Magnitude of electric field 32m from a rod

AI Thread Summary
The discussion centers on calculating the electric field produced by a uniformly charged rod at two distances: 4.0mm from the surface and 32m away. The calculated electric field at 4.0mm is approximately 1.0 x 10^7 N/C, which is confirmed as correct. However, the calculated field at 32m, found to be 4.5 x 10^3 N/C, is questioned for its accuracy. Participants note that the formula used for both calculations should remain consistent, but the influence of distance on the electric field's behavior is highlighted, suggesting that the approximation methods for near and far fields may differ. The discussion emphasizes the importance of considering the geometry of the rod and the distribution of charge when calculating electric fields at varying distances.
Les talons
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Homework Statement


"A rod 55cm long and 1.0cm in radius carries a 4.4μC charge distributed uniformly over its length. What is the approximate magnitude of the electric field 4.0mm from the rod surface, not near either end? What is the approximate magnitude of the electric field 32m from the rod?"

Homework Equations


E = q_enclosed/(2∏*ε_0*r*L)


The Attempt at a Solution


For the field at 4.0mm, I calculated the magnitude as: (4.4/1*10^6)N/(2∏*8.85*10^(-12)*0.014*0.55)C = 1.0*10^7 N/C
For the field at 32m, I calculated it the same way: (4.4/1*10^6)N/(2∏*8.85*10^(-12)*32.01*0.55)C = 4.5*10^3 N/C
The answer for the field I got at 4.0mm is correct, but the magnitude found at 32m is wrong. I don't see how the formula for the magnitude would change in the two situations. The change in magnitude looks good to me since the distance from the rod increases by a factor of 10^3. There is still not any flux through the ends of the rod, so that would not change the calculation, either. I must be missing something really obvious. Thanks for reading!
 
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The difference is between close up and a long way away.
Did you use an approximation in the near-field calculation?
Are you expected to use one in the far-field calculation?

What makes you think there is no flux through the ends of the rod?
 
At 32 m each part of the rod is approximately equidistant from every other part of the rod, in a percentage sense. Use that fact.
 
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