Two point charges `q_(1)= 20 muC` and `q_(2)=25 muC` are placed at (-1, 1, 1) m and (3, 1, -2)m, with respect to a coordinate system. Find the magnitude and unit vector along electrostatic force on `q_(2)` ?

Answer

To solve the problem of finding the magnitude and unit vector along the electrostatic force on charge \( q_2 \) due to charge \( q_1 \), we can follow these steps:

### Step 1: Identify the coordinates and charges
The coordinates of the charges are:
- \( q_1 = 20 \, \mu C \) at \( (-1, 1, 1) \, m \)
- \( q_2 = 25 \, \mu C \) at \( (3, 1, -2) \, m \)

### Step 2: Calculate the distance \( r \) between the charges
The distance \( r \) between the two charges can be calculated using the distance formula in 3D:
\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting the coordinates:
\[ r = \sqrt{(3 - (-1))^2 + (1 - 1)^2 + (-2 - 1)^2} \]
\[ = \sqrt{(3 + 1)^2 + (0)^2 + (-3)^2} \]
\[ = \sqrt{4^2 + 0 + (-3)^2} \]
\[ = \sqrt{16 + 0 + 9} = \sqrt{25} = 5 \, m \]

### Step 3: Calculate the magnitude of the electrostatic force \( F \)
Using Coulomb's law, the magnitude of the electrostatic force \( F \) 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:
\[ F = 9 \times 10^9 \frac{|20 \times 10^{-6} \times 25 \times 10^{-6}|}{5^2} \]
\[ = 9 \times 10^9 \frac{500 \times 10^{-12}}{25} \]
\[ = 9 \times 10^9 \times 20 \times 10^{-12} \]
\[ = 0.18 \, N \]

### Step 4: Determine the direction of the force
The direction of the force can be represented by the vector \( \vec{r} \) from \( q_1 \) to \( q_2 \):
\[ \vec{r} = (3 - (-1)) \hat{i} + (1 - 1) \hat{j} + (-2 - 1) \hat{k} \]
\[ = (4) \hat{i} + (0) \hat{j} + (-3) \hat{k} \]
Thus, \( \vec{r} = 4 \hat{i} + 0 \hat{j} - 3 \hat{k} \).

### Step 5: Calculate the unit vector \( \hat{r} \)
The unit vector \( \hat{r} \) in the direction of \( \vec{r} \) is given by:
\[ \hat{r} = \frac{\vec{r}}{|\vec{r}|} \]
Where \( |\vec{r}| = r = 5 \, m \):
\[ \hat{r} = \frac{4 \hat{i} + 0 \hat{j} - 3 \hat{k}}{5} \]
\[ = \frac{4}{5} \hat{i} + 0 \hat{j} - \frac{3}{5} \hat{k} \]

### Final Results
- The magnitude of the electrostatic force on \( q_2 \) is \( 0.18 \, N \).
- The unit vector along the electrostatic force on \( q_2 \) is \( \frac{4}{5} \hat{i} - \frac{3}{5} \hat{k} \).