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(D) The electric field $\vec{E}$ produced by an infinitely long charged wire at a distance $r$ is given by $E = \frac{\lambda}{2 \pi \varepsilon_0 r}$

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(D) The electric field $\vec{E}$ produced by an infinitely long charged wire at a distance $r$ is given by $E = \frac{\lambda}{2 \pi \varepsilon_0 r}$,directed radially outward.Since the dipole moment $\vec{p}$ is oriented along the thread (parallel to the wire),the force on the positive charge $+q$ is $\vec{F}_1 = q\vec{E}$ and the force on the negative charge $-q$ is $\vec{F}_2 = -q\vec{E}$.Because the electric field $\vec{E}$ is uniform along the direction of the wire at a constant distance $r$,the magnitude of the field at both charges of the dipole is the same.Therefore,the forces $\vec{F}_1$ and $\vec{F}_2$ are equal in magnitude and opposite in direction,resulting in a net force $F = F_1 - F_2 = 0$.

$A$ dipole with an electric moment $\vec{p}$ is located at a distance $r$ from a long thread charged uniformly with a linear charge density $\lambda$. Find the force $F$ acting on the dipole if the vector $\vec{p}$ is oriented along the thread.

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