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

Options

A. `(9p)/(32 pi epsilon_(0))hat(k)`
B. `(-7p)/(32pi epsilon_(0))hat(k)`
C. `(7p)/(32pi epsilon_(0))hat(k)`
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, \( \vec{p} \), is located at point \( (0, 0, 0) \) and has a dipole moment \( \vec{p} = p \hat{k} \).
- The second dipole, \( \frac{p}{2} \), is located at point \( (1, 0, 2) \) and has a dipole moment \( \vec{p_2} = \frac{p}{2} \hat{k} \).

### Step 2: Calculate the Electric Field Due to the First Dipole at (1, 0, 0)
- The distance \( r \) from the first dipole to the point (1, 0, 0) is:
\[ r_1 = \sqrt{(1-0)^2 + (0-0)^2 + (0-0)^2} = 1 \text{ m} \]
- The electric field \( \vec{E_1} \) due to a dipole at the equatorial position is given by:
\[ \vec{E_1} = \frac{k \cdot \vec{p}}{r^3} \]
Since we are at the equatorial position, the direction of the electric field will be opposite to the dipole moment:
\[ \vec{E_1} = -\frac{k \cdot p}{1^3} \hat{k} = -k p \hat{k} \]

### Step 3: Calculate the Electric Field Due to the Second Dipole at (1, 0, 0)
- The distance \( r \) from the second dipole (1, 0, 2) to the point (1, 0, 0) is:
\[ r_2 = \sqrt{(1-1)^2 + (0-0)^2 + (0-2)^2} = 2 \text{ m} \]
- The electric field \( \vec{E_2} \) due to the second dipole at the axial position is given by:
\[ \vec{E_2} = \frac{2k \cdot \vec{p_2}}{r^3} \]
Here, \( \vec{p_2} = \frac{p}{2} \hat{k} \):
\[ \vec{E_2} = \frac{2k \cdot \left(\frac{p}{2}\right)}{2^3} \hat{k} = \frac{kp}{4} \hat{k} \]

### Step 4: Combine the Electric Fields
- The resultant electric field \( \vec{E} \) at point (1, 0, 0) is the vector sum of \( \vec{E_1} \) and \( \vec{E_2} \):
\[ \vec{E} = \vec{E_1} + \vec{E_2} = -k p \hat{k} + \frac{kp}{4} \hat{k} \]
- Simplifying this gives:
\[ \vec{E} = \left(-kp + \frac{kp}{4}\right) \hat{k} = \left(-\frac{4kp}{4} + \frac{kp}{4}\right) \hat{k} = -\frac{3kp}{4} \hat{k} \]

### Step 5: Express in Terms of \( \epsilon_0 \)
- Recall that \( k = \frac{1}{4\pi\epsilon_0} \):
\[ \vec{E} = -\frac{3}{4} \cdot \frac{p}{4\pi\epsilon_0} \hat{k} = -\frac{3p}{16\pi\epsilon_0} \hat{k} \]

### Final Result
The resultant electric field at the point (1m, 0, 0) due to the two dipoles is:
\[ \vec{E} = -\frac{3p}{16\pi\epsilon_0} \hat{k} \]