The equal point charges each of `3 muC` are separated by a centain distance in metres. If they are located at `(hat(i) + hat(j) + hat(k)) and (2 hat(i) + 3 hat(j) + hat(k))` , then the electrostatic force between them is .

Options

A. ` 9 xx 10^(3)` N
B. `16 xx 10^(-3) N`
C. `10^(-3)` N
D. `9 xx 10^(-2) N`

Explanation

To solve the problem of finding the electrostatic force between two equal point charges of \(3 \, \mu C\) located at the points \((\hat{i} + \hat{j} + \hat{k})\) and \((2\hat{i} + 3\hat{j} + \hat{k})\), we will follow these steps:

### Step 1: Identify the Charges
The charges are given as:
\[ q_1 = q_2 = 3 \, \mu C = 3 \times 10^{-6} \, C \]

### Step 2: Determine the Position Vectors
The position vectors of the charges are:
\[ \vec{r_1} = \hat{i} + \hat{j} + \hat{k} \]
\[ \vec{r_2} = 2\hat{i} + 3\hat{j} + \hat{k} \]

### Step 3: Calculate the Displacement Vector
The displacement vector \(\vec{R}\) between the two charges is given by:
\[ \vec{R} = \vec{r_2} - \vec{r_1} \]
Calculating this:
\[ \vec{R} = (2\hat{i} + 3\hat{j} + \hat{k}) - (\hat{i} + \hat{j} + \hat{k}) = (2 - 1)\hat{i} + (3 - 1)\hat{j} + (1 - 1)\hat{k} = \hat{i} + 2\hat{j} \]

### Step 4: Calculate the Magnitude of the Displacement Vector
The magnitude of the displacement vector \(\vec{R}\) is:
\[ R = |\vec{R}| = \sqrt{(1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5} \]

### Step 5: Apply Coulomb's Law
According to Coulomb's Law, the electrostatic force \(F\) between two point charges is given by:
\[ F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{R^2} \]
Substituting the values:
\[ F = \frac{1}{4\pi \epsilon_0} \frac{(3 \times 10^{-6})(3 \times 10^{-6})}{(\sqrt{5})^2} \]
\[ = \frac{1}{4\pi \epsilon_0} \frac{9 \times 10^{-12}}{5} \]

### Step 6: Substitute the Value of \(\epsilon_0\)
The value of \(\epsilon_0\) (the permittivity of free space) is approximately:
\[ \epsilon_0 \approx 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \]
Substituting this into the equation:
\[ F = \frac{9 \times 10^{-12}}{4\pi (8.85 \times 10^{-12}) \cdot 5} \]

### Step 7: Calculate the Force
Calculating the denominator:
\[ 4\pi (8.85 \times 10^{-12}) \cdot 5 \approx 5.54 \times 10^{-10} \]
Thus,
\[ F \approx \frac{9 \times 10^{-12}}{5.54 \times 10^{-10}} \approx 0.0162 \, N \]

### Final Answer
The electrostatic force between the two charges is approximately:
\[ F \approx 16.2 \, mN \quad \text{or} \quad 16.2 \times 10^{-3} \, N \]