# Electric Potential of Dipole

## 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?

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

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tiny-tim
Homework Helper
hi roman15! (have a mu: µ )

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

then move the infinity charge into position! hmmm im not sure i understand what you mean

tiny-tim
Homework Helper
hi roman15! (just got up :zzz: …)

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

so move that charge from infinity to 2 cm 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

tiny-tim
Homework Helper
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) 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

tiny-tim
Homework Helper
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

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?

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?

tiny-tim
Homework Helper
so Va-Vb=(q/4piε)[1/ra-1/rb]
so rb approaches infinity
then V=(q/4piεra) right?
yes! and ra here is given as 2 cm

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?

tiny-tim
Homework Helper
just two last question, why would ra be the 2cm?
i don't understand … 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?

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

tiny-tim
yes that should do it 