Two point charges `3 muC` and `2.5 muC` are placed at point A (1, 1, 2)m and B (0, 3, -1)m respectively. Find out electric field intensity at point C(3, 3, 3)m.

Answer

To find the electric field intensity at point C due to the two point charges located at points A and B, we will follow these steps:

### Step 1: Identify the Charges and Their Positions
- Charge \( Q_1 = 3 \, \mu C = 3 \times 10^{-6} \, C \) is located at point A (1, 1, 2) m.
- Charge \( Q_2 = 2.5 \, \mu C = 2.5 \times 10^{-6} \, C \) is located at point B (0, 3, -1) m.
- We need to find the electric field at point C (3, 3, 3) m.

### Step 2: Calculate the Position Vectors
- Position vector from A to C:
\[ \vec{r_A} = \vec{C} - \vec{A} = (3 - 1) \hat{i} + (3 - 1) \hat{j} + (3 - 2) \hat{k} = 2 \hat{i} + 2 \hat{j} + 1 \hat{k} \]

- Position vector from B to C:
\[ \vec{r_B} = \vec{C} - \vec{B} = (3 - 0) \hat{i} + (3 - 3) \hat{j} + (3 - (-1)) \hat{k} = 3 \hat{i} + 0 \hat{j} + 4 \hat{k} \]

### Step 3: Calculate the Magnitudes of the Position Vectors
- Magnitude of \( \vec{r_A} \):
\[ r_A = \sqrt{(2)^2 + (2)^2 + (1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \, m \]

- Magnitude of \( \vec{r_B} \):
\[ r_B = \sqrt{(3)^2 + (0)^2 + (4)^2} = \sqrt{9 + 0 + 16} = \sqrt{25} = 5 \, m \]

### Step 4: Calculate the Electric Fields at Point C
- Electric field due to charge \( Q_1 \) at point C:
\[ \vec{E_A} = k \frac{Q_1}{r_A^2} \hat{r_A} \]
where \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \) and \( \hat{r_A} \) is the unit vector in the direction of \( \vec{r_A} \):
\[ \hat{r_A} = \frac{\vec{r_A}}{r_A} = \frac{2 \hat{i} + 2 \hat{j} + 1 \hat{k}}{3} \]
Thus,
\[ \vec{E_A} = 9 \times 10^9 \cdot \frac{3 \times 10^{-6}}{3^2} \left(\frac{2}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{1}{3} \hat{k}\right) \]
\[ = 9 \times 10^9 \cdot \frac{3 \times 10^{-6}}{9} \left(\frac{2}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{1}{3} \hat{k}\right) \]
\[ = 10^3 \left(2 \hat{i} + 2 \hat{j} + 1 \hat{k}\right) = 2000 \hat{i} + 2000 \hat{j} + 1000 \hat{k} \, N/C \]

- Electric field due to charge \( Q_2 \) at point C:
\[ \vec{E_B} = k \frac{Q_2}{r_B^2} \hat{r_B} \]
where \( \hat{r_B} \) is the unit vector in the direction of \( \vec{r_B} \):
\[ \hat{r_B} = \frac{\vec{r_B}}{r_B} = \frac{3 \hat{i} + 0 \hat{j} + 4 \hat{k}}{5} \]
Thus,
\[ \vec{E_B} = 9 \times 10^9 \cdot \frac{2.5 \times 10^{-6}}{5^2} \left(\frac{3}{5} \hat{i} + 0 \hat{j} + \frac{4}{5} \hat{k}\right) \]
\[ = 9 \times 10^9 \cdot \frac{2.5 \times 10^{-6}}{25} \left(\frac{3}{5} \hat{i} + 0 \hat{j} + \frac{4}{5} \hat{k}\right) \]
\[ = 10^3 \cdot \left(0.3 \hat{i} + 0 + 0.4 \hat{k}\right) = 720 \hat{i} + 0 \hat{j} + 720 \hat{k} \, N/C \]

### Step 5: Calculate the Resultant Electric Field at Point C
- Adding the electric fields \( \vec{E_A} \) and \( \vec{E_B} \):
\[ \vec{E_{net}} = \vec{E_A} + \vec{E_B} = (2000 + 720) \hat{i} + (2000 + 0) \hat{j} + (1000 + 720) \hat{k} \]
\[ = 2720 \hat{i} + 2000 \hat{j} + 1720 \hat{k} \, N/C \]

### Final Result
The electric field intensity at point C is:
\[ \vec{E_{net}} = 2720 \hat{i} + 2000 \hat{j} + 1720 \hat{k} \, N/C \]