Two small electric dipoles each of dipole moment `p hati` are situated at `(0,0,0)` and `(r,0,0)`. The electric potential at a point `((r)/(2),(sqrt(3r))/(2),0)` is:

Options

A. `(p)/(4pi in_(0)r^(2))`
B. 0
C. `(p)/(2pi in_(0)r^(2))`
D. `(p)/(8pi in_(0)r^(2))`

Explanation

To solve the problem, we need to calculate the electric potential at the point \(\left(\frac{r}{2}, \frac{\sqrt{3r}}{2}, 0\right)\) due to two electric dipoles located at the points \((0, 0, 0)\) and \((r, 0, 0)\) with dipole moments \(\mathbf{p} = p \hat{i}\).

### Step-by-Step Solution:

1. **Identify the Positions of the Dipoles:**
- Dipole 1 is at the origin: \( \mathbf{r_1} = (0, 0, 0) \)
- Dipole 2 is at \( \mathbf{r_2} = (r, 0, 0) \)

2. **Identify the Point of Interest:**
- The point where we need to calculate the potential is \( \mathbf{r_n} = \left(\frac{r}{2}, \frac{\sqrt{3r}}{2}, 0\right) \)

3. **Calculate the Distances from Each Dipole to the Point:**
- Distance from Dipole 1 to point \( n \):
\[ r_1 = \sqrt{\left(\frac{r}{2} - 0\right)^2 + \left(\frac{\sqrt{3r}}{2} - 0\right)^2} = \sqrt{\frac{r^2}{4} + \frac{3r}{4}} = \sqrt{r^2} = r \]
- Distance from Dipole 2 to point \( n \):
\[ r_2 = \sqrt{\left(\frac{r}{2} - r\right)^2 + \left(\frac{\sqrt{3r}}{2} - 0\right)^2} = \sqrt{\left(-\frac{r}{2}\right)^2 + \left(\frac{\sqrt{3r}}{2}\right)^2} = \sqrt{\frac{r^2}{4} + \frac{3r}{4}} = \sqrt{r^2} = r \]

4. **Calculate the Angles:**
- For Dipole 1:
\[ \cos \theta_1 = \frac{\mathbf{p} \cdot \mathbf{r_n}}{|\mathbf{p}| |\mathbf{r_n}|} = \frac{p \cdot \frac{r}{2}}{p \cdot r} = \frac{1}{2} \]
- For Dipole 2:
\[ \cos \theta_2 = \frac{\mathbf{p} \cdot \mathbf{r_n}}{|\mathbf{p}| |\mathbf{r_n}|} = \frac{p \cdot \left(-\frac{r}{2}\right)}{p \cdot r} = -\frac{1}{2} \]

5. **Calculate the Electric Potential due to Each Dipole:**
- The formula for the electric potential \( V \) due to a dipole is given by:
\[ V = k \frac{\mathbf{p} \cdot \hat{r}}{r^2} \]
- For Dipole 1:
\[ V_1 = k \frac{p \cdot \frac{1}{2}}{r^2} = \frac{kp}{2r^2} \]
- For Dipole 2:
\[ V_2 = k \frac{p \cdot \left(-\frac{1}{2}\right)}{r^2} = -\frac{kp}{2r^2} \]

6. **Total Electric Potential at Point \( n \):**
- The total potential \( V_n \) is the sum of the potentials due to both dipoles:
\[ V_n = V_1 + V_2 = \frac{kp}{2r^2} - \frac{kp}{2r^2} = 0 \]

### Final Answer:
The electric potential at the point \(\left(\frac{r}{2}, \frac{\sqrt{3r}}{2}, 0\right)\) is \( V_n = 0 \).