Originally posted by suyver
I can think of two other (anti)quarks that also will have charge -2/3. Can you too? Why can/can't it be one of those? Also... who said that it had to be just one quark? There are also quarks with charge of -1/3, so why can't you have two of them? [I know the answers to these questions, I am just trying to get you to think about it... 
]
I just wanted to elaborate on this for future reference. The hypercharge of the quarks and gluon is also just as important in determining the unknown as the electrical charges are. From the five choices given in the problem, it is clear which one is correct, and so far you are all doing great at that. But the question by suyver does raise some issues. In the real world, where your choices are not limited to the given five, anti-charm and anti-top would satisfy the charge-conservation requirement, yet they would not be correct. Here's why; hypercharge-conservation and isospin-conservation.
In the electroweak-regime, the properties that determine the flavours of particles are related to the electric charge by;
<br />
Q = I_z + Y_{weak}<br />
where Q is the electrical charge, I_z is the isospin eigenvalue, and Y_{weak} is the weak hypercharge associated with the flavour of the particle. In conserving I_z and Y_{weak}, you automatically conserve electric charge. So I will run through the math of it here since you have already found the correct answer.
The hypercharge is, itself, a composite of several inmportant quantum numbers;
<br />
Y_{weak} = \frac {b + S + C + B + T}{2}<br />
obtains for all quarks and their composites, with b the baryon number, S[/itex] the strangeness, C the charmness, B the bottomness, and T the topness for a given particle. For the known quarks, and in this case the ones we are looking at, the properties in each of these is as follows;<br />
<br />
for the anti-up quark;<br />
<br />
<br />
I_z = -\frac {1}{2}, b = -\frac {1}{3}, S = C = B = T = 0<br /><br />
<br />
for the anti-charm quark;<br />
<br />
<br />
I_z = 0, b = -\frac {1}{3}, S = B = T = 0, C = -1<br /><br />
<br />
for the anti-top quark;<br />
<br />
<br />
I_z = 0, b = -\frac {1}{3}, S = C = B = 0, T = -1<br /><br />
<br />
for the up quark;<br />
<br />
<br />
I_z = +\frac {1}{2}, b = +\frac {1}{3}, S = C = B = T = 0<br /><br />
<br />
for the gluon;<br />
<br />
<br />
I_z = 0, b = 0, S = C = B = T = 0<br /><br />
<br />
Putting these figures of the known objects into a conservation equation gives us the appropriate numbers for the unknown;<br />
<br />
<br />
I^{unknown}_z = I^{g}_z - I^{u}_z = 0 - \frac {1}{2} = -\frac {1}{2}<br /><br />
<br />
<br />
Y^{unknown}_{weak} = Y^{g}_{weak} - Y^{u}_{weak} = 0 - \frac {1}{6} = -\frac {1}{6}<br /><br />
<br />
Which quark or antiquark has Y_{weak} = -\frac {1}{6} and I_z = -\frac {1}{2} ? It cannot be either the anti-charm or anti-top quarks because of the isospin number. This leaves only the anti-up quark, which has both I_z = -\frac {1}{2} and Y_{weak} = -\frac {1}{6}. This is also in accordance with a general rule that photons, Z particles, and gluons can only produce particle-antiparticle pairs. So the anti-up quark is clearly the identity of the unknown, all others being excluded in one way or another.
Just out of curiosity, what are your possible answers?
