A point dipole 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. `(|vec(P)|)/(4pi in_(0) d^(2))`
B. `0,(|vec(p)|)/( 4 pi in_(0) d^(3))`
C. `(|vec(p)|)/( 4 x in_(0) d^(2)),(vec(p))/(4 pi in_(0) d^(3))`
D. `0, (vec(p))/(4 pi in_(0) d^(3))`

Explanation

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

### Step 1: Understand the Dipole Configuration
A dipole consists of two equal and opposite charges, \(+q\) and \(-q\), separated by a distance \(2a\). The dipole moment \(\vec{p}\) is given by:
\[ \vec{p} = q \cdot 2a \]
This dipole is placed at the origin, and we are interested in the potential and electric field at a point \(P\) located on the y-axis at a distance \(d\) from the origin.

### Step 2: Calculate the Electric Potential \(V\)
The electric potential \(V\) at a point due to a dipole is given by:
\[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{\vec{p} \cdot \hat{r}}{r^2} \]
where \(\hat{r}\) is the unit vector in the direction from the dipole to the point where the potential is being calculated, and \(r\) is the distance from the dipole to that point.

For a point on the y-axis at distance \(d\), the distance from the dipole to point \(P\) is:
\[ r = \sqrt{d^2 + a^2} \]
Since the dipole is symmetric and the point \(P\) is on the equatorial line (y-axis), the potential due to the dipole at point \(P\) is:
\[ V = 0 \]
This is because the contributions from the positive and negative charges cancel each other out at this point.

### Step 3: Calculate the Electric Field \(\vec{E}\)
The electric field \(\vec{E}\) due to a dipole at a point in space is given by:
\[ \vec{E} = \frac{1}{4\pi \epsilon_0} \cdot \left( \frac{3(\vec{p} \cdot \hat{r})\hat{r} - \vec{p}}{r^3} \right) \]
For our case, since we are looking at a point on the y-axis, we can simplify the expression. The electric field at point \(P\) can be derived as follows:

1. The distance \(r\) from the dipole to point \(P\) is still \(\sqrt{d^2 + a^2}\).
2. The unit vector \(\hat{r}\) points from the dipole to point \(P\), which is along the y-axis.

Thus, the electric field can be expressed as:
\[ E = \frac{1}{4\pi \epsilon_0} \cdot \frac{2p}{d^3} \]
where we have neglected the term involving \(a\) because \(a\) is much smaller than \(d\) (i.e., \(a \ll d\)).

### Final Results
- The electric potential \(V\) at point \(P\) on the y-axis is:
\[ V = 0 \]
- The electric field \(\vec{E}\) at point \(P\) on the y-axis is:
\[ E = \frac{1}{4\pi \epsilon_0} \cdot \frac{2p}{d^3} \]