From 35fae812efd734b10b3e74c4c615565d0f3d947c Mon Sep 17 00:00:00 2001
From: Arian Ubuntu <ariandervishaj@icloud.com>
Date: Wed, 21 Feb 2024 16:03:45 +0100
Subject: [PATCH] Correction equation solution 17

---
 03_charge_electrique_champs_electrique.md | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/03_charge_electrique_champs_electrique.md b/03_charge_electrique_champs_electrique.md
index cb7ac7f..9b06035 100644
--- a/03_charge_electrique_champs_electrique.md
+++ b/03_charge_electrique_champs_electrique.md
@@ -456,8 +456,8 @@ F=F_{4\rightarrow 3},\\
 \end{align}
 Nous connaissons déjà la direction de la force et donc dans quelle direction se trouvera $Q_4$ (la direction pointée par l'angle $\theta_4$). La première condition nous permet finalement de calculer la distance entre $Q_3$ et $Q_4$, $r_{34}$
 \begin{align}
-F&=k\frac{Q_3Q_4}{r_{34}^2}=\frac{9\cdot 10^9\cdot 50\cdot 10^{-6}\cdot 65\cdot 10^{6-}}{r_{34}^2},\nonumber\\
-r_{34}^2&=k\frac{Q_3Q_4}{r_{34}^2}=\frac{29.2}{280},\nonumber\\
+F&=k\frac{Q_3Q_4}{r_{34}^2}=\frac{9\cdot 10^9\cdot 50\cdot 10^{-6}\cdot 65\cdot 10^{-6}}{r_{34}^2},\nonumber\\
+r_{34}^2&=k\frac{Q_3Q_4}{F}=\frac{29.2}{280},\nonumber\\
 r_{34}&=0.32\mathrm{m}.
 \end{align}
 
-- 
GitLab