The electric field at a point on the equatorial plane at a distance r from the centre of a dipole having dipole moment `vecp` is given by (r >> seperation of two charges forming dipole , `epsilon_0` = permittivity of free space

Options

A. `vecE = vecP /4pi epsilon_0 r^3`
B. `vecE = 2 vecP /4pi epsilon_0 r^3`
C. `vecE = - vecP /4pi epsilon_0 r^2`
D. `vecE = - vecP /4pi epsilon_0 r^3`

Answer

To find the electric field at a point on the equatorial plane of an electric dipole, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Dipole Configuration**:
- A dipole consists of two equal and opposite charges (+q and -q) separated by a distance \(2L\). The dipole moment \(\vec{p}\) is defined as \( \vec{p} = q \cdot 2L \).

2. **Identifying the Point of Interest**:
- We are interested in the electric field at a point on the equatorial plane at a distance \(r\) from the center of the dipole. Here, \(r\) is much greater than the separation \(2L\) of the dipole charges.

3. **Electric Field Components**:
- At point P on the equatorial plane, the electric field due to the dipole has two components: a radial component \(E_r\) and a transverse component \(E_\theta\).

4. **Calculating the Potential**:
- The electric potential \(V\) at point P due to the dipole can be expressed as:
\[ V = \frac{kp \cos \theta}{r^2} \]
- Here, \(k\) is the Coulomb's constant, \(p\) is the dipole moment, and \(\theta\) is the angle with respect to the dipole axis.

5. **Finding the Electric Field**:
- The electric field \(E\) is related to the potential by the relation:
\[ E = -\frac{dV}{dr} \]
- For the radial component \(E_r\):
\[ E_r = -\frac{dV}{dr} = -\frac{d}{dr}\left(\frac{kp \cos \theta}{r^2}\right) \]
- After differentiating, we find:
\[ E_r = \frac{2kp \cos \theta}{r^3} \]

6. **Calculating the Transverse Component**:
- For the transverse component \(E_\theta\):
\[ E_\theta = -\frac{1}{r}\frac{dV}{d\theta} \]
- After differentiating, we find:
\[ E_\theta = \frac{kp \sin \theta}{r^3} \]

7. **Resultant Electric Field**:
- The resultant electric field \(E\) at the equatorial point is given by:
\[ E = \sqrt{E_r^2 + E_\theta^2} \]
- At the equatorial plane, \(\theta = 90^\circ\) implies \(\cos(90^\circ) = 0\) and \(\sin(90^\circ) = 1\). Therefore:
\[ E_r = 0 \quad \text{and} \quad E_\theta = \frac{kp}{r^3} \]
- Thus, the resultant electric field at the equatorial point is:
\[ E = \frac{kp}{r^3} \]

8. **Expressing in Vector Form**:
- The constant \(k\) can be expressed as \(k = \frac{1}{4\pi \epsilon_0}\), leading to:
\[ \vec{E} = -\frac{1}{4\pi \epsilon_0} \frac{\vec{p}}{r^3} \]

### Final Result:
The electric field at a point on the equatorial plane at a distance \(r\) from the center of the dipole is given by:
\[ \vec{E} = -\frac{1}{4\pi \epsilon_0} \frac{\vec{p}}{r^3} \]