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Field of a charge

  1. Apr 25, 2012 #1
    Suppose you want to find the field of the charge q relative to the coorinate system on the picture. That wont be very hard to do and the field will in general have both a vertical and horizontal component. Now suppose we let the charge get very very far out on the x-axis, such that the horizontal distance to the origin is much much larger than the vertical distance from the origin to q.
    My question is:
    Will the vertical component of the force then be negligible compared to the horizontal? Because geometry does seem to suggest that, but on the other hand it is not really intuitive to me why. Evidently it is not like you are moving further away vertically than horizontally, it's rather the opposite.
     

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  2. jcsd
  3. Apr 25, 2012 #2
    But, you are moving farther away from the charge. Hence, both components of the field get smaller, but the ratio of of vertical to horizontal component goes as d/x (which is the tangent of the angle) where d is the vertical offset. So, it's clear that the ratio drops linearly with distance, while the field magnitude drops as distance is squared.

    For this reason I would say that the vertical component is negligible in the far field limit. In the limit as x goes to infinity, the vertical component of field goes to zero as 1/r^3 and the horizontal component goes to zero as 1/r^2. At some point along the way, you may want to neglect the vertical component.
     
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