Solution :
\( E = k \frac{|q|}{r^2} = 9 \times 10^9 \times \frac{2 \times 10^{-6}}{(0,1)^2} = 1,8 \times 10^6 \, \text{N/C} \)

Solution :
\( \vec{F} = q’ \vec{E} \) ⇒ \( F = 3 \times 10^{-6} \times 1,8 \times 10^6 = 5,4 \, \text{N} \)
Direction : vers la charge positive (attraction)

Solution :
Champ de \( q_1 \): \( E_1 = k \frac{|q_1|}{r^2} = 9 \times 10^9 \times \frac{10^{-6}}{(0,1)^2} = 9 \times 10^5 \, \text{N/C} \) (→)
Champ de \( q_2 \): \( E_2 = 9 \times 10^5 \, \text{N/C} \) (→ même direction)
Champ total: \( E = E_1 + E_2 = 1,8 \times 10^6 \, \text{N/C} \) vers \( q_2 \)

Solution :
1. Lignes radiales sortant de la charge (+)
2. Lignes sortant de la + vers la – (courbes)

Solution :
\( F_e = k \frac{q^2}{r^2} \), \( F_g = G \frac{m^2}{r^2} \)
\( \frac{F_e}{F_g} = \frac{k q^2}{G m^2} = \frac{9 \times 10^9 \times (1,6 \times 10^{-19})^2}{6,67 \times 10^{-11} \times (9,1 \times 10^{-31})^2} \approx 4,2 \times 10^{42} \)

Solution :
\( W = q E d = 2 \times 10^{-6} \times 1000 \times 0,05 = 10^{-4} \, \text{J} \)

Solution :
Équilibre ⇒ \( m \vec{g} + q \vec{E} = \vec{0} \)
\( E = \frac{m g}{|q|} = \frac{10^{-6} \times 9,81}{10^{-9}} = 9810 \, \text{N/C} \) vers le bas

Solution :
\( V = k \frac{q}{r} = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{0,05} = 9 \times 10^5 \, \text{V} \)

Solution :
\( E_p = k \frac{q_1 q_2}{r} = 9 \times 10^9 \times \frac{2 \times 10^{-6} \times (-3 \times 10^{-6})}{0,1} = -0,54 \, \text{J} \)

Solution :
1. \( E = k \frac{|Q|}{r^2} = 9 \times 10^9 \times \frac{10^{-6}}{(0,1)^2} = 9 \times 10^5 \, \text{N/C} \) (radial)
2. \( F = |q|E = 0,5 \times 10^{-6} \times 9 \times 10^5 = 0,45 \, \text{N} \) (attractive)
3. \( V = k \frac{Q}{r} = 9 \times 10^9 \times \frac{10^{-6}}{0,1} = 9 \times 10^4 \, \text{V} \)