# Homework Help: Stupid Charges

1. Feb 2, 2008

Stupid Charges!!!

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of .108 N when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is disconnected, the spheres repel each other with an electrostatic force of .036 N. Of the initial charges on the spheres, with a positive net charge, what was the (a) negative charge of one of them and (b) the positive charge of the other?

Well I know that I need to use Coulomb's Law since that is all we have studied. I know that I have one equation

$$F_{12}=\frac{k|q_1||q_2|}{r^2}$$

$$\Rightarrow \frac{k|q_1||q_2|}{.5^2}=.108$$

But I am having a hard time writing the second equation in terms of $q_1$ and $q_2$.

I know that $q_1+q_2$ is a positive number, that should help.

Any hints??

Thanks,
Casey

Last edited: Feb 2, 2008
2. Feb 2, 2008

Would my second equation be $$\frac{k|\frac{q_1}{2}||\frac{q_2}{2}}{.5^2}=.036$$

since the charge must distribute uniformly between the spheres? My prof gave me a hint that the second equation should be $$F=k*\frac{(q_1+q_2)^2}{4r^2}$$ But I have no idea where he got that?!

Last edited: Feb 2, 2008
3. Feb 2, 2008

### Shooting Star

When the spheres are connected, the potential of the two spheres become equal and the charge is divided according to that, not evenly as you are assuming. You have to use the formula for the capacitance of a sphere, and calculate the charge on each. The only thing that remains constant is the sum of the charges.

4. Feb 2, 2008

Thanks Shooting Star, but we have not learned anything about capacitance in the text yet; the chapter consists of ONLY Coulomb's Law. The charge should distribute evenly. There is an example problem that is somewhat similar in which say explain that since the spheres are identical, the charge will be uniform.

Basically, I can only use Coulomb's Law for this.

Can anyone see where my professor got the hint $$F=k*\frac{(q_1+q_2)^2}{4r^2}$$ from?

5. Feb 2, 2008

### Gear300

Since the spheres are identical, the charge should distribute uniformly. While q1 and q2 are before the wire and q1' and q2' after, take the change in charge of q1 is equal to the change of charge of q2. Since they attract at first, take q1 to be negative and q2 to be positive. Since the net charge is positive initially, q1 + q2 is positive. And since charge is conserved, q1' + q2' = q1 + q2, which is positive on both sides. Since the charges repel afterward, with the net charge as positive, then both q1' and q2' should be positive.

Last edited: Feb 2, 2008
6. Feb 2, 2008

Okay. And since the charges distribute evenly $q_1'=q_2'=\frac{q_1+q_2}{2}$ and that looks to me like it should do it. Charge is conserved, nice call. Thanks!!!