Two short dipoles `phat(k)` and `P/2 hat(k)` are located at `(0,0,0)` & `(1m, 0,2m)` respectivley. The resultant electric field due to the two dipoles at the point `(1m, 0,0)` is

Options

A. `(9p)/(32 pi in_(0))`
B. `(-7 p)/(32 pi in_(0))`
C. `(7p)/(32 pi in_(0))`
D. none of these

Explanation

To find the resultant electric field at the point (1m, 0, 0) due to the two dipoles located at (0, 0, 0) and (1m, 0, 2m), we will follow these steps:

### Step 1: Identify the dipoles and their positions
- The first dipole, \( \mathbf{p} = \hat{k} \), is located at the origin (0, 0, 0).
- The second dipole, \( \mathbf{p_2} = \frac{P}{2} \hat{k} \), is located at (1m, 0, 2m).

### Step 2: Calculate the electric field due to the first dipole at the point (1m, 0, 0)
The formula for the electric field \( \mathbf{E} \) due to a dipole at a point in space is given by:

\[ \mathbf{E} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2\mathbf{p} \cdot \hat{r}}{r^3} - \frac{\mathbf{p}}{r^3} \]

Where:
- \( \mathbf{p} \) is the dipole moment,
- \( \hat{r} \) is the unit vector from the dipole to the point,
- \( r \) is the distance from the dipole to the point.

For the first dipole located at (0, 0, 0):
- Position vector to the point (1, 0, 0) is \( \mathbf{r_1} = (1, 0, 0) \).
- Distance \( r_1 = \sqrt{(1-0)^2 + (0-0)^2 + (0-0)^2} = 1 \, \text{m} \).
- Unit vector \( \hat{r_1} = \frac{\mathbf{r_1}}{r_1} = (1, 0, 0) \).

Thus, the electric field \( \mathbf{E_1} \) due to the first dipole is:

\[ \mathbf{E_1} = \frac{1}{4 \pi \epsilon_0} \cdot \left( \frac{2 \mathbf{p} \cdot \hat{r_1}}{r_1^3} - \frac{\mathbf{p}}{r_1^3} \right) \]

Substituting \( \mathbf{p} = \hat{k} \):

\[ \mathbf{E_1} = \frac{1}{4 \pi \epsilon_0} \cdot \left( \frac{2 \cdot \hat{k} \cdot (1, 0, 0)}{1^3} - \frac{\hat{k}}{1^3} \right) = \frac{1}{4 \pi \epsilon_0} \cdot \left( 0 - \hat{k} \right) = -\frac{1}{4 \pi \epsilon_0} \hat{k} \]

### Step 3: Calculate the electric field due to the second dipole at the point (1m, 0, 0)
For the second dipole located at (1, 0, 2):
- Position vector to the point (1, 0, 0) is \( \mathbf{r_2} = (0, 0, -2) \).
- Distance \( r_2 = \sqrt{(1-1)^2 + (0-0)^2 + (0-2)^2} = 2 \, \text{m} \).
- Unit vector \( \hat{r_2} = (0, 0, -1) \).

Thus, the electric field \( \mathbf{E_2} \) due to the second dipole is:

\[ \mathbf{E_2} = \frac{1}{4 \pi \epsilon_0} \cdot \left( \frac{2 \cdot \frac{P}{2} \cdot \hat{r_2}}{r_2^3} - \frac{\frac{P}{2}}{r_2^3} \right) \]

Substituting \( \mathbf{p_2} = \frac{P}{2} \hat{k} \):

\[ \mathbf{E_2} = \frac{1}{4 \pi \epsilon_0} \cdot \left( \frac{P \cdot (0, 0, -1)}{2^3} - \frac{\frac{P}{2}}{2^3} \right) = \frac{1}{4 \pi \epsilon_0} \cdot \left( \frac{P \cdot (0, 0, -1)}{8} - \frac{P/2}{8} \right) \]

\[ = \frac{1}{4 \pi \epsilon_0} \cdot \left( \frac{(0, 0, -P)}{8} - \frac{(P/2)}{8} \right) = \frac{1}{4 \pi \epsilon_0} \cdot \left( \frac{(0, 0, -P - P/2)}{8} \right) = -\frac{3P}{32 \pi \epsilon_0} \hat{k} \]

### Step 4: Calculate the resultant electric field
Now, we can find the resultant electric field \( \mathbf{E} \) at the point (1m, 0, 0):

\[ \mathbf{E} = \mathbf{E_1} + \mathbf{E_2} \]

Substituting the values:

\[ \mathbf{E} = -\frac{1}{4 \pi \epsilon_0} \hat{k} - \frac{3P}{32 \pi \epsilon_0} \hat{k} = -\left( \frac{8}{32 \pi \epsilon_0} + \frac{3P}{32 \pi \epsilon_0} \right) \hat{k} \]

\[ = -\frac{(8 + 3P)}{32 \pi \epsilon_0} \hat{k} \]

Thus, the resultant electric field at the point (1m, 0, 0) is:

\[ \mathbf{E} = -\frac{(8 + 3P)}{32 \pi \epsilon_0} \hat{k} \]