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}} =