I think I am brain dead today. I can't even solve this simple problem

  • Thread starter flyingpig
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  • #1
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Homework Statement



A uniform electric field of magnitude 325 V/m is directed in the negative y direction in Figure P25.9. The coordinates of point A are (-0.200, -0.300) m, and those of point B are (0.400, 0.500) m. Calculate the potential difference VB - VA, using the blue path.

P-M-E-P%20%289%29.PNG





The Attempt at a Solution



[tex]\Delta V = E \cdot \Delta d [/tex]

[tex]\Delta d = \sqrt{(0.4 - (-0.2))^2 + (0.5 - (-0.2))^2} = 1[/tex]

[tex]\Delta V = V_b - V_a = (-325V/m)(1m) = -325V[/tex]

Apparently the actual answer is just the vertical path, that is (-325V/m) * (-0.8m) = 260V

Now my question is, just what am I doing? I did the path from a to b and found the length to be 1.
 

Answers and Replies

  • #2
tiny-tim
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hi flyingpig! :smile:

(have a delta: ∆ and a square-root: √ :wink:)

work done is a dot product of two vectors

∆d is a vector, not a scalar

(you can only treat force and displacement as scalars in a purely 1D situation)
 
  • #3
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Wait you know what I just overlooked

[tex]-E \cdot \Delta d = <0,-325> \cdot <0,0.8> = -260V[/tex]

But how did the book get +260V?
 
  • #4
tiny-tim
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you've miscounted the minuses :wink:
 
  • #5
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What do you mean?
 
  • #6
tiny-tim
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you wrote -E in letters, but then you wrote +E in numbers :wink:
 
  • #7
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No I had <0,-325>
 
  • #8
gneill
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[tex] dV = - \vec{E} \cdot \vec{ds} [/tex]
 
  • #9
tiny-tim
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  • #10
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Is it just me or is PF laggin really hard the last two days? It almost always pops up a dead link for me.
 
  • #11
gneill
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Is it just me or is PF laggin really hard the last two days? It almost always pops up a dead link for me.

Yeah, its very, very slow, and often only gets only partway through loading a page before the browser gives up. It's almost unusable.

It looks for all the world like a denial of service attack is going on on their IP address.
 

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