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)/(32pi epsilon_(0))hat(k)`
B. `(-7P)/(32pi epsilon_(0))hat(k)`
C. `(7P)/(32pi epsilon_(0))hat(k)`
D. none of these

Answer

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, \( \vec{p_1} = P \hat{k} \), is located at the origin (0, 0, 0).
- The second dipole, \( \vec{p_2} = \frac{P}{2} \hat{k} \), is located at (1m, 0, 2m).

### Step 2: Calculate the Electric Field due to the First Dipole
The electric field \( \vec{E_1} \) due to a dipole at a point in space is given by the formula:
\[ \vec{E} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2 \vec{p}}{r^3} \]
where \( \vec{p} \) is the dipole moment and \( r \) is the distance from the dipole to the point of interest.

For the first dipole at (0, 0, 0):
- The distance \( r_1 \) from (0, 0, 0) to (1, 0, 0) is 1m.
- Therefore, the electric field \( \vec{E_1} \) at (1, 0, 0) is:
\[ \vec{E_1} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2P \hat{k}}{(1)^3} = \frac{2P}{4 \pi \epsilon_0} \hat{k} = \frac{P}{2 \pi \epsilon_0} \hat{k} \]

### Step 3: Calculate the Electric Field due to the Second Dipole
For the second dipole at (1m, 0, 2m):
- The distance \( r_2 \) from (1, 0, 2) to (1, 0, 0) is 2m.
- Therefore, the electric field \( \vec{E_2} \) at (1, 0, 0) is:
\[ \vec{E_2} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2 \cdot \frac{P}{2} \hat{k}}{(2)^3} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{P \hat{k}}{8} = \frac{P}{32 \pi \epsilon_0} \hat{k} \]

### Step 4: Determine the Direction of Each Electric Field
- The electric field \( \vec{E_1} \) from the first dipole points in the positive \( \hat{k} \) direction.
- The electric field \( \vec{E_2} \) from the second dipole also points in the positive \( \hat{k} \) direction.

### Step 5: Calculate the Resultant Electric Field
The resultant electric field \( \vec{E_{net}} \) at point (1, 0, 0) is the vector sum of \( \vec{E_1} \) and \( \vec{E_2} \):
\[ \vec{E_{net}} = \vec{E_1} + \vec{E_2} = \left(\frac{P}{2 \pi \epsilon_0} + \frac{P}{32 \pi \epsilon_0}\right) \hat{k} \]
To combine these, we need a common denominator:
\[ \vec{E_{net}} = \left(\frac{16P}{32 \pi \epsilon_0} + \frac{P}{32 \pi \epsilon_0}\right) \hat{k} = \frac{17P}{32 \pi \epsilon_0} \hat{k} \]

### Final Result
The resultant electric field at the point (1m, 0, 0) due to the two dipoles is:
\[ \vec{E_{net}} = \frac{17P}{32 \pi \epsilon_0} \hat{k} \]