Electrostatics, ONE TEENSY little problemo please HELPPPPP thanks

AI Thread Summary
The discussion centers on calculating the work required to move a 3.2 µC charge from infinity to a point P in the presence of two other charges, q1 = 1.3 µC and q2 = -2.6 µC, positioned 0.60 m apart. Participants clarify that the electric potential energy (PE) should be calculated for the pairs involving the 3.2 µC charge, specifically between q1 & q3 and q2 & q3, while noting that the PE between q1 & q2 remains constant and can be excluded. The initial PE at infinity is zero, leading to the conclusion that the work required equals the change in PE as the charge is moved to point P. Ultimately, the discussion emphasizes understanding the relationship between electric potential energy and work in electrostatics.
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[SOLVED] electrostatics, ONE TEENSY little problemo! please HELPPPPP thanks!

Homework Statement



Two charges of q1 = 1.3 µC and q2 = -2.6 µC are d = 0.60 m apart at two vertices of an equilateral triangle as in Figure P16.56.



(d) What is the work required to move a 3.2 µC charge from infinity to point P?


Homework Equations



kq1q2 / r = Ue

The Attempt at a Solution



i tried 0 and used 19,478.33 V as voltage, but i don't really understand how to do this problem, like you don't need to tell me the answer, i just want to be able to understand how to find something when the distance is infinity.
 
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How does the electric potential energy change as the particle goes from infinity to its final location? (Consider the charges in pairs.)
 
i don't think i understand what you are saying?
 
What's the electric potential energy of the system when the 3.2 µC charge is in its initial position? (Just consider the charge pairs that include that 3.2 µC charge, plug into your formula, and add up the energies.)

Then figure the electric potential energy when the charge is at infinity. (Same formula.)

The work required will equal the change in potential energy.
 
okay so do i do kq1q2 / r?
 
physicsbhelp said:
okay so do i do kq1q2 / r?
Yes. That's the potential energy between two charges separated by a distance r.
 
are the q's the two charges in the problem of 3.2e-6?
 
physicsbhelp said:
are the q's the two charges in the problem of 3.2e-6?

Call the charges q1, q2, and q3. You'll have PE between: q1 & q2, q1 & q3, and q2 & q3. (Note that the PE between q1 & q2 doesn't change, so you can skip that one.)

So you have two pairs that you need to use to calculate the total electric PE.
 
for q1 & q3
 
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  • #10
for q2 & q3


IS THAT RIGHT?
 
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  • #11
Looks pretty good to me. (Don't forget the units.) Add them up to get a total.

What's the PE when q3 is at infinity?
 
  • #12
so i just add up: q1andq3 + q2andq3
 
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  • #13
or do i need to add q1andq2 to that also?
 
  • #14
physicsbhelp said:
so i just add up: q1andq3 + q2andq3 = -0.06233
Yep. (Don't forget units.)
 
  • #15
okay thanks. so then how would i find the work to move the charge from infinity to point P?
 
  • #16
physicsbhelp said:
or do i need to add q1andq2 to that also?
You could add the PE from q1&q2 to get the total. But since all we care about is the change in PE, that part will not count anyway. (q3 is the only charge that's being moved.)
 
  • #17
physicsbhelp said:
okay thanks. so then how would i find the work to move the charge from infinity to point P?
Find the PE when q3 is at infinity. The work needed will equal the change in PE.
 
  • #18
but i don't get what the R for teh equation would be for q3
 
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  • #19
physicsbhelp said:
but i don't get what the R for teh equation would be for p3
Hint: q3 is infinitely far away. So what happens to the PE?
 
  • #20
the PE becomes 0?
 
  • #21
The initial PE, when q3 is at infinity, is zero--since 1/infinity goes to zero.

You are moving q3 from infinity to point P. The work needed to do that will be the change in electric PE, which equals final PE - initial PE.
 
  • #22
That's what I'd say.
 
  • #23
Thank You Thank You Thank You!
 
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