8. Charge $$ Q_{A}=-20 \mu C $$ is located at $$ A(-6,4,7) $$ and change $$ Q_{B}=50 \mu C $$ is at $$ B(5,8-2) $$ free space The vector force excented on $$ Q_{A} $$ by $$ Q_{B} $$ is $$ \text { Columb laws }: F=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} R^{2}} \cdot \overrightarrow{q^{R}} $$ Where $$ K $$ is coulomb constant $$ \left(8.99 \times 10^{9} \mathrm{~N} \mathrm{M}\right. $$

Question

UpStudy AI · Accepted Answer

To determine the vector force exerted on charge $$ Q_A $$ by charge $$ Q_B $$ using Coulomb's Law, we'll follow these steps:

### Given:
- **Charge $$ Q_A $$**: $$ -20 \, \mu C $$ located at point $$ A(-6, 4, 7) $$
- **Charge $$ Q_B $$**: $$ 50 \, \mu C $$ located at point $$ B(5, 8, -2) $$
- **Coulomb's Constant $$ k $$**: $$ 8.99 \times 10^{9} \, \text{N} \cdot \text{m}^2/\text{C}^2 $$

### Coulomb's Law:
$$
\vec{F} = k \frac{Q_A Q_B}{R^3} \vec{R}
$$
where:
- $$ \vec{R} $$ is the displacement vector from $$ Q_A $$ to $$ Q_B $$
- $$ R $$ is the magnitude of $$ \vec{R} $$

### Step 1: Calculate the Displacement Vector $$ \vec{R} $$
$$
\vec{R} = B - A = (5 - (-6), \, 8 - 4, \, -2 - 7) = (11, \, 4, \, -9)
$$

### Step 2: Calculate the Distance $$ R $$
$$
R = \sqrt{11^2 + 4^2 + (-9)^2} = \sqrt{121 + 16 + 81} = \sqrt{218} \approx 14.7648 \, \text{meters}
$$

### Step 3: Apply Coulomb's Law
$$
\vec{F} = k \frac{Q_A Q_B}{R^3} \vec{R}
$$
Substituting the known values:
$$
Q_A Q_B = (-20 \times 10^{-6} \, C)(50 \times 10^{-6} \, C) = -1 \times 10^{-9} \, \text{C}^2
$$
$$
R^3 = (14.7648)^3 \approx 3228.4 \, \text{m}^3
$$
$$
\vec{F} = \frac{8.99 \times 10^{9} \times (-1 \times 10^{-9})}{3228.4} \times (11, \, 4, \, -9)
$$
$$
\vec{F} \approx \frac{-8.99}{3228.4} \times (11, \, 4, \, -9) \approx -0.002783 \times (11, \, 4, \, -9) \, \text{N}
$$

### Step 4: Calculate the Components
$$
F_x = -0.002783 \times 11 \approx -0.0306 \, \text{N}
$$
$$
F_y = -0.002783 \times 4 \approx -0.0111 \, \text{N}
$$
$$
F_z = -0.002783 \times (-9) \approx 0.0250 \, \text{N}
$$

### Final Vector Force:
$$
\vec{F} \approx (-0.0306 \, \mathbf{i} - 0.0111 \, \mathbf{j} + 0.0250 \, \mathbf{k}) \, \text{N}
$$

### Summary:
The vector force exerted on $$ Q_A $$ by $$ Q_B $$ is approximately:
$$
\vec{F} \approx -0.0306\, \mathbf{i} - 0.0111\, \mathbf{j} + 0.0250\, \mathbf{k} \, \text{N}
$$
The vector force on $$ Q_A $$ by $$ Q_B $$ is approximately $$ -0.0306\, \mathbf{i} - 0.0111\, \mathbf{j} + 0.0250\, \mathbf{k} \, \text{N} $$.

Real Tutor Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Physics Questions