Three point charges, q_(1),q_(2) and q3, lie along the x-axis at x=0,x=3.0cm and x=5.0 cm, respectively. Calculate the magnitude and the direction of the electric force on each of the three point charges wher q_(1)=+6.0mu C,q_(2)=+1.5mu C and q_(3)=-2.0mu C a. Draw the force diagram for this problem.

To calculate the magnitude and direction of the electric force on each of the three point charges, we can use Coulomb's law, which states that the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.<br /><br />The formula for Coulomb's law is:<br /><br />$F = k \cdot \frac{|q_1 \cdot q_2|}{r^2}$<br /><br />where:<br />- F is the electric force between the charges<br />- k is the Coulomb constant, which is approximately $8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$<br />- $q_1$ and $q_2$ are the charges<br />- r is the distance between the charges<br /><br />Let's calculate the electric force on each charge:<br /><br />1. Electric force on $q_1$:<br /> - Distance between $q_1$ and $q_2$: $r_{12} = 3.0 \, \text{cm} = 0.03 \, \text{m}$<br /> - Distance between $q_1$ and $q_3$: $r_{13} = 5.0 \, \text{cm} = 0.05 \, \text{m}$<br /> - Electric force on $q_1$ due to $q_2$: $F_{12} = k \cdot \frac{|q_1 \cdot q_2|}{r_{12}^2} = 8.99 \times 10^9 \cdot \frac{6.0 \times 10^{-6} \cdot 1.5 \times 10^{-6}}{(0.03)^2} \approx 1.65 \, \text{N}$<br /> - Electric force on $q_1$ due to $q_3$: $F_{13} = k \cdot \frac{|q_1 \cdot q_3|}{r_{13}^2} = 8.99 \times 10^9 \cdot \frac{6.0 \times 10^{-6} \cdot (-2.0 \times 10^{-6})}{(0.05)^2} \approx -1.41 \, \text{N}$<br /><br />2. Electric force on $q_2$:<br /> - Distance between $q_2$ and $q_3$: $r_{23} = 2.0 \, \text{cm} = 0.02 \, \text{m}$<br /> - Electric force on $q_2$ due to $q_3$: $F_{23} = k \cdot \frac{|q_2 \cdot q_3|}{r_{23}^2} = 8.99 \times 10^9 \cdot \frac{1.5 \times 10^{-6} \cdot 2.0 \times 10^{-6}}{(0.02)^2} \approx 2.25 \, \text{N}$<br /><br />3. Electric force on $q_3$:<br /> - Distance between $q_3$ and $q_1$: $r_{31} = 5.0 \, \text{cm} = 0.05 \, \text{m}$<br /> - Distance between $q_3$ and $q_2$: $r_{32} = 2.0 \, \text{cm} = 0.02 \, \text{m}$<br /> - Electric force on $q_3$ due to $q_1$: $F_{31} = k \cdot \frac{|q_1 \cdot q_3|}{r_{31}^2} = 8.99 \times 10^9 \cdot \frac{6.0 \times 10^{-6} \cdot (-2.0 \times 10^{-6})}{(0.05)^2} \approx -1.41 \, \text{N}$<br /> - Electric force on $q_3$ due to $q_2$: $F_{32} = k \cdot \frac{|q_2 \cdot q_3|}{r_{32}^2} = 8.99 \times 10^9 \cdot \frac{1.5 \times 10^{-6} \cdot 2.0 \times 10^{-6}}{(0.02)^2} \approx 2.25 \, \text{N}$<br /><br />Now, let's draw the force diagram for this problem:<br /><br />```<br /> q1 (+6.0 μC) q2 (+1.5 μC) q3 (-2.0 μC)<br /> ↑ ↑ ↑<br />