Coulombs Law of two metal balls

[jayz618]

Homework Statement

Two metal balls A and B of negligible radius are floating at rest on Space Station Freedom between two metal bulkheads, connected by a taut nonconducting thread of length 1.30 m. Ball A carries charge q, and ball B carries charge 2q. Each ball is 1.66 m away from a bulkhead.
(a) If the tension in the string is 4.00 N, what is the magnitude of q?
(b)What happens to this system as time passes ?

Homework Equations

F=Ke q₁xq₂ / x²

The Attempt at a Solution

I have gotten as far as finding the magnitude of q, which i got 1.94 x 10⁵ C

Is that right ?

I am confused as to what this system might do over time ?

 

[kuruman]

I haven't done the calculation but 1.94x10⁵ C sounds like a bit too much. Can you show the details of your calculation?

 

[jayz618]

I had somebody help me (hope they didnt mess me up ! )

I was told that Frepulsion is the same as tension, so therefore

4=8.9875x10⁹ = q1xq2 / x²

From here , we can use algebra to get the following equation.

q²= 4N(1.3M)² / 2 (8.987x10⁹)

??

 

[kuruman]

No, they didn't mess you up. The numbers as you have them are correct. If you redo the calculation you should get a different number from the one you quoted initially with negative powers of ten for the charge. So you need to redo it.

 

[jayz618]

I keep coming up with 194333.1241

by multiplying out the numbers, i get to 6.76 / 1.797 x 10 10

Then the square root of that

 

[kuruman]

What you are calculating is

6.76/1.791*10¹⁰ which your calculating program interprets as 6.76*10¹⁰/1.791

What you should be calculating is

6.76/(1.791*10¹⁰)

The first expression puts 10¹⁰ in the numerator. the second expression puts it in the denominator where it belongs.

 

[jayz618]

I sort of figured that out, but thanks for confirming that. So do I need to store the 1.1791 x 10 to the tenth ? then divide it out ?

 

[kuruman]

Yes, or take the inverse of the number first and then multiply what you get with 6.76, then take the square root.

 

[jayz618]

Ok, I now come up with 3.77 x 10 ^-10

You know what's sad ? Is that I used a free trial of one of those tutor services and he came up with the same 1.94 x 10 10

 

[jayz618]

So what will this system do as time passes ? I can't even fathom a response to that part.

 

[jayz618]

So is the 1.66 M away from each bulkead basically a non used number thrown into screw with us ?

 

[kuruman]

Ok, I now come up with 3.77 x 10 ^-10

You know what's sad ? Is that I used a free trial of one of those tutor services and he came up with the same 1.94 x 10 10

You forgot to take the square root of that.

What do you think happens when time goes on? These balls are near a conducting surface. What happens to the conducting surface when when you bring a whole of charge near it?

 

[jayz618]

1.9 x 10 -5 ?

Will the charge of the balls be transferred to the bulkhead ?

When a conducter is charged in a small region, the charge will be distributed across the entire surface.

 

[jayz618]

Solved !

 

[kuruman]

1.9 x 10 -5 ?

Will the charge of the balls be transferred to the bulkhead ?

When a conducter is charged in a small region, the charge will be distributed across the entire surface.

The charge on a ball induces charges of opposite sign on the bulkhead nearest it. Because one bulkhead pulls one way and the other the opposite way, the non-conducting string that connects the balls is under tension. This is my question: Is there a net force on the two-charged-balls system? If "yes" the balls will accelerate and hit a bulkhead; if no, the balls will just sit there.

 

[jayz618]

Well, because the string is under tension and is non conducting, wouldn't there have to be a net force acting upon it ? Its what is pushing the balls apart to begin with right ?

 

[kuruman]

The tension is internal to the system. If ball A pulls on ball B with force F, then ball B pulls on ball A with force -F by Newton's 3rd Law. Therefore the tension does not affect the two-ball system. As I said, the tension is generated by the bulkheads pulling on the balls nearest them. How do the forces exerted by the bulkheads compare?

 

[jayz618]

I think I see where you are going, because opposites attract, the forces on each bulkhead are the same, which is pulling each ball towards the bulkheads. Meaning the forces on the ball are like forces ? I don't know if I am confusing myself more.

 

[kuruman]

The fact that there is tension in the string does not mean that there is a net force on the balls. If you don't believe me, grab your shirt collar and pull up with your hand. There is tension in your shirt as you pull, but there is no net force acting on you. If there were a net force, you will be flying. What external forces do you think generate the tension in the string?

 

[jayz618]

I really don't know.

I just sat down with a tutor at school for 45 minutes and he couldn't explain any of my homework problems, so quite honestly, I give up.

You can close this thread if you would like.

 

[kuruman]

Give yourself one more chance. The charges on the balls induce opposite charges on the bulkheads nearest them. These opposite charges attract the balls nearest them so the string is under tension. If the forces exerted by the bulkheads are equal and opposite, the net force will be zero, but if one bulkhead attracts the the ball nearest it more than the other bulkhead attracts the other ball, then the net force will not be zero. So, are the bulkhead forces balanced or are they not? What do you think?

 

[jayz618]

I was extremely happy when I read what you just wrote again.

Since the distance between both bulkeads is 1.66m on both sides !, that means the object is centered inside the bulkheads, meaning that the charges on each side are equal, causing them to stay the same distance apart on each side, so therefore over time, the balls are going to stay in the same position ?

I really hope that's right.

 

[kuruman]

But the balls don't have equal charges, one has twice the charge of the other so it induces more charge on the bulkhead closest to it, so ...

 

[jayz618]

So as time passes, the system will move towards ball b's bulkhead ?

 

[kuruman]

Yes. Can you explain to me in your own words why?