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

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

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 $$r_{1} = .871m$$ and $$r_{2} = .889m$$.

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

Plugging into 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.

 

[rl.bhat]

Which charge is nearer to the point P?

 

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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.