Show that for the spherical wave shown above, the asymptotic electric field as r
ID: 2303633 • Letter: S
Question
Show that for the spherical wave shown above, the asymptotic electric field as r --> 0, i.e. the near field, is an electric dipole field, as shown below. Additionally, what is the corresponding dipole moment?
FIGURE 11.11 The asymptotic spherical wave. At any time and position the magnetic field is azimuthal-parallel to a line of latitude; and the electric field is orthogonal toB- parallel to a line of longitude. At the time and radius shown, B and E are in the +? and directions, respectively. One half period later at this radius B and E will be in the _? and -0 directions. The Poynting vector points radially outward.Explanation / Answer
An ideal dipole consists of two opposite charges with infinitesimal separation. The potential and field of such an ideal dipole are found next as a limiting case of an example of two opposite charges at non-zero separation.
Two closely spaced opposite charges have a potential of the form:
{displaystyle phi (mathbf {r} )={ rac {q}{4pi arepsilon _{0}left|mathbf {r} -mathbf {r} _{+} ight|}}-{ rac {q}{4pi arepsilon _{0}left|mathbf {r} -mathbf {r} _{-} ight|}} ,} [{displaystyle phi (mathbf {r} )={ rac {q}{4pi arepsilon _{0}left|mathbf {r} -mathbf {r} _{+} ight|}}-{ rac {q}{4pi arepsilon _{0}left|mathbf {r} -mathbf {r} _{-} ight|}} ,}]
with charge separation, d, defined as
{displaystyle mathbf {d} =mathbf {r} _{+}-mathbf {r} _{-} ,} [{displaystyle mathbf {d} =mathbf {r} _{+}-mathbf {r} _{-} ,}]
The position relative to their center of mass (assuming equal masses), R, and the unit vector in the direction of R are given by:
{displaystyle mathbf {R} =mathbf {r} -{ rac {mathbf {r} _{+}+mathbf {r} _{-}}{2}},quad {hat {mathbf {R} }}={ rac {mathbf {R} }{R}} ,} [{displaystyle mathbf {R} =mathbf {r} -{ rac {mathbf {r} _{+}+mathbf {r} _{-}}{2}},quad {hat {mathbf {R} }}={ rac {mathbf {R} }{R}} ,}]
Taylor expansion in d/R (see multipole expansion and quadrupole) allows this potential to be expressed as a series.[4][5]
{displaystyle phi (mathbf {R} )={ rac {1}{4pi arepsilon _{0}}}{ rac {qmathbf {d} cdot {hat {mathbf {R} }}}{R^{2}}}+Oleft({ rac {d^{2}}{R^{2}}} ight)pprox { rac {1}{4pi arepsilon _{0}}}{ rac {mathbf {p} cdot {hat {mathbf {R} }}}{R^{2}}} ,} [{displaystyle phi (mathbf {R} )={ rac {1}{4pi arepsilon _{0}}}{ rac {qmathbf {d} cdot {hat {mathbf {R} }}}{R^{2}}}+Oleft({ rac {d^{2}}{R^{2}}} ight)pprox { rac {1}{4pi arepsilon _{0}}}{ rac {mathbf {p} cdot {hat {mathbf {R} }}}{R^{2}}} ,}]
where higher order terms in the series are vanishing at large distances, R, compared to d.[6] Here, the electric dipole moment p is, as above:
{displaystyle mathbf {p} =qmathbf {d} .} [{displaystyle mathbf {p} =qmathbf {d} .}]
The result for the dipole potential also can be expressed as:[7]
{displaystyle phi (mathbf {R} )=-mathbf {p} cdot mathbf { abla } { rac {1}{4pi arepsilon _{0}R}} ,} [{displaystyle phi (mathbf {R} )=-mathbf {p} cdot mathbf { abla } { rac {1}{4pi arepsilon _{0}R}} ,}]
which relates the dipole potential to that of a point charge. A key point is that the potential of the dipole falls off faster with distance Rthan that of the point charge.
The electric field of the dipole is the negative gradient of the potential, leading to:[7]
{displaystyle mathbf {E} left(mathbf {R} ight)={ rac {3left(mathbf {p} cdot {hat {mathbf {R} }} ight){hat {mathbf {R} }}-mathbf {p} }{4pi arepsilon _{0}R^{3}}} .} [{displaystyle mathbf {E} left(mathbf {R} ight)={ rac {3left(mathbf {p} cdot {hat {mathbf {R} }} ight){hat {mathbf {R} }}-mathbf {p} }{4pi arepsilon _{0}R^{3}}} .}]
Thus, although two closely spaced opposite charges are not quite an ideal electric dipole (because their potential at short distances is not that of a dipole), at distances much larger than their separation, their dipole moment pappears directly in their potential and field.
As the two charges are brought closer together (d is made smaller), the dipole term in the multipole expansion based on the ratio d/R becomes the only significant term at ever closer distances R, and in the limit of infinitesimal separation the dipole term in this expansion is all that matters. As d is made infinitesimal, however, the dipole charge must be made to increase to hold p constant. This limiting process results in a "point dipole
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.