From 1ca41b959e3e5cd61deec044571f22c0ad8ddf05 Mon Sep 17 00:00:00 2001
From: Orestis <orestis.malaspinas@pm.me>
Date: Mon, 7 Mar 2022 12:59:44 +0100
Subject: [PATCH] correction
---
03_charge_electrique_champs_electrique.md | 8 +++++---
1 file changed, 5 insertions(+), 3 deletions(-)
diff --git a/03_charge_electrique_champs_electrique.md b/03_charge_electrique_champs_electrique.md
index 7754ded..12c15c1 100644
--- a/03_charge_electrique_champs_electrique.md
+++ b/03_charge_electrique_champs_electrique.md
@@ -389,13 +389,15 @@ $$
$$
La loi de Coulomb nous donne immédiatement $F_{2\rightarrow 3}$ et $F_{1\rightarrow 3}$
\begin{align}
-F_{2\rightarrow 3}&=k\frac{Q_2Q_3}{r_23^2}\cong 1.2\mathrm{N},\\
-F_{1\rightarrow 3}&=k\frac{Q_1Q_3}{r_13^2}\cong 2.7\mathrm{N}.
+F_{2\rightarrow 3}&=k\frac{Q_2Q_3}{r_{23}^2}\cong
+2.02\mathrm{N},\\
+F_{1\rightarrow 3}&=k\frac{Q_1Q_3}{r_{13}^2}\cong
+0.86\mathrm{N}.
\end{align}
La charge $Q_3$ étant négative la force de $Q_1$ est répulsive, alors que celle de $Q_2$ est attractive.
On a donc finalement
$$
-F=-F_{2\rightarrow 3}+F_{1\rightarrow 3}=-1.5\mathrm{N},
+F=-F_{2\rightarrow 3}+F_{1\rightarrow 3}=-1.16\mathrm{N},
$$
et la force pointe vers la gauche.
--
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