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Electric Potential of Dipole

  • Thread starter roman15
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  • #1
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Homework Statement


Charges + and -q are placed at -1cm and 1cm along the y axis. If their electric potential energy is -45microJ, what is q?


Homework Equations


U=qV
V=Es


The Attempt at a Solution


I got an equation for the electric potential between the two charges, but it was relative to some point charge that was placed between them, so I didnt know how to use that equation to what I had to solve for
 

Answers and Replies

  • #2
tiny-tim
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hi roman15! :smile:

(have a mu: µ :wink:)

start with one charge at the origin, and the other at infinity …

then move the infinity charge into position! :wink:
 
  • #3
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hmmm im not sure i understand what you mean
 
  • #4
tiny-tim
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hi roman15! :smile:

(just got up :zzz: …)

when one charge is at infinity, the PE is zero …

so move that charge from infinity to 2 cm :wink:
 
  • #5
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hmm well i derived an equation for the electric potential for a point within the dipole, that is for points where -1cm<y<1cm
the change in electric potential= EP(final)-EP(initial)
but wouldnt i use the initial point as -0.01m and the final point as 0.01m, or 0 and 2cm

V(y)=(q/4piε)[(1/|y-0.01|)-(1/|y+0.01|)]
thats the equation i got for the electric potential between the point charges of the dipole
 
  • #6
tiny-tim
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Charges + and -q are placed at -1cm and 1cm along the y axis. If their electric potential energy is -45microJ, what is q?
this isn't a question about electric potential, but about electric potential energy …

you're assuming that it means the electric potential at the origin is -45µJ/C (= -45µV) :redface:
 
  • #7
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how can you assume that the electric potential at the origin is that?
and i know this is a question about electric potential energy, but i need the electric potential to solve it dont I?
Im honestly so confused by this question because i dont know how to solve for the charge
 
  • #8
tiny-tim
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how can you assume that the electric potential at the origin is that?
we can't! but that's what you've done!
and i know this is a question about electric potential energy, but i need the electric potential to solve it dont I?
no, you can find the PE directly, using the method i've already mentioned
 
  • #9
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ok so...moving the point charge from infinity to the origin, 0cm on the y axis then?
at infinity the EP is zero, but then how do i get the EP at the origin?
 
  • #10
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ok wait...sorry
ok so Va-Vb=(q/4piε)[1/ra-1/rb]
so rb approaches infinity
then V=(q/4piεra) right?
i think the only thing confusing me is what would ra be in the situation of the dipole?
 
  • #11
tiny-tim
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so Va-Vb=(q/4piε)[1/ra-1/rb]
so rb approaches infinity
then V=(q/4piεra) right?
yes! :smile:

and ra here is given as 2 cm
 
  • #12
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just two last question, why would ra be the 2cm? from what infinity is the point charge coming from? is it infinity along the y axis?
 
  • #13
tiny-tim
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just two last question, why would ra be the 2cm?
i don't understand :confused: … the 2 cm is given in the question
from what infinity is the point charge coming from? is it infinity along the y axis?
does it matter?

what formula do you intend to use?
 
  • #14
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i think im just having trouble visualizing the question...but i understand how to solve it
i intend on using q/4piεr where r would be 0.02m
 
  • #15
tiny-tim
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i think im just having trouble visualizing the question...but i understand how to solve it
i intend on using q/4piεr where r would be 0.02m
yes that should do it :smile:
 

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