1. **Problem statement:** Three particles q1, q2, and q3 are placed along the x-axis with distances 2 cm between q1 and q2, and 2 cm between q2 and q3. Charges: $q_1 = 5\ \text{nC}$, $q_2 = -5\ \text{nC}$, $q_3 = 5\ \text{nC}$. Calculate the forces between them and the net forces on each. 2. **Formula used:** The electrostatic force between two point charges is given by Coulomb's law: $$F = k_e \frac{|q_1 q_2|}{r^2}$$ where $k_e = 8.99 \times 10^9\ \text{N m}^2/\text{C}^2$, $q_1$ and $q_2$ are charges, and $r$ is the distance between them. 3. **Important notes:** - Charges with opposite signs attract; same signs repel. - Distances must be in meters for SI units. 4. **Convert distances:** $2\ \text{cm} = 0.02\ \text{m}$ 5. **Calculate forces:** **a. Force $F_{1\to2}$ exerted by $q_1$ on $q_2$:** $$F_{1\to2} = k_e \frac{|q_1 q_2|}{r^2} = 8.99 \times 10^9 \times \frac{|5 \times 10^{-9} \times (-5) \times 10^{-9}|}{(0.02)^2}$$ $$= 8.99 \times 10^9 \times \frac{25 \times 10^{-18}}{0.0004} = 8.99 \times 10^9 \times 6.25 \times 10^{-14} = 5.62 \times 10^{-4} \text{N}$$ Direction: Since $q_1$ is positive and $q_2$ is negative, force is attractive, so $q_2$ is pulled toward $q_1$ (to the left). **b. Force $F_{2\to3}$ exerted by $q_2$ on $q_3$:** Same magnitude as above because charges and distance are same: $$F_{2\to3} = 5.62 \times 10^{-4} \text{N}$$ Direction: $q_2$ negative, $q_3$ positive, force attractive, so $q_3$ pulled toward $q_2$ (to the left). **c. Force $F_{3\to1}$ exerted by $q_3$ on $q_1$:** Distance between $q_3$ and $q_1$ is $4\ \text{cm} = 0.04\ \text{m}$. $$F_{3\to1} = k_e \frac{|q_3 q_1|}{r^2} = 8.99 \times 10^9 \times \frac{25 \times 10^{-18}}{(0.04)^2} = 8.99 \times 10^9 \times \frac{25 \times 10^{-18}}{0.0016} = 8.99 \times 10^9 \times 1.5625 \times 10^{-14} = 1.40 \times 10^{-4} \text{N}$$ Direction: Both positive charges repel, so force on $q_1$ is to the left (away from $q_3$). 6. **Net forces:** **d. Net force on $q_1$:** Forces on $q_1$ are from $q_2$ and $q_3$. Force from $q_2$ on $q_1$ is equal in magnitude to $F_{1\to2}$ but opposite direction (Newton's third law), so $5.62 \times 10^{-4} \text{N}$ to the right (since $q_2$ attracts $q_1$). Force from $q_3$ on $q_1$ is $1.40 \times 10^{-4} \text{N}$ to the left. Net force: $$F_{net,1} = 5.62 \times 10^{-4} - 1.40 \times 10^{-4} = 4.22 \times 10^{-4} \text{N}$$ Direction: to the right. **e. Net force on $q_2$:** Forces from $q_1$ and $q_3$. Force from $q_1$ on $q_2$ is $5.62 \times 10^{-4} \text{N}$ to the left (attraction). Force from $q_3$ on $q_2$ is $5.62 \times 10^{-4} \text{N}$ to the right (attraction). Net force: $$F_{net,2} = 5.62 \times 10^{-4} - 5.62 \times 10^{-4} = 0$$ No net force. **f. Net force on $q_3$:** Forces from $q_2$ and $q_1$. Force from $q_2$ on $q_3$ is $5.62 \times 10^{-4} \text{N}$ to the left (attraction). Force from $q_1$ on $q_3$ is equal in magnitude to $F_{3\to1}$ but opposite direction, so $1.40 \times 10^{-4} \text{N}$ to the right (repulsion). Net force: $$F_{net,3} = 5.62 \times 10^{-4} - 1.40 \times 10^{-4} = 4.22 \times 10^{-4} \text{N}$$ Direction: to the left. **Final answers:** - a. $5.62 \times 10^{-4} \text{N}$ to the left - b. $5.62 \times 10^{-4} \text{N}$ to the left - c. $1.40 \times 10^{-4} \text{N}$ to the left - d. $4.22 \times 10^{-4} \text{N}$ to the right - e. $0 \text{N}$ - f. $4.22 \times 10^{-4} \text{N}$ to the left