Questions: What is the magnitude and direction net force on qi.?

Solution Steps

\(q_1 = +3.0 \mu C = +3.0 \times 10^{-6} C\) \(q_2 = -4.0 \mu C = -4.0 \times 10^{-6} C\) \(q_3 = -7.0 \mu C = -7.0 \times 10^{-6} C\) \(r_{12} = 0.20 m\) \(r_{13} = 0.15 m\)

The force between \(q_1\) and \(q_2\) is given by Coulomb's law: \(F_{12} = k \frac{|q_1q_2|}{r_{12}^2}\) \(F_{12} = (8.99 \times 10^9 N m^2/C^2) \frac{|(3.0 \times 10^{-6} C)(-4.0 \times 10^{-6} C)|}{(0.20 m)^2}\) \(F_{12} = 2.697 N\) Since \(q_1\) and \(q_2\) have opposite signs, the force is attractive, so \(q_2\) pulls \(q_1\) to the left.

The force between \(q_1\) and \(q_3\) is given by Coulomb's law: \(F_{13} = k \frac{|q_1q_3|}{r_{13}^2}\) \(F_{13} = (8.99 \times 10^9 N m^2/C^2) \frac{|(3.0 \times 10^{-6} C)(-7.0 \times 10^{-6} C)|}{(0.15 m)^2}\) \(F_{13} = 8.391 N\) Since \(q_1\) and \(q_3\) have opposite signs, the force is attractive, so \(q_3\) pulls \(q_1\) to the right.

The net force on \(q_1\) is the vector sum of the forces \(F_{12}\) and \(F_{13}\). Since \(F_{12}\) is to the left and \(F_{13}\) is to the right, we subtract the magnitudes: \(F_{net} = F_{13} - F_{12}\) \(F_{net} = 8.391 N - 2.697 N\) \(F_{net} = 5.694 N\)

Since \(F_{13}\) is greater than \(F_{12}\), the net force is directed to the right.

Final Answer