How Do You Calculate Charge from Electric Potential Energy in a Dipole?

[roman15]

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

 

[tiny-tim]

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! ​

 

[roman15]

hmmm I am not sure i understand what you mean

 

[tiny-tim]

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

 

[roman15]

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 wouldn't 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]

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)

 

[roman15]

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 don't I?
Im honestly so confused by this question because i don't know how to solve for the charge

 

[tiny-tim]

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 don't I?

no, you can find the PE directly, using the method I've already mentioned

 

[roman15]

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?

 

[roman15]

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]

so Va-Vb=(q/4piε)[1/ra-1/rb]
so rb approaches infinity
then V=(q/4piεra) right?

yes!

and r_a here is given as 2 cm

 

[roman15]

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]

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?

 

[roman15]

i think I am 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]

i think I am 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