Equilateral Triangle Point Charge Problem: Solving for the Unknown Charge

[Romperstomper]

Question: Three point charges, which are initially at infinity, are placed at the corners of an equilateral triangle with sides d. Two of the point charges have a charge of q. If zero net work is required to place the three charges at the corners of the triangle, what must the value be of the third charge?

What I did:
q = 2 of the charges
z = the third charge
d = the final distance between each charge

0 = Uelec_q + Uelec_q + Uelec_z

0 = kq(\frac{-1}{d}) + kq(\frac{-1}{d}) + kz(\frac{-1}{d})

0 = \frac{-2kq}{d} - \frac{kz}{d}

-2q = z

The correct answer is \frac{-q}{2} = z
Can anyone see what I did wrong? I've tried this problem a few times and have gotten the same answer each time.

 

[ehild]

Question: Three point charges, which are initially at infinity, are placed at the corners of an equilateral triangle with sides d. Two of the point charges have a charge of q. If zero net work is required to place the three charges at the corners of the triangle, what must the value be of the third charge?

What I did:
q = 2 of the charges
z = the third charge
d = the final distance between each charge

0 = Uelec_q + Uelec_q + Uelec_z

0 = kq(\frac{-1}{d}) + kq(\frac{-1}{d}) + kz(\frac{-1}{d})

No work is needed to bring in the first charge (q).
The second charge (q) feels the field of the first one. The work needed to place it at a distance d from the first charge is
W_1 = k(\frac{q^2}{d}).
The third charge feels the force from both charges already present.The potential at the third corner of the triangle is
U=2k(\frac{q}{d}).
So the work done when the charge z is brought in from infinity is z times this potential,
W_2 = 2k(\frac{qz}{d}).
And the total work is zero.


ehild

 

[Romperstomper]

Ahh. I see where I messed up now. I wasn't taking into account the second charge in the Uelec formula and was leaving out the Uelec of the other 'z' on each. I was looking at it as all points moving together at the same time instead of each one moving individually.

Thanks a bunch!