Question 5 A \( 2.0 \times 10-6 \mathrm{C} \) charge is located halfway between an \( 8.0 \times 10-6 \mathrm{C} \) charge and \( \mathrm{a}-5.0 \times \mathrm{x} \) \( 10-6 \mathrm{C} \). Find the net force on the \( 2.0 \times 10-6 \mathrm{C} \) charge.

To find the net force on the \( 2.0 \times 10^{-6} \, \mathrm{C} \) charge, we need to consider the forces exerted on it by the other two charges: \( 8.0 \times 10^{-6} \, \mathrm{C} \) and \( -5.0 \times 10^{-6} \, \mathrm{C} \). Let's denote the charges as follows: - \( q_1 = 8.0 \times 10^{-6} \, \mathrm{C} \) (positive charge) - \( q_2 = -5.0 \times 10^{-6} \, \mathrm{C} \) (negative charge) - \( q_3 = 2.0 \times 10^{-6} \, \mathrm{C} \) (the charge we are analyzing) Assume the positions of the charges are as follows: - \( q_1 \) is at position \( x = 0 \, \mathrm{m} \) - \( q_3 \) is at position \( x = d \, \mathrm{m} \) (halfway between \( q_1 \) and \( q_2 \)) - \( q_2 \) is at position \( x = 2d \, \mathrm{m} \) Since \( q_3 \) is halfway between \( q_1 \) and \( q_2 \), we can set \( d = 1 \, \mathrm{m} \) for simplicity. Thus: - \( q_1 \) is at \( x = 0 \, \mathrm{m} \) - \( q_3 \) is at \( x = 1 \, \mathrm{m} \) - \( q_2 \) is at \( x = 2 \, \mathrm{m} \) ### Step 1: Calculate the force between \( q_1 \) and \( q_3 \) Using Coulomb's law, the force \( F_{13} \) between \( q_1 \) and \( q_3 \) is given by: \[ F_{13} = k \frac{|q_1 q_3|}{r_{13}^2} \] where: - \( k = 8.99 \times 10^9 \, \mathrm{N \cdot m^2/C^2} \) (Coulomb's constant) - \( r_{13} = 1 \, \mathrm{m} \) (the distance between \( q_1 \) and \( q_3 \)) Substituting the values: \[ F_{13} = 8.99 \times 10^9 \frac{|(8.0 \times 10^{-6})(2.0 \times 10^{-6})|}{(1)^2} \] Calculating \( F_{13} \): \[ F_{13} = 8.99 \times 10^9 \frac{16.0 \times 10^{-12}}{1} = 143.84 \, \mathrm{N} \] Since both charges \( q_1 \) and \( q_3 \) are positive, the force \( F_{13} \) is repulsive, acting to the right (away from \( q_1 \)). ### Step 2: Calculate the force between \( q_2 \) and \( q_3 \) Now, we calculate the force \( F_{23} \) between \( q_2 \) and \( q_3 \): \[ F_{23} = k \frac{|q_2 q_3|}{r_{23}^2} \] where: - \( r_{23} = 1 \, \mathrm{m} \) (the distance between \( q_2 \) and \( q_3 \)) Substituting the values: \[ F_{23} = 8.99 \times 10^9 \frac{|(-5.0 \times 10^{-6})(2.0 \times 10^{-6})|}{(1)^2} \] Calculating \( F_{23} \): \[ F_{23} = 8.99 \times 10^9 \frac{10.0 \times 10^{-12}}{1} = 89.9 \, \mathrm{N} \] Since \( q_2 \) is negative and \( q_3 \) is positive, the force \( F_{23} \) is attractive, acting to the left (toward \( q_2 \)). ### Step 3: Determine the net force on \( q_3 \) Now we can find the net force \( F_{\text{net}} \) on \( q_3 \): \[ F_{\text{net}} = F_{13} - F_{23} \] Substituting the values: \[ F_{\text{net}} = 143.84 \, \mathrm{N} - 89.9 \, \mathrm{N} = 53.94 \, \mathrm{N} \] ### Conclusion The net force on the \( 2.0 \times 10^{-6} \, \mathrm{C} \) charge is approximately \( 53.94 \, \mathrm{N} \) to the right.