Question

A small electric dipole of dipole moment P is placed perpendicular at a distance r from an infinitely long rod of linear charge density lambda. The net electric force on the dipole is?

Answer 1

\frac{P\lambda}{2\pi\epsilon_0 r^2}

Answer 2

The electric field due to an infinitely long charged rod at a distance $r$ is $E = \frac{\lambda}{2\pi\epsilon_0 r}$ in the radial direction. The force on an electric dipole $\vec{P}$ in a non-uniform electric field $\vec{E}$ is given by $\vec{F} = (\vec{P} \cdot \nabla) \vec{E}$.

If the dipole moment $\vec{P}$ is oriented radially (along $\hat{r}$), then $\vec{P} = P \hat{r}$. The force is calculated as $\vec{F} = P \frac{\partial \vec{E}}{\partial r}$. Since $\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \hat{r}$, and assuming $\hat{r}$ is constant in direction for the partial derivative (which is true locally in Cartesian coordinates), we get $\vec{F} = P \frac{\partial}{\partial r} \left( \frac{\lambda}{2\pi\epsilon_0 r} \right) \hat{r} = P \frac{\lambda}{2\pi\epsilon_0} \left( -\frac{1}{r^2} \right) \hat{r} = -\frac{P\lambda}{2\pi\epsilon_0 r^2} \hat{r}$. The negative sign indicates an attractive force if $\vec{P}$ is directed radially outwards (for positive $\lambda$).

If the dipole moment $\vec{P}$ is oriented tangentially (along $\hat{\theta}$), then $\vec{P} = P \hat{\theta}$. The force is calculated using the gradient operator in cylindrical coordinates. This yields a tangential force of magnitude $\frac{P\lambda}{2\pi\epsilon_0 r^2}$.

In both common interpretations of the dipole's orientation, the magnitude of the force is the same.