Question

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.
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

Answer 1

To find the resultant vector \(\mathbf{R}\) from the three positive charges, we'll first need to calculate the forces that each charge exerts on the other.

### Given:
- \(q_1 = 2 \, \mu C = 2 \times 10^{-6} \, C\)
- \(q_2 = 2 \, \mu C = 2 \times 10^{-6} \, C\)
- \(q_3 = 4 \, \mu C = 4 \times 10^{-6} \, C\)
- Positions of charges:
- \(q_1\) at \((0, 0.3)\) m
- \(q_2\) at \((0, -0.3)\) m
- \(q_3\) at \((0.4, 0)\) m
- Distances \(d_{13} = d_{23} = 0.5 \, m\)

### Step 1: Calculate Forces

Using Coulomb's Law, the force between two charges is given by:

\[
F = k \frac{|q_1 q_2|}{r^2}
\]

where:
- \(k \approx 8.99 \times 10^9 \, N \cdot m^2/C^2\)
- \(r\) is the distance between the charges.

#### For \(F_{13}\) (force between \(q_1\) and \(q_3\)):
\[
F_{13} = k \frac{|q_1 q_3|}{d_{13}^2} = (8.99 \times 10^9) \frac{(2 \times 10^{-6})(4 \times 10^{-6})}{(0.5)^2}
\]
\[
F_{13} = (8.99 \times 10^9) \frac{8 \times 10^{-12}}{0.25} = (8.99 \times 10^9) \times (3.2 \times 10^{-11}) \approx 287.68 \, N
\]

#### For \(F_{23}\) (force between \(q_2\) and \(q_3\)):
\[
F_{23} = k \frac{|q_2 q_3|}{d_{23}^2} = (8.99 \times 10^9) \frac{(2 \times 10^{-6})(4 \times 10^{-6})}{(0.5)^2} = 287.68 \, N
\]

### Step 2: Direction of Forces

#### \(\mathbf{F_{13}}\):
- This force acts to the left (towards \(q_1\)).
- Angle with horizontal = \(\theta_A = \tan^{-1}\left(\frac{0.3}{0.4}\right) = \tan^{-1}(0.75) \approx 36.87^\circ\)

#### \(\mathbf{F_{23}}\):
- This force acts to the right (towards \(q_2\)).
- Angle with horizontal = \(\theta_B = \tan^{-1}\left(\frac{-0.3}{0.4}\right) = \tan^{-1}(-0.75) \approx -36.87^\circ\)

### Step 3: Component Calculation

Now calculate the components of these forces.

- For \(F_{13}\):
- \(F_{13,x} = F_{13} \cos(36.87^\circ) = 287.68 \cdot \frac{4}{5} = 230.14 \, N\)
- \(F_{13,y} = F_{13} \sin(36.87^\circ) = 287.68 \cdot \frac{3}{5} = 172.61 \, N\)

- For \(F_{23}\):
- \(F_{23,x} = F_{23} \cos(-36.87^\circ) = 287.68 \cdot \frac{4}{5} = 230.14 \, N\)
- \(F_{23,y} = F_{23} \sin(-36.87^\circ) = 287.68 \cdot -\frac{3}{5} = -172.61 \, N\)

### Step 4: Resultant Force Components

Now we can find the resulting components:

\[
R_x = F_{13,x} + F_{23,x} = 230.14 + 230.14 = 460.28 \, N
\]
\[
R_y = F_{13,y} + F_{23,y} = 172.61 - 172.61 = 0 \, N
\]

### Step 5: Resultant Magnitude

Finally, the magnitude of the resultant vector \(\mathbf{R}\):

\[
R = \sqrt{R_x^2 + R_y^2} = \sqrt{(460.28)^2 + (0)^2} = 460.28 \, N
\]

### Final Answer
\(\boxed{460.3 \, N}\)