Solution
Let the interaction energy between two parallel dipoles side-by-side at distance $r$ be $u$. Since parallel side-by-side dipoles repel, $u = \frac{kp^2}{r^3}$.
1. Initial Energy ($U_0$):
Dipoles are at positions 1, 2, and 3. All parallel.
- Pair (1, 2): Distance $r$. Energy $= +u$.
- Pair (2, 3): Distance $r$. Energy $= +u$.
- Pair (1, 3): Distance $2r$. Energy $= \frac{kp^2}{(2r)^3} = \frac{u}{8}$.
2. Final Energy ($U_f$):
Dipole 1 is reversed. Now it is anti-parallel to 2 and 3.
- Pair (1, 2): Anti-parallel at $r$. Energy $= -u$.
- Pair (2, 3): Parallel at $r$. Energy $= +u$.
- Pair (1, 3): Anti-parallel at $2r$. Energy $= -\frac{u}{8}$.
3. Work Done ($W$):
$$W = U_f – U_0 = \left(-\frac{u}{8}\right) – \left(\frac{17u}{8}\right) = -\frac{18u}{8}$$We need to express $W$ in terms of $U_0$. From the initial step, $u = \frac{8 U_0}{17}$. Substituting this into the work equation:
$$W = -\frac{18}{8} \left( \frac{8 U_0}{17} \right) = -\frac{18 U_0}{17}$$Answer: (c) $-\frac{18 U_0}{17}$