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(C) The electric field due to an infinitely long charged wire at a distance $r$ is given by $E = \frac{\lambda}{2\pi\epsilon_0 r}$.The force on a smal

Vedclass · Accepted Answer

$F_3 > F_2 = F_4 > F_1 = F_5$

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(C) The electric field due to an infinitely long charged wire at a distance $r$ is given by $E = \frac{\lambda}{2\pi\epsilon_0 r}$.The force on a small electric dipole $\vec{p}$ in a non-uniform electric field is given by $\vec{F} = (\vec{p} \cdot 
abla) \vec{E}$.For a dipole oriented at an angle $	heta$ with the radial direction,the magnitude of the force is $F = |p \cos 	heta \frac{dE}{dr}|$.Since $\frac{dE}{dr} = -\frac{\lambda}{2\pi\epsilon_0 r^2}$,the magnitude of the force is $F = p |\cos 	heta| \frac{\lambda}{2\pi\epsilon_0 r^2}$.Here,$r$ is the same for all dipoles. The force depends on $|\cos 	heta|$,where $	heta$ is the angle between the dipole moment $\vec{p}$ and the radial electric field direction.For $p_3$,$	heta = 0^\circ$,so $|\cos 0^\circ| = 1$.For $p_2$ and $p_4$,$	heta = 45^\circ$,so $|\cos 45^\circ| = \frac{1}{\sqrt{2}} \approx 0.707$.For $p_1$ and $p_5$,$	heta = 90^\circ$,so $|\cos 90^\circ| = 0$.Thus,$F_3 > F_2 = F_4 > F_1 = F_5$.

Five identical small electric dipoles are placed at a distance $r$ from an infinitely long straight wire carrying a uniform charge density $\lambda$. If they experience forces of magnitude $F_1, F_2, F_3, F_4$ and $F_5$ respectively,then:

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