An electric dipole having charges `+q` and `-q` at a separation r. At distance `d gt gt r` along the axis of the dipole , the field is proportional to

Options

A. `(q)/(d^(2))`
B. `(qr)/(d^(2))`
C. `(q)/(d^(3))`
D. `(qr)/(d^(3))`

Answer

To solve the problem of finding the electric field due to an electric dipole at a point along its axis at a distance much greater than the separation of the charges, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Electric Dipole**:
An electric dipole consists of two equal and opposite charges, +q and -q, separated by a distance r. The dipole moment \( p \) is defined as:
\[ p = q \cdot d \]
where \( d \) is the separation between the charges.

2. **Positioning the Dipole**:
Place the dipole along the x-axis with the positive charge +q at \( x = -\frac{r}{2} \) and the negative charge -q at \( x = \frac{r}{2} \). The point P where we want to find the electric field is located at a distance \( d \) from the center of the dipole (which is at the origin).

3. **Calculating the Electric Field**:
The electric field \( E \) at point P due to the positive charge +q is given by:
\[ E_1 = \frac{k \cdot q}{(d - \frac{r}{2})^2} \]
and the electric field due to the negative charge -q is:
\[ E_2 = \frac{k \cdot q}{(d + \frac{r}{2})^2} \]

4. **Net Electric Field**:
Since the electric fields due to the two charges are in opposite directions, the net electric field \( E \) at point P is:
\[ E = E_1 - E_2 = \frac{k \cdot q}{(d - \frac{r}{2})^2} - \frac{k \cdot q}{(d + \frac{r}{2})^2} \]

5. **Simplifying the Expression**:
Factor out \( k \cdot q \):
\[ E = k \cdot q \left( \frac{1}{(d - \frac{r}{2})^2} - \frac{1}{(d + \frac{r}{2})^2} \right) \]
To combine the fractions, find a common denominator:
\[ E = k \cdot q \cdot \frac{(d + \frac{r}{2})^2 - (d - \frac{r}{2})^2}{(d - \frac{r}{2})^2 (d + \frac{r}{2})^2} \]

6. **Using the Difference of Squares**:
The numerator simplifies using the difference of squares:
\[ (a^2 - b^2) = (a - b)(a + b) \]
where \( a = d + \frac{r}{2} \) and \( b = d - \frac{r}{2} \):
\[ E = k \cdot q \cdot \frac{(r)(r)}{(d - \frac{r}{2})^2 (d + \frac{r}{2})^2} \]

7. **Considering \( d \gg r \)**:
When \( d \) is much greater than \( r \), we can approximate:
\[ (d - \frac{r}{2})^2 \approx d^2 \quad \text{and} \quad (d + \frac{r}{2})^2 \approx d^2 \]
Thus, the electric field simplifies to:
\[ E \approx \frac{2kqr^2}{d^3} \]

8. **Final Result**:
The electric field at a distance \( d \) along the axis of the dipole is proportional to:
\[ E \propto \frac{p}{d^3} \]
where \( p = q \cdot r \) is the dipole moment.