# What is the electrical potential at point P?

## 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 [Broken]

## 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

Last edited by a moderator:

Arman777
Gold Member
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
Homework Helper
Gold Member
2020 Award
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
Homework Helper
Gold Member
2020 Award
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?

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

Yes I got the right answer. Thanks.