, "unite": "cm", "explication": "d² = k × q² / F = 9,0×10⁹ × (4,0×10⁻⁸)² / (3,6×10⁻²) = 4,0×10⁻⁴ m², donc d = 2,0×10⁻² m = 2,0 cm.", "donnees": [ "k", "conversions" ] }, { "enonce": "Deux charges identiques sont placées à 10 cm l'une de l'autre. Elles se repoussent avec une force F = 9,0×10⁻³ N. Combien vaut chaque charge ? Réponds en nC.", "aide": "Comme les deux charges sont égales, |q₁ × q₂| vaut q². Isole q², puis prends la racine.", "reponse": 100, "tolerancePct": 5, "unite": "nC", "explication": "q² = F × d² / k = 9,0×10⁻³ × (0,10)² / (9,0×10⁹) = 1,0×10⁻¹⁴, donc q = 1,0×10⁻⁷ C = 100 nC.", "donnees": [ "k", "conversions" ] }, { "enonce": "Sans calculatrice : on triple la distance entre deux charges. Par combien la force est-elle divisée ?", "aide": "La force varie en 1/d². Si la distance est multipliée par un nombre, la force est divisée par le carré de ce nombre.", "reponse": 9, "tolerancePct": 2, "explication": "La force est divisée par 3² = 9. Doubler la distance divise déjà la force par 4.", "donnees": [] }, { "enonce": "Une charge de 1,6×10⁻⁵ C subit une force de 4,8×10⁻³ N. Quelle relation donne la valeur du champ à cet endroit ?", "choix": [ "E = \\dfrac{F}{q}", "E = F \\times q", "E = \\dfrac{q}{F}" ], "bonne": 0, "explication": "Le champ est la force par unité de charge. On divise donc la force par la charge qui subit cette force.", "donnees": [] }, { "enonce": "Quelle est la valeur de ce champ ?", "reponse": 300, "tolerancePct": 3, "unite": "N.C⁻¹|N/C", "explication": "E = 4,8×10⁻³ / (1,6×10⁻⁵) = 3,0×10² N.C⁻¹.", "donnees": [] }, { "enonce": "Quelle force subit une charge de 2,0 nC placée dans un champ de 5,0×10⁴ N.C⁻¹ ?", "aide": "Pars de E = F / q et isole F.", "reponse": 0.0001, "tolerancePct": 3, "unite": "N", "explication": "F = q × E = 2,0×10⁻⁹ × 5,0×10⁴ = 1,0×10⁻⁴ N.", "donnees": [ "conversions" ] }, { "enonce": "Une charge ponctuelle de 5,0 nC crée un champ autour d'elle. Quelle relation donne la valeur de ce champ à la distance d ?", "choix": [ "E = k \\times |q| \\times d^2", "E = k \\times \\dfrac{|q|}{d^2}", "E = \\dfrac{k}{|q| \\times d^2}" ], "bonne": 1, "explication": "Le champ varie en 1/d², comme la force. On obtient cette relation en divisant la force de Coulomb par la charge test.", "donnees": [] }, { "enonce": "Quelle est la valeur de ce champ à 15 cm de la charge ?", "reponse": 2000, "tolerancePct": 4, "unite": "N.C⁻¹|N/C", "explication": "E = 9,0×10⁹ × 5,0×10⁻⁹ / (0,15)² = 45 / 0,0225 = 2,0×10³ N.C⁻¹.", "donnees": [ "k", "conversions" ] }, { "enonce": "À quelle distance de cette même charge le champ est-il 4 fois plus faible qu'à 15 cm ? Réponds en cm.", "aide": "Le champ varie en 1/d². Pour diviser le champ par 4, il faut donc doubler la distance.", "reponse": 30, "tolerancePct": 4, "unite": "cm", "explication": "Pour diviser le champ par 4, il faut multiplier la distance par 2, donc d = 30 cm.", "donnees": [] }, { "enonce": "Mars a une masse de 6,4×10²³ kg et un rayon de 3,4×10⁶ m. Que vaut le champ de gravitation à sa surface ?", "aide": "Utilise l'expression du champ de gravitation. Ici, la distance d est le rayon de la planète.", "reponse": 3.69, "tolerancePct": 4, "unite": "N.kg⁻¹|N/kg", "explication": "g = 6,67×10⁻¹¹ × 6,4×10²³ / (3,4×10⁶)² = 3,7 N.kg⁻¹, soit un peu plus du tiers de la pesanteur terrestre.", "donnees": [ "G" ] }, { "enonce": "Quelle est l'unité du champ électrique ?", "aide": "E = F / q : une force divisée par une charge.", "unite": "N.C⁻¹|N/C", "explication": "Le champ est une force par unité de charge, donc il s'exprime en newtons par coulomb, N.C⁻¹.", "donnees": [] } ] } 19:Taef,

k = 9{,}0 \times 10^{9}\ \mathrm{N.m^2.C^{-2}}

1a:Tb6d,

G = 6{,}67 \times 10^{-11}\ \mathrm{N.m^2.kg^{-2}}

1b:Ta90,

1\ \mathrm{nC} = 10^{-9}\ \mathrm{C} \quad 1\ \mu\mathrm{C} = 10^{-6}\ \mathrm{C}

1c:Td68,

F = k \times \dfrac{|q_1 \times q_2|}{d}

1d:Tcef,

F = k \times |q_1 \times q_2| \times d^2

1e:Tee6,

F = k \times \dfrac{|q_1 \times q_2|}{d^2}

1f:T67f,

E = \dfrac{F}{q}

20:T450,

E = F \times q

21:T67d,

E = \dfrac{q}{F}

22:T76a,

E = k \times |q| \times d^2

23:T9c1,

E = k \times \dfrac{|q|}{d^2}

24:T95c,

E = \dfrac{k}{|q| \times d^2}

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