Questions: Q1=8.00 nC is at (0.800 m, 0) ; Q2=-5.00 nC is at (0,0.300 m); Q 3=2.00 nC is at (0,0). What is the direction in degrees of the net force on Q3 counter- clockwise from the +x-direction?

Solution Steps

• Calculate the force between \( Q_3 \) and \( Q_1 \) using Coulomb's Law: \[ F_{31} = k \frac{|Q_3 \cdot Q_1|}{r_{31}^2} \] where \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \), \( Q_3 = 2.00 \, \text{nC} \), \( Q_1 = 8.00 \, \text{nC} \), and \( r_{31} = 0.800 \, \text{m} \).

• Calculate the force between \( Q_3 \) and \( Q_2 \): \[ F_{32} = k \frac{|Q_3 \cdot Q_2|}{r_{32}^2} \] where \( Q_2 = -5.00 \, \text{nC} \) and \( r_{32} = 0.300 \, \text{m} \).

• The force \( F_{31} \) is along the line connecting \( Q_3 \) and \( Q_1 \), which is along the positive \( x \)-axis.
• The force \( F_{32} \) is along the line connecting \( Q_3 \) and \( Q_2 \), which is along the negative \( y \)-axis.

• Calculate the \( x \)-component of the net force: \[ F_{x} = F_{31} \] since \( F_{32} \) has no \( x \)-component.

• Calculate the \( y \)-component of the net force: \[ F_{y} = -F_{32} \] since \( F_{32} \) is directed along the negative \( y \)-axis.

• Use the tangent function to find the angle \( \theta \) of the net force with respect to the positive \( x \)-axis: \[ \theta = \tan^{-1}\left(\frac{F_{y}}{F_{x}}\right) \]
• Since \( F_{y} \) is negative, the angle will be in the fourth quadrant, measured counter-clockwise from the positive \( x \)-axis.

Final Answer