To find the magnitude and unit vector along the electrostatic force on charge \( q_2 \), we will follow these steps:
### Step 1: Identify the Charges and Their Positions
Given:
- Charge \( q_1 = 20 \, \mu C = 20 \times 10^{-6} \, C \) located at point \( (-1, 1, 1) \, m \)
- Charge \( q_2 = 25 \, \mu C = 25 \times 10^{-6} \, C \) located at point \( (3, 1, -2) \, m \)
### Step 2: Calculate the Distance Between the Charges
The distance \( R \) between the two charges can be calculated using the distance formula in three-dimensional space:
\[
R = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}
\]
Substituting the coordinates:
- \( x_1 = -1, y_1 = 1, z_1 = 1 \)
- \( x_2 = 3, y_2 = 1, z_2 = -2 \)
\[
R = \sqrt{(3 - (-1))^2 + (1 - 1)^2 + (-2 - 1)^2}
\]
\[
R = \sqrt{(3 + 1)^2 + 0^2 + (-3)^2}
\]
\[
R = \sqrt{4^2 + 0 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, m
\]
### Step 3: Calculate the Electrostatic Force
Using Coulomb's Law, the magnitude of the electrostatic force \( F \) between two point charges is given by:
\[
F = k \frac{|q_1 q_2|}{R^2}
\]
Where \( k = 9 \times 10^9 \, N m^2/C^2 \).
Substituting the values:
\[
F = 9 \times 10^9 \frac{|20 \times 10^{-6} \times 25 \times 10^{-6}|}{5^2}
\]
\[
F = 9 \times 10^9 \frac{500 \times 10^{-12}}{25}
\]
\[
F = 9 \times 10^9 \times 20 \times 10^{-12}
\]
\[
F = 180 \times 10^{-3} \, N = 0.18 \, N
\]
### Step 4: Determine the Direction of the Force
The force on \( q_2 \) due to \( q_1 \) will be directed along the line connecting the two charges. We need to find the position vector \( \mathbf{R} \) from \( q_1 \) to \( q_2 \):
\[
\mathbf{R} = (x_2 - x_1) \hat{i} + (y_2 - y_1) \hat{j} + (z_2 - z_1) \hat{k}
\]
\[
\mathbf{R} = (3 - (-1)) \hat{i} + (1 - 1) \hat{j} + (-2 - 1) \hat{k}
\]
\[
\mathbf{R} = 4 \hat{i} + 0 \hat{j} - 3 \hat{k}
\]
### Step 5: Calculate the Unit Vector
To find the unit vector \( \hat{R} \), we divide \( \mathbf{R} \) by its magnitude:
\[
|\mathbf{R}| = \sqrt{(4)^2 + (0)^2 + (-3)^2} = \sqrt{16 + 0 + 9} = \sqrt{25} = 5
\]
Thus, the unit vector \( \hat{R} \) is:
\[
\hat{R} = \frac{\mathbf{R}}{|\mathbf{R}|} = \frac{4 \hat{i} + 0 \hat{j} - 3 \hat{k}}{5} = \frac{4}{5} \hat{i} + 0 \hat{j} - \frac{3}{5} \hat{k}
\]
### Final Result
- The magnitude of the electrostatic force on \( q_2 \) is \( 0.18 \, N \).
- The unit vector along the direction of the force is \( \frac{4}{5} \hat{i} - \frac{3}{5} \hat{k} \).
