Questions: Find F1 and F2, then calculate the net force on q3. F1=33.5 N F2= F2 is the force exerted onto q3 by q2. Forces directed left are negative (-); right are positive (+).

Solution Steps

Using Coulomb's Law: \[ F = k \frac{|q_2 q_3|}{r^2} \]

Where:

• \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
• \( q_2 = +90.6 \times 10^{-6} \, \text{C} \)
• \( q_3 = -84.2 \times 10^{-6} \, \text{C} \)
• \( r = 0.432 \, \text{m} \)

\[ F_2 = 8.99 \times 10^9 \frac{(90.6 \times 10^{-6})(84.2 \times 10^{-6})}{(0.432)^2} \]

\[ F_2 = 8.99 \times 10^9 \frac{(90.6 \times 10^{-6})(84.2 \times 10^{-6})}{0.186624} \]

\[ F_2 = 8.99 \times 10^9 \frac{7.62612 \times 10^{-9}}{0.186624} \]

\[ F_2 = 8.99 \times 10^9 \times 4.087 \times 10^{-8} \]

\[ F_2 = 367.5 \, \text{N} \]

Since \( q_2 \) is positive and \( q_3 \) is negative, the force \( F_2 \) will be attractive and directed to the right (positive direction).

Given:

• \( F_1 = 33.5 \, \text{N} \) (directed to the left, negative direction)
• \( F_2 = 367.5 \, \text{N} \) (directed to the right, positive direction)

Net force \( \vec{F} \): \[ \vec{F} = F_2 - F_1 \] \[ \vec{F} = 367.5 \, \text{N} - 33.5 \, \text{N} \] \[ \vec{F} = 334 \, \text{N} \]

Final Answer