What is the electrical potential at point P?

[ikihi]

Homework Statement

Three point charges of +6.00 μC, +4.00 [/B]μC, and +6.00 μC are placed along the x-axis 0.200 m above the +4.00 μC charge as shown in the figure below. What is the electrical potential at point P (relative to infinity) due to these charges?

Diagram:
https://pbs.twimg.com/media/C6cPxGfU0AA9QMB.jpg:large

Homework Equations

Vp = V1 + V2 + V3
Vp= $\frac {kQ} {r^2}$ + $\frac {kQ} {r^2}$ + $\frac {kQ} {r^2}$

The Attempt at a Solution

Vp= $\frac {kQ} {r^2}$ + $\frac {kQ} {r^2}$ + $\frac {kQ} {r^2}$

Vp= 8.99 x10⁹ ⋅ 6 x 10^-6 / 0.283² + 8.99 x10⁹ ⋅ 4.00 x 10^-6 / 0.200² + 8.99 x10⁹ ⋅ 6 x 10^-6 / 0.283²

Vp = 2.24 × 10⁶ V

 

[Arman777]

Your potential Energy defition is wrong It cannot be $V_p=\frac {kQ} {r^2}$.Think about it.

 

[ikihi]

Your potential Energy definition is wrong It cannot be $V_p=\frac {kQ} {r^2}$.Think about it.

I'm not sure what other equation it would be. This is electrical potential (voltage). I know there are 3 charges and there is a distance r for each of them to the point P.

 

[haruspex]

I'm not sure what other equation it would be. This is electrical potential (voltage). I know there are 3 charges and there is a distance r for each of them to the point P.

You are confusing it with the formula for the field. Your equation is dimensionally wrong.

 

[ikihi]

You are confusing it with the formula for the field. Your equation is dimensionally wrong.

Oh okay I see my error. I squared the r, when it shouldn't be.

 

[haruspex]

Oh okay I see my error. I squared the r, when it shouldn't be.

Right.
Do you get the right answer now, or can't you tell yet?

 

[ikihi]

Right.
Do you get the right answer now, or can't you tell yet?

Yes I got the right answer. Thanks.