A point dipole `vecp=-p_(0)xhatx` is kept at the origin. The potential and electric field due to this dipole on the y-axis at a distance d are respectively (take `V=0` at infinity)

Options

A. `(|vecp|)/(4piin_(0)d^(2)),(vecp)/(4piin_(0)d^(3))`
B. `0,(-vecp)/(piin_(0)d^(3))`
C. `0,(vecp)/(4piin_(0)d^(3))`
D. `(|vecp|)/(4piin_(o)d^(2)),(-vecp)/(4piin_(0)d^(3))`

Answer

To solve the problem of finding the potential and electric field due to a point dipole at the origin on the y-axis at a distance \( d \), we can follow these steps:

### Step 1: Understand the Dipole Moment
The dipole moment is given as \( \vec{p} = -p_0 \hat{x} \). This means the dipole is oriented along the negative x-direction.

### Step 2: Define the Coordinates
We are interested in the potential and electric field at a point on the y-axis at a distance \( d \) from the origin. Thus, the coordinates of the point of interest are \( (0, d) \).

### Step 3: Potential Due to a Dipole
The potential \( V \) due to a dipole at a point in space is given by the formula:
\[ V(\vec{r}) = \frac{1}{4\pi \epsilon_0} \frac{\vec{p} \cdot \hat{r}}{r^2} \]
where \( \hat{r} \) is the unit vector in the direction of the point where we are calculating the potential, and \( r \) is the distance from the dipole to that point.

### Step 4: Calculate \( \hat{r} \) and \( r \)
In our case, the distance \( r \) from the origin to the point \( (0, d) \) is:
\[ r = d \]
The unit vector \( \hat{r} \) pointing from the dipole to the point \( (0, d) \) is:
\[ \hat{r} = \frac{(0, d)}{d} = (0, 1) \]

### Step 5: Calculate \( \vec{p} \cdot \hat{r} \)
Now we calculate the dot product \( \vec{p} \cdot \hat{r} \):
\[ \vec{p} = -p_0 \hat{x} = (-p_0, 0) \]
\[ \hat{r} = (0, 1) \]
Thus,
\[ \vec{p} \cdot \hat{r} = (-p_0, 0) \cdot (0, 1) = 0 \]

### Step 6: Substitute into the Potential Formula
Since \( \vec{p} \cdot \hat{r} = 0 \), we find:
\[ V = \frac{1}{4\pi \epsilon_0} \frac{0}{d^2} = 0 \]

### Step 7: Electric Field Due to a Dipole
The electric field \( \vec{E} \) due to a dipole is given by:
\[ \vec{E} = \frac{1}{4\pi \epsilon_0} \left( \frac{2\vec{p}}{r^3} + \frac{\vec{p}}{r^3} \right) \]
For points along the axis perpendicular to the dipole moment, the electric field is given by:
\[ E = \frac{1}{4\pi \epsilon_0} \frac{2p_0}{d^3} \]
The direction of the electric field will be along the negative y-axis due to the negative dipole moment.

### Final Results
Thus, the potential \( V \) and electric field \( \vec{E} \) at a distance \( d \) on the y-axis are:
\[ V = 0 \]
\[ E = -\frac{2p_0}{4\pi \epsilon_0 d^3} \hat{y} \]