Potential energy of three equal charges at triangle corners

[stylez03]

The potential to consider for the third charge does work out to be twice that of the single charge in this case. That is because the charges have the same size and sign, and the first two charges are equidistant from the third charge. So please don't think that the new potential is always twice the first one.
In general, $$Vnew = \Sigma k Q_{i}/r_{i}$$. I know that may look confusing but it just says to add up the potential from each charge to find the total potential. The distances to each charge may be different, the size or sign may differ as well.

Okay the more I understand this, the more I get confused. The charge value of all 3 three charges are the SAME, the signs are all the same, the distance between all three charges are the SAME. I'm sorry maybe it's just not going into my head, but I'm still not getting what you're saying. You said the potential of charge 3 is just the summation of c1/c2, okay I get that. The potential for c2 is $$V= kQ/r$$, the only change is that we now have another charge being brought in, there is no sign differences, all the values are still the same. So I don't know what you're asking to sum up because if c1/c2 produce the same potential individually, then they should be 2*V when C3 comes in. From what you're saying that's incorrect, I wish there was a solid example in the book but there isn't.

 

[robb_]

No, the total potential for charge three is just twice that of the single charge alone in this problem. Sorry if it didnt come out like that.

 

[stylez03]

No, the total potential for charge three is just twice that of the single charge alone in this problem. Sorry if it didnt come out like that.

So ... $$V= k(2Q)/r$$ ??

 

[robb_]

Yes that is Vnew for the third charge.
I found a worked example: http://farside.ph.utexas.edu/teaching/302l/lectures/node27.html
This is a confusing topic so don't fret. I need a nap now. good luck.:zzz:

 

[stylez03]

Yes that is Vnew for the third charge.
I found a worked example: http://farside.ph.utexas.edu/teaching/302l/lectures/node27.html
This is a confusing topic so don't fret. I need a nap now. good luck.:zzz:

Yea, I haven't found this chapter to be so easy. Thank you barring with me.

 

[stylez03]

Just to clarify

Total work for Charge 3 => U = Q*Vnew

Where Vnew = $$\frac {k(2Q)} {r}$$

$$U = Q * \frac {k(2Q)} {r}$$

I only ask because I'm low on tries since I've been working on this problem earlier.

 

[robb_]

Yes, now add that energy to the energy to bring in the second and first charges. (first one is zero)

 

[stylez03]

Yes, now add that energy to the energy to bring in the second and first charges. (first one is zero)

That's just $$U_{tot}$$ for Charge 3 right?

 

[robb_]

That will be the total energy associated with the charges-> the potential energy.

 

[stylez03]

That will be the total energy associated with the charges-> the potential energy.

Okay, potential energy, for the entire system am I correct?

 

[robb_]

yuppers if we are both talking tomatoes!