An illustration with 3 positively charged spheres. q Subscript 1 baseline and q Subscript 2 baseline are on the y-axis and equally spaced from the x-axis. The third, labeled q Subscript 3 baseline, is on the x-axis. The diagonal distance from the charge in the positive y-axis to the one on the x-axis is labeled d Subscript 13 baseline. The distance from the charge in on the negative y-axis to the one on the x-axis is labeled d Subscript 23 baseline. On the other side of q Subscript 3 baseline the angles are labeled theta Subscript B baseline and theta Subscript A baseline. All 3 charges are positive.

Question

A 2 µC charge q1 and a 2 µC charge q2 are 0.3 m from the x-axis. A 4 µC charge q3 is 0.4 m from the y-axis. The distances d13 and d23 are 0.5 m. Find the magnitude and direction of the resulting vector R. Round your answer to the nearest tenth.

A 2 µC charge q1 and a 2 µC charge q2 are 0.3 m from the x-axis. A 4 µC charge q3 is 0.4 m from the y-axis. The distances d13 and d23 are 0.5 m. Find the magnitude and direction of the resulting vector R. Round your answer to the nearest tenth.

The magnitude is
460.3
N.

The direction is

Answer 1

To find the direction of the resulting force vector \( \mathbf{R} \) exerted on charge \( q_3 \) due to charges \( q_1 \) and \( q_2 \), we need to consider the forces acting on \( q_3 \) from both \( q_1 \) and \( q_2 \).

Finding the Forces

1. Force due to \( q_1 \): \[ F_{13} = k \frac{|q_1 \cdot q_3|}{d_{13}^2} \] Where \( k \) is Coulomb's constant, \( k \approx 8.99 \times 10^9 , \text{N m}^2/\text{C}^2 \), \( q_1 = 2 \times 10^{-6} , \text{C} \), and \( q_3 = 4 \times 10^{-6} , \text{C} \). \[ d_{13} = 0.5 , \text{m} \] So, \[ F_{13} = (8.99 \times 10^9) \frac{(2 \times 10^{-6})(4 \times 10^{-6})}{(0.5)^2} \] \[ = (8.99 \times 10^9) \frac{8 \times 10^{-12}}{0.25} = (8.99 \times 10^9) \cdot 3.2 \times 10^{-11} = 287.68 , \text{N} \]

   The direction of this force is along the line connecting \( q_1 \) and \( q_3 \). The coordinates of \( q_1 \) are \( (0, 0.3) \), and those of \( q_3 \) are \( (0.4, 0) \). The angle \( \theta_1 \) formed with the x-axis can be calculated using: \[ \theta_1 = \tan^{-1} \left(\frac{0.3}{0.4}\right) \approx 36.87^\circ \]

2. Force due to \( q_2 \): Similarly, \[ F_{23} = k \frac{|q_2 \cdot q_3|}{d_{23}^2} \] Again, \( q_2 = 2 \times 10^{-6} , \text{C} \). \[ F_{23} = (8.99 \times 10^9) \frac{(2 \times 10^{-6})(4 \times 10^{-6})}{(0.5)^2} = 287.68 , \text{N} \] The direction of this force is along the line connecting \( q_2 \) and \( q_3 \). The coordinates of \( q_2 \) are \( (0, -0.3) \), and the direction angle \( \theta_2 \): \[ \theta_2 = \tan^{-1} \left(\frac{-0.3}{0.4}\right) \approx -36.87^\circ \]

Resultant Force Vector \( \mathbf{R} \)

We can resolve each force into its components.

• For \( F_{13} \):
  • \( F_{13x} = F_{13} \cos(36.87^\circ) \)
  • \( F_{13y} = F_{13} \sin(36.87^\circ) \)

Substituting, \[ F_{13x} = 287.68 \cos(36.87^\circ) = 287.68 \times 0.8 \approx 230.1 , \text{N} \] \[ F_{13y} = 287.68 \sin(36.87^\circ) = 287.68 \times 0.6 \approx 172.6 , \text{N} \]

• For \( F_{23} \):
  • \( F_{23x} = F_{23} \cos(-36.87^\circ) = 287.68 \times 0.8 \approx 230.1 , \text{N} \)
  • \( F_{23y} = F_{23} \sin(-36.87^\circ) = 287.68 \times (-0.6) \approx -172.6 , \text{N} \)

Now combine the forces: \[ R_x = F_{13x} + F_{23x} = 230.1 + 230.1 = 460.2 , \text{N} \] \[ R_y = F_{13y} + F_{23y} = 172.6 - 172.6 = 0 , \text{N} \]

Magnitude of Resulting Vector

The magnitude: \[ R = \sqrt{R_x^2 + R_y^2} = \sqrt{(460.2)^2 + (0)^2} = 460.2 , \text{N} \]

Direction of Resultant Vector

Since \( R_y = 0 \): \[ \text{Direction} = \tan^{-1} \left(\frac{R_y}{R_x}\right) = \tan^{-1}(0) = 0^\circ \]

Conclusion

The resulting vector \( \mathbf{R} \) has a magnitude of 460.2 N and is directed along the positive x-axis (0°).

Final answer: Magnitude: 460.2 N, Direction: 0°.