GREPhysics.NET: GR8677 #53

GR8677 #53

Problem

GREPhysics.NET Official Solution

Electromagnetism $\Rightarrow$ }Accelerating Charges

One doesn't really need to understand the mess Griffiths has on accelerating charges to solve this problem. In fact, common sense works. Since the particle spends the most time about the origin, the field there should be maximal. The origin corresponds to $\theta=90^\circ$ , via the system used in the diagram. This is choice (C).

Alternate Solutions

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. $\theta = 0$ (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 $\sin \theta$ to narrow it further down to (C).

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Comments

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

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ee7klt
2005-11-11 04:54:58

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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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Post A Comment!

You are replying to:

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. $\theta = 0$ (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 $\sin \theta$ to narrow it further down to (C).

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