A dipole of dipole moment 'p' is placed in a non-uniform electric field along x-axis. Electric field is increasing at the rate of `1 V//m^(2)` then the force on dipole is :-

Options

A. 0
B. 2p
C. p/2
D. p

Answer

To find the force on a dipole placed in a non-uniform electric field, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Dipole Moment**:
The dipole moment \( p \) is defined as the product of the charge \( Q \) and the separation distance \( d \) between the charges:
\[ p = Q \cdot d \]

2. **Electric Field Variation**:
We are given that the electric field \( E \) is increasing at the rate of \( \lambda = 1 \, \text{V/m}^2 \). This means that for a small distance \( d \), the change in electric field can be expressed as:
\[ \Delta E = \lambda \cdot d \]

3. **Electric Field at Two Points**:
Let’s denote the electric field at point 2 as \( E_2 \). The electric field at point 1, which is a distance \( d \) away from point 2, can be expressed as:
\[ E_1 = E_2 + \Delta E = E_2 + \lambda \cdot d \]

4. **Calculating Forces on Charges**:
The force \( F_1 \) on the positive charge \( +Q \) at point 1 is:
\[ F_1 = Q \cdot E_1 = Q \cdot (E_2 + \lambda \cdot d) \]
The force \( F_2 \) on the negative charge \( -Q \) at point 2 is:
\[ F_2 = -Q \cdot E_2 \]

5. **Net Force on the Dipole**:
The net force \( F \) on the dipole is the difference between the forces on the two charges:
\[ F = F_1 - F_2 = Q \cdot (E_2 + \lambda \cdot d) - (-Q \cdot E_2) \]
Simplifying this gives:
\[ F = Q \cdot (E_2 + \lambda \cdot d + E_2) = Q \cdot (2E_2 + \lambda \cdot d) \]

6. **Substituting for \( \lambda \)**:
Since \( \lambda = 1 \, \text{V/m}^2 \), we can substitute this into our equation:
\[ F = Q \cdot (2E_2 + d) \]

7. **Expressing in terms of Dipole Moment**:
Since \( p = Q \cdot d \), we can express \( Q \) in terms of \( p \) and \( d \):
\[ Q = \frac{p}{d} \]
Substituting this into our force equation gives:
\[ F = \frac{p}{d} \cdot (2E_2 + d) \]

8. **Final Expression for Force**:
As \( d \) approaches zero (in the limit for small dipoles), the dominant term becomes:
\[ F = p \cdot \lambda \]
Since \( \lambda = 1 \, \text{V/m}^2 \), we have:
\[ F = p \]

### Conclusion:
The force on the dipole is equal to the dipole moment \( p \).