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GR8677 #53
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Alternate Solutions |
Herminso
2009-09-22 13:56:33 | For a electric dipole =qd\cos {(\omega t)}\hat{x}) we have:
![\vec{E}=-\frac{\mu_0qd\omega^2}{4\pi}(\frac{\sin {\theta}}{r})\cos {[\omega(t-r/c)]}\hat{\theta}](/cgi-bin/mimetex.cgi?\vec{E}=-\frac{\mu_0qd\omega^2}{4\pi}(\frac{\sin {\theta}}{r})\cos {[\omega(t-r/c)]}\hat{\theta}) and ![\vec{B}=-\frac{\mu_0qd\omega^2}{4\pi c}(\frac{\sin {\theta}}{r})\cos {[\omega(t-r/c)]}\hat{\phi}](/cgi-bin/mimetex.cgi?\vec{B}=-\frac{\mu_0qd\omega^2}{4\pi c}(\frac{\sin {\theta}}{r})\cos {[\omega(t-r/c)]}\hat{\phi})
Thus the oscilation of the electric field is in the xy-plane and the maximum is at  , just the y-axes.
|  | ee7klt
2005-11-11 04:54:58 | hi,
since oscillations happen only in the x-y plane, the E-field vector thus cannot have components in the z-direction. This eliminates (A) and (B).
The minimum occurs when you're looking directly down the x-axis i.e.  (all you see is the infinitesimal tip of the field vector which doesn't look like it'll generate any waves coming towards you) - this eliminates (D). From here, I guess you'll need to remember that the field goes as  to narrow it further down to (C). | |
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Comments |
wittensdog
2009-09-28 21:43:42 | The one thing I'm seeing over and over again is that every physics GRE problem has a quick and simple way to do it. Sometimes it's a calculation trick, sometimes it's just knowing something by heart.
In this case, I strongly recommend remembering two basic facts which have a good chance of coming up on any test:
1.) an oscillating charge never radiates in the direction of its oscillation axis
2.) the polarization of an oscillation charge is parallel to the oscillation axis
Don't bother wondering why (at least not for the sake of the GRE), just memorize that. If you use those two pieces of information, you immediately see that the E-field should be in the xy plane from the restriction on the polarization, and also that the maximum field strength should be at 90 degrees, since along the x axis it has no magnitude, and increases as you move away from the x axis.
So far, in all of my studying, I've never come along a GRE problem that required a formula with more than 3 or 4 terms in it. Maybe that's a slight exaggeration, but you get the point. |  | Herminso
2009-09-22 13:56:33 | For a electric dipole we have:
and
Thus the oscilation of the electric field is in the xy-plane and the maximum is at , just the y-axes.
Herminso
2009-09-22 14:22:02 |
Where  is measured from the positive x-axes on the xy-plane.
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| | a19grey2
2008-11-03 22:14:06 | I recommend just learning what the picture of an oscillating dipole looks like. rnhttp://en.wikipedia.org/wiki/DipolernrnIt'll help you later in your physics life anyway...
physicsisgod
2008-11-05 20:32:44 |
Here's a cool video of it:
http://www.vis.uni-stuttgart.de/ufac/dipole/
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| | jw111
2008-09-14 11:09:42 | You see the maximum OSCILATION of charge when you stand on y axis, and minimum oscilation when you stand on x axis.
=> max field take place on y axis
the ocilation is on x-y plane, the field line wave mainly on x-y plane.
=>oscilating field on x-y plane | | ee7klt
2005-11-11 04:54:58 | hi,
since oscillations happen only in the x-y plane, the E-field vector thus cannot have components in the z-direction. This eliminates (A) and (B).
The minimum occurs when you're looking directly down the x-axis i.e. (all you see is the infinitesimal tip of the field vector which doesn't look like it'll generate any waves coming towards you) - this eliminates (D). From here, I guess you'll need to remember that the field goes as to narrow it further down to (C).
Simplicio
2009-03-31 17:13:47 |
It's oscillating along the x-axis so you can't just say "it is in the x-y plane". It might as well been the x-z plane ...
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| | physicsDen
2005-11-09 22:04:58 | is it me, or is this a poor attempt at a solution?
alpha
2005-11-09 22:15:19 |
it's a quick and dirty way to arrive at the right answer.
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