An ideal electric dipole is situated at the origin, and points in the z direction, as in Fig. 3.36. An electric charge is released from rest at a point in the x y plane. Show that it swings back and forth in a semi-circular arc, as though it were a pendulum supported at the origin.^28

Question

An ideal electric dipole is situated at the origin, and points in the z direction, as in Fig. 3.36. An electric charge is released from rest at a point in the x y plane. Show that it swings back and forth in a semi-circular arc, as though it were a pendulum supported at the [latex] ext { origin. }{ }^{28}[/latex]

Accepted Answer

[latex]F =q E =\frac{q p}{4 \pi \epsilon_{0} r^{3}}(2 \cos heta \hat{ r }+\sin heta \hat{ heta })[/latex] .

Now consider the pendulum: [latex]F =-m g \hat{ z }-T \hat{ r }, ext { where } T-m g \cos \phi=m v^{2} / l[/latex] and (by conservation of energy) [latex] m g l \cos \phi=(1 / 2) m v^{2} \Rightarrow v^{2}=2 g l \cos \phi[/latex] (assuming it started from rest at [latex]\phi=90^{\circ}[/latex] , as stipulated). But [latex]\cos \phi=-\cos heta, ext { so } T=m g(-\cos heta)+(m / l)(-2 g l \cos heta)=-3 m g \cos heta[/latex] , and hence

[latex]F =-m g(\cos heta \hat{ r }-\sin heta \hat{ heta })+3 m g \cos heta \hat{ r }=m g(2 \cos heta \hat{ r }+\sin heta \hat{ heta })[/latex] .

This total force is such as to keep the pendulum on a circular arc, and it is identical to the force on q in the field of a dipole, with [latex]m g\leftrightarrow q p / 4 \pi \epsilon_{0} l^{3}[/latex] . Evidently q also executes semicircular motion, as though it were on a tether of fixed length l.

Question 3.56: An ideal electric dipole is situated at the origin, and poin...

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