1. **Problem statement:** Three charges $q_1 = 5$ nC, $q_2 = -5$ nC, and $q_3 = 5$ nC are placed along the x-axis with $2$ cm between $q_1$ and $q_2$, and $2$ cm between $q_2$ and $q_3$. We need to find the forces between them and the net forces on each charge. 2. **Formula used:** The electrostatic force between two point charges is given by Coulomb's law: $$F = k \frac{|q_a q_b|}{r^2}$$ where $k = 8.99 \times 10^9$ N m$^2$/C$^2$, $q_a$ and $q_b$ are the charges, and $r$ is the distance between them. 3. **Convert units:** Distances: $2$ cm = $0.02$ m Charges: $5$ nC = $5 \times 10^{-9}$ C 4. **Calculate forces:** **a. Force $q_1$ exerts on $q_2$:** $$F_{12} = k \frac{|q_1 q_2|}{r^2} = 8.99 \times 10^9 \times \frac{5 \times 10^{-9} \times 5 \times 10^{-9}}{(0.02)^2}$$ $$= 8.99 \times 10^9 \times \frac{25 \times 10^{-18}}{0.0004} = 8.99 \times 10^9 \times 6.25 \times 10^{-14} = 5.62 \times 10^{-4} \text{ N}$$ Since $q_1$ is positive and $q_2$ is negative, the force is attractive, so $q_1$ pulls $q_2$ towards it (to the left). **b. Force $q_2$ exerts on $q_3$:** Same magnitude as above because charges and distance are the same: $$F_{23} = 5.62 \times 10^{-4} \text{ N}$$ Since $q_2$ is negative and $q_3$ positive, force is attractive, so $q_2$ pulls $q_3$ towards it (to the left). **c. Force $q_3$ exerts on $q_1$:** Distance between $q_1$ and $q_3$ is $4$ cm = $0.04$ m. $$F_{31} = k \frac{|q_3 q_1|}{r^2} = 8.99 \times 10^9 \times \frac{25 \times 10^{-18}}{(0.04)^2} = 8.99 \times 10^9 \times \frac{25 \times 10^{-18}}{0.0016} = 8.99 \times 10^9 \times 1.5625 \times 10^{-14} = 1.40 \times 10^{-4} \text{ N}$$ Both charges positive, so force is repulsive; $q_3$ pushes $q_1$ to the left. 5. **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 $F_{12}$ but opposite direction (since action-reaction), so $5.62 \times 10^{-4}$ N to the right (because $q_2$ attracts $q_1$). - Force from $q_3$ on $q_1$ is $1.40 \times 10^{-4}$ N to the left (repulsive). Net force: $$F_{net,1} = 5.62 \times 10^{-4} - 1.40 \times 10^{-4} = 4.22 \times 10^{-4} \text{ N to the right}$$ **e. Net force on $q_2$:** Forces from $q_1$ and $q_3$. - From $q_1$: $5.62 \times 10^{-4}$ N to the right (attraction). - From $q_3$: $5.62 \times 10^{-4}$ N to the left (attraction). They are equal and opposite, so: $$F_{net,2} = 0 \text{ N}$$ **f. Net force on $q_3$:** Forces from $q_2$ and $q_1$. - From $q_2$: $5.62 \times 10^{-4}$ N to the left (attraction). - From $q_1$: $1.40 \times 10^{-4}$ N to the right (repulsion). Net force: $$F_{net,3} = 5.62 \times 10^{-4} - 1.40 \times 10^{-4} = 4.22 \times 10^{-4} \text{ N to the left}$$