Calculating Work to Move a Charge in an Equilateral Triangle

[rgold]

Homework Statement

two particles with charges 4e and -4e are fixed at the vertices of an equilateral triangle with sides of length a. If k=1/4 pi Ԑ what quantity of work is required to move a particle with a charge q from the other vertex to the center of the line joining the fixed charges?

Homework Equations

W = -delta U

The Attempt at a Solution

-(4kQq / a - 2kQq / a) = -2kQq / a => 2kQq / a
I know this is for when the charges are equal (both are Q) but I am not sure how to translate that into my problem...

 

[rgold]

would it be (( 4*1/4 pi Ԑ *4) - (2*1/4 pi Ԑ *-4))/a?

 

[haruspex]

I know this is for when the charges are equal (both are Q) but I am not sure how to translate that into my problem...

No, it's for one charge of q and one of Q.

would it be (( 4*1/4 pi Ԑ *4) - (2*1/4 pi Ԑ *-4))/a?

No. Please post your reasoning and working.

 

[rgold]

No, it's for one charge of q and one of Q.

No. Please post your reasoning and working.

I think I am just getting more confused and that I should start from the beginning... What is the best way to go about this problem?

 

[haruspex]

I think I am just getting more confused and that I should start from the beginning... What is the best way to go about this problem?

What is the potential where the charge starts?

 

[rgold]

What is the potential where the charge starts?

kQq/r

 

[Qwertywerty]

kQq/r

And the potential energy, at the time it starts?

You should also be using the brackets appropriately - 1/4πε₀ is read as (1/4)*πε₀.

 

[Qwertywerty]

W = -delta U

Also, which work is this?

 

[rgold]

Also, which work is this?

Wouldn't that be the work that I'm looking for?

 

[rgold]

And the potential energy, at the time it starts?

You should also be using the brackets appropriately - 1/4πε₀ is read as (1/4)*πε₀.

So would it be ((4*1/4πε)-(-4*1/4πε))/a?

 

[Qwertywerty]

Wouldn't that be the work that I'm looking for?

Is that the work done by the external force, or the electric force?

So would it be ((4*1/4πε)-(-4*1/4πε))/a?

As hurspex pointed out, that is wrong. I was merely requesting you to write in a more, correct fashion. You will need to show your working.

 

[rgold]

Is that the work done by the external force, or the electric force?

As hurspex pointed out, that is wrong. I was merely requesting you to write in a more, correct fashion. You will need to show your working.

at first i thought i should be using (1/(4*πε₀))*((q₁q₁₂)/r) is this correct? and should my answer have an 'a' is it?

 

[rgold]

Is that the work done by the external force, or the electric force?

As hurspex pointed out, that is wrong. I was merely requesting you to write in a more, correct fashion. You will need to show your working.

or do i need to look at it as a dipole moment?

 

[haruspex]

kQq/r

I meant the total potential, due to the two fixed charges.

 

[haruspex]

or do i need to look at it as a dipole moment?

You do not care about forces. This is just about potentials.

 

[Qwertywerty]

or do i need to look at it as a dipole moment?

Can you consider a +4e and -4e as a dipole for any distance from the dipole? What is their constraint?