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Electric Potential due to a dipole

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Homework Statement



NOTE: Coordinates are in centimeters.

A dipole consists of two point charges: +q = 0.911 μC at (-2.53,0) and -q at (2.53,0).

a) Calculate V, the potential created by the dipole at (31.9,82).

Homework Equations


[tex]
V = k \frac{q}{r} \\
\\ r_{1} = \sqrt{(x-.0253)^2 + (y)^2}
\\ r_{2} = \sqrt{(x+.0253)^2 + (y)^2}
\\ V = \sum_{q} V_{due \ to \ q} = V_{+} + V_{-}
\\ \ \ \ = k \frac{q_{+}}{r_{2}} + k \frac{q_{-}}{r_{1}}
[/tex]


The Attempt at a Solution


Using the equations above, where x = .319 meters (distance from (0,0) to point for potential) and y = .82 meters. I got [tex]r_{1} = .871m[/tex] and [tex]r_{2} = .889m[/tex].

I converted the microColumbs to Columbs so I know it can't be that.

Plugging in to the electric potential formula, I got:
V=(8.99e9)*(9.11e-7)/(.871) - (8.99e9)*(9.11e-7)/(.889)
V=9403-9213
V=190 Volts

I put this answer into the assignment and it's wrong. I'm on my last trial, I can't seem to figure out what I'm doing wrong. The school is closed tommorow so I can't go to tutoring.

Please help in whatever way you can, and any hints would be appreciated.
 
Last edited:

Answers and Replies

  • #2
rl.bhat
Homework Helper
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Which charge is nearer to the point P?
 
  • #3
127
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I see. I redone the calculations and just got the negative of my previous answer:
-190.38.

I'm unsure of whether to put in the negative, isn't electric potential a scalar?

EDIT: I put in the negative and I still got it wrong. :_(

I still need to figure this out though. Any more help you can give will be appreciated.
 
Last edited:

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