How Do Identical Conducting Spheres Behave With Different Initial Charges?

[rocomath]

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.108 N when separated by 50.0 cm, center-to-center. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0360 N. What were the intial charges on the spheres?

[tex]F_1=\frac{k\cdot q_1 q_2}{r^2}[/tex]

[tex]F_2=\frac{k\cdot (q_1 +q_2)}{r^2}[/tex]

Is my approach for situation 2 correct, since they are connected by a wire. I eventually set them equal to each other, I'm stuck on solving for one of the charges. Just a hint please, appreciate it.

edit: Hmm, re-reading it over, it tells me that [tex]q_1=q_2[/tex]?

If [tex]q_1=q_2[/tex], then my equation reduces to a quadratic.

[tex]F_1=\frac{k\cdot q^2}{r^2}[/tex]

[tex]F_2=\frac{k\cdot 2q}{r^2}[/tex]

 

[Kurdt]

You wouldn't add them. If the charge is equal you would have a charge multiplied by the same charge and thus you could just write it as the charge squared. Of course the total charge in both expressions must be the same which will allow you to eliminate a variable.

 

[rocomath]

You wouldn't add them. If the charge is equal you would have a charge multiplied by the same charge and thus you could just write it as the charge squared. Of course the total charge in both expressions must be the same which will allow you to eliminate a variable.

Argh! So I finally caved. The solution says I do add the charges, strange.

http://img207.imageshack.us/img207/4336/185cd3.jpg

 

[Kurdt]

Yes, but not adding in the way you were trying it. You will notice it says the total charge is conserved which I said you can use to eliminate a variable. What they've done is say for part two both charges are equal and are given by [itex]q=(q_1+q_2)/2[/itex], therefore

[tex] F_b = \frac{kq^2}{r^2} = \frac{k(q_1+q_2)^2}{4r^2}[/tex]

I just thought it would be easier to solve for q and then eliminate a variable.

 

[Saladsamurai]

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.108 N when separated by 50.0 cm, center-to-center. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0360 N. What were the intial charges on the spheres?

[tex]F_1=\frac{k\cdot q_1 q_2}{r^2}[/tex]

[tex]F_2=\frac{k\cdot (q_1 +q_2)}{r^2}[/tex]

Is my approach for situation 2 correct, since they are connected by a wire. I eventually set them equal to each other, I'm stuck on solving for one of the charges. Just a hint please, appreciate it.

edit: Hmm, re-reading it over, it tells me that [tex]q_1=q_2[/tex]?

If [tex]q_1=q_2[/tex], then my equation reduces to a quadratic.

[tex]F_1=\frac{k\cdot q^2}{r^2}[/tex]

[tex]F_2=\frac{k\cdot 2q}{r^2}[/tex]

Is this from Halliday and Resnick by chance?

Just curious if we are using the same text for the same course. Then I know I should be following your threads!

 

[rocomath]

Is this from Halliday and Resnick by chance?

Just curious if we are using the same text for the same course. Then I know I should be following your threads!

Lol yes. My course uses University Physics though, I hate it and find this one better.

 

[majimusprime]

i think i solved it...had the same question...im not very good at latex..so forgive my hastyness...and lots of text...this is my first time actually..was looking up the question on google


[tex] (q_1+q_2)^2 = \frac{F_b 4r^2}{k}[/tex]


okay...find the square roots of either side and that will be

||[tex] (q_1+q_2)[/tex] ||= [tex]\sqrt {\frac{F_b 4r^2}{k}}[/tex]

but they were opposite so actually

[tex] (q_1-q_2)= \sqrt {\frac{F_b 4r^2}{k}}[/tex]


so

[tex] q_1= q_2 + \sqrt {\frac{F_b 4r^2}{k}}[/tex]


substitute that value for [tex] q_1 [/tex] in the [tex] F_a=\frac{k\cdot q_1 q_2}{r^2}[/tex]


you will get a cuadratic equation...that you will solve for Q2 and the rest is fairly trivial...thanks for the inspiration...