An electron is released from rest on the axis of an electric dipole is 25 nm from the center of the dipole. What is the magnitude of the electron's acceleration if the dipole moment is `3.6 xx 10^(-29)C.m`? Assume that 25 nm is much larger than the separation of the charged particles that form the dipole ?

Explanation

To solve the problem, we need to find the acceleration of an electron placed at a distance of 25 nm from the center of an electric dipole with a dipole moment of \(3.6 \times 10^{-29} \, \text{C.m}\).

### Step-by-Step Solution:

1. **Understand the Electric Field of a Dipole**:
The electric field \(E\) at a point along the axis of a dipole at a distance \(r\) from the center of the dipole, where \(r\) is much larger than the separation of the charges forming the dipole, is given by the formula:
\[
E = \frac{2kp}{r^3}
...