1. Problem Statement & Approximations

We need to find the electrostatic force between a point charge $Q$ and a neutral cylinder of volume $V$ placed at a distance $x$ from the charge.

Conditions: $x$ is much larger than the linear dimensions of the cylinder.

Approximation: Since the distance is large, the cylinder acts as a small dipole. The hint suggests considering charges appearing on the “flat faces”. This implies we treat the cylinder effectively as a small dielectric or conducting slab perpendicular to the field, where the depolarization field is uniform (similar to a parallel plate capacitor geometry). The induced dipole moment $\vec{p}$ is directed along the field.

2. General Formulation

The force on a small induced dipole $\vec{p}$ in a non-uniform field $\vec{E}$ is:

$$ F = p \frac{dE}{dx} $$

The electric field $E$ due to the point charge $Q$ at distance $x$ is:

$$ E = \frac{Q}{4\pi\epsilon_0 x^2} $$

The gradient of the field is:

$$ \frac{dE}{dx} = \frac{Q}{4\pi\epsilon_0} \left( -\frac{2}{x^3} \right) = -\frac{2Q}{4\pi\epsilon_0 x^3} $$

3. Case (i): Insulating Cylinder (Permittivity $\varepsilon_r$)

Inside the material, the macroscopic electric field $E_{in}$ is related to the external field $E_{ext}$ by the boundary conditions. For a shape dominated by flat faces perpendicular to the field (depolarization factor $\approx 1$):

$$ E_{in} = \frac{E_{ext}}{\varepsilon_r} $$

The polarization density (dipole moment per unit volume) is:

$$ \vec{P} = (\varepsilon_r – 1)\epsilon_0 \vec{E}_{in} = (\varepsilon_r – 1)\epsilon_0 \frac{\vec{E}_{ext}}{\varepsilon_r} $$

The total induced dipole moment $p$ for volume $V$ is:

$$ p = P \cdot V = \frac{\varepsilon_r – 1}{\varepsilon_r} \epsilon_0 V E $$

Substituting this into the force equation:

$$ F = \left( \frac{\varepsilon_r – 1}{\varepsilon_r} \epsilon_0 V E \right) \frac{dE}{dx} $$ $$ F = \frac{\varepsilon_r – 1}{\varepsilon_r} \epsilon_0 V \left( \frac{Q}{4\pi\epsilon_0 x^2} \right) \left( -\frac{2Q}{4\pi\epsilon_0 x^3} \right) $$ $$ F = – \frac{(\varepsilon_r – 1) 2 \epsilon_0 V Q^2}{\varepsilon_r (4\pi\epsilon_0)^2 x^5} = – \frac{(\varepsilon_r – 1) Q^2 V}{\varepsilon_r 8 \pi^2 \epsilon_0 x^5} $$

Magnitude:

4. Case (ii): Conducting Cylinder

For a conductor, the relative permittivity $\varepsilon_r \to \infty$. We can take the limit of the insulator result:

$$ \lim_{\varepsilon_r \to \infty} \frac{\varepsilon_r – 1}{\varepsilon_r} = 1 $$

Alternatively, for a conductor, the internal field must be zero, so the induced surface charges create a field that perfectly cancels $E_{ext}$, leading to the maximum possible polarization for this geometry ($p = \epsilon_0 V E_{ext}$).

Substituting the limit into the force expression:

Note: The force is attractive (indicated by the negative sign in derivation) because the induced dipole is aligned with the field.