What is the electrical potential at point P?

  • Thread starter ikihi
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  • #1
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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
 
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Answers and Replies

  • #2
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Your potential Energy defition is wrong It cannot be ##V_p=\frac {kQ} {r^2}##.Think about it.
 
  • #3
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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.
 
  • #4
haruspex
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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.
 
  • #5
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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.
 
  • #6
haruspex
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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?
 
  • #7
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Right.
Do you get the right answer now, or can't you tell yet?

Yes I got the right answer. Thanks.
 

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