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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.

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