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

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Homework Help Overview

The problem involves calculating the potential difference in a uniform electric field of 325 V/m directed in the negative y direction, between two points A and B with given coordinates. The original poster attempts to find the potential difference using a specific path but questions their approach after discovering a different method yielding a different result.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of potential difference using different paths and question the validity of the original poster's approach. There are discussions about the vector nature of displacement and the dot product involved in the calculations.

Discussion Status

The discussion is ongoing, with participants providing clarifications on vector calculations and addressing potential misunderstandings. There is no explicit consensus on the correct interpretation of the calculations yet.

Contextual Notes

Participants are navigating through the implications of using different paths for calculating potential difference and the associated vector mathematics. There are also mentions of technical issues with the forum itself, which may affect participation.

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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.
 
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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)
 
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?
 
you've miscounted the minuses :wink:
 
What do you mean?
 
you wrote -E in letters, but then you wrote +E in numbers :wink:
 
No I had <0,-325>
 
[tex]dV = - \vec{E} \cdot \vec{ds}[/tex]
 
flyingpig said:
No I had <0,-325>

that is +E :wink:
 
  • #10
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
flyingpig said:
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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