Files
CITISE1/TSE/Electrostatique/exemple2.md
2026-04-08 20:11:20 +02:00

14 lines
346 B
Markdown

# Large Exemple 2
$q_a = -1.80\mu C$, $q_b = 3.50 \mu C$, $q_c = -1.40 \mu C$
$a = 5.20$ cm
---
$\vec{F_A} = \vec{F_{B,A}} + \vec{F_{C,A}}$
$q_b q_c = 2a$ et $q_c q_a = a$, ainsi $q_bq_a = \sqrt{(2a)^2 + a^2} = a\sqrt{5}$
$\vec{F_{B,A}} = \frac{q_a q_b \cdot 9\cdot 10^9}{r^2}$, avec $\frac{1}{4\pi \epsilon}=9\cdot 10^9$
$\vec{F_{C,A}} = $