# Electric fields et al.

1. Feb 6, 2005

### thisisfudd

Hi, I am lazy so I am going to post two questions in the same thread. One of them, I think, has been posted before but I wasn't satisfied with the conclusions in the thread.

1. Suppose that electrical attraction, rather than gravity, were responsible for holding the Moon in orbit around the Earth. If equal and opposite charges Q were placed on the Earth and the Moon, what should be the value of Q to maintain the present orbit? Use these data: mass of Earth = 5.98 x 10^24 kg, mass of Moon = 7.35 x 10^22 kg, radius of orbit = 3.84 x 10^8 meters. Treat the Earth and Moon as point particles.

At first, since it referred to "the present orbit" I set E = kQ/r^2 equal to Gm1m2/r^2. That doesn't really work because you wind up with final units of C^2, not just C, which you want for Q. Otherwise, this makes the most sense to me, but that probably means it's wrong.

Then I was thinking about what it means that they stay in the same orbit. Does that mean that they are at "equilibrium"? I don't really know, and if I did, I'm still not sure what I would do. Also, do I have to use centripetal acceleration? I thought maybe i would. Anyway, at this point I am kind of stuck here.

2. A large electroscope is made with "leaves" that are 78-cm-long wires with tiny 24-g spheres at the ends. When charged, nearly all the charge resides on the spheres. If the wires each make a 30 degree angle with the vertical, what total charge Q must have been applied to the electroscope? Ignore the mass of the wires.

OK, firstly, do I consider the spheres as really small point particles and ignore their mass? I want to, but I bet that's wrong. Second, do I have to set up some kind of force diagram, with tension and such? I think that is where I am stuck. Actually, I just find this problem to be extremely obnoxious. My sense is that the net force is equal to zero ... I have to admit, I have been kind of staring idly at this problem so if you have at least a hint, I would appreciate it.

BTW, these are problems 60 and 64, respectively, in Giancoli 6th edition.

2. Feb 6, 2005

### christinono

For the first question, you can use the formulae

$$F_g = \frac{Gm_1m_2}{r^2}$$

$$F_e = \frac{kq_1q_2}{r^2}$$

3. Feb 6, 2005

### leolaw

Hey thisisfudd, i think we are taking the same course.
I have posted your first quesiton before:

And after examining the equation, they do end up the same units, $$C^2$$
notice that you are multiplying $$q_1$$ by $$q_2$$ so you will get $$q^2$$, which is why you will have $$C^2$$ unit at then end

Last edited: Feb 6, 2005
4. Feb 6, 2005

### thisisfudd

Oh, good call, that helps a lot. I knew it made sense but I couldn't figure out the C^2 thing. So, score. Thanks!

Yeah, I'm taking physics electricity and magnetism (second-semester easy physics, if you will) up here in sunny Boston.

5. Feb 7, 2005

### leolaw

i dont know where are you from , but for me, this is cold as hell

6. Feb 7, 2005

### thisisfudd

A large electroscope is made with "leaves" that are 78-cm-long wires with tiny 24-g spheres at the ends. When charged, nearly all the charge resides on the spheres. If the wires each make a 30 degree angle with the vertical, what total charge Q must have been applied to the electroscope? Ignore the mass of the wires.

Any ideas?