1. Conceptual Analysis

The problem asks for the electrostatic force between a fixed dipole $\vec{p}_0$ and a neutral molecule with polarizability $\alpha$ located at a distance $r$ on the dipole’s axis.

The physics principles involved are:

• Electric Field of a Dipole: The fixed dipole creates a non-uniform electric field $\vec{E}$ at the location of the molecule.
• Induced Dipole Moment: The external field polarizes the molecule, inducing a dipole moment $\vec{p} = \alpha \vec{E}$.
• Force on a Dipole: A dipole in a non-uniform electric field experiences a net force given by $\vec{F} = (\vec{p} \cdot \nabla)\vec{E}$.

2. Electric Field of the Source Dipole

Consider the source dipole $\vec{p}_0$ placed at the origin, aligned along the $z$-axis (the axis of symmetry). The electric field on the axis at a distance $r$ is given by:

$$ \vec{E}(r) = \frac{2p_0}{4\pi\epsilon_0 r^3} \hat{r} $$

Let $k = \frac{1}{4\pi\epsilon_0}$. Then,

$$ \vec{E}(r) = \frac{2k p_0}{r^3} \hat{r} $$

3. Induced Dipole Moment

The molecule is polarized by this local field. Since $\alpha$ is positive, the induced dipole moment $\vec{p}$ is parallel to $\vec{E}$:

$$ \vec{p} = \alpha \vec{E} = \alpha \left( \frac{2k p_0}{r^3} \right) \hat{r} $$

4. Calculation of the Force

The force on the induced dipole is the gradient of its potential energy, or more directly:

$$ \vec{F} = (\vec{p} \cdot \nabla) \vec{E} $$

Since the setup is one-dimensional (radial), this simplifies to:

$$ F_r = p \frac{dE}{dr} $$

Substituting the expressions for $p$ and $E$:

$$ F_r = \left( \frac{2\alpha k p_0}{r^3} \right) \cdot \frac{d}{dr} \left( \frac{2k p_0}{r^3} \right) $$

Differentiating $E$ with respect to $r$:

$$ \frac{dE}{dr} = 2k p_0 \frac{d}{dr}(r^{-3}) = 2k p_0 (-3 r^{-4}) = -\frac{6k p_0}{r^4} $$

Now, calculating the force:

$$ F_r = \left( \frac{2\alpha k p_0}{r^3} \right) \left( -\frac{6k p_0}{r^4} \right) $$ $$ F_r = – \frac{12 \alpha k^2 p_0^2}{r^7} $$

The negative sign indicates that the force is attractive.

5. Final Result

Substituting $k = \frac{1}{4\pi\epsilon_0}$ back into the expression:

$$ |F| = 12 \alpha p_0^2 \left( \frac{1}{4\pi\epsilon_0} \right)^2 \frac{1}{r^7} $$ $$ |F| = \frac{12 \alpha p_0^2}{16 \pi^2 \epsilon_0^2 r^7} $$