How to rearrange coulomb's law equation to make q2 the subject

1. Aug 9, 2004

jillmus

have been trying this for 8 hours. have finally come up with q = Fe X r squared, divided by Ke.

In the problem I have been told the value for:
Fe = 5.5 x 10-9, r = 5.0 x 10 -10 and of course Ke (8.99 x 10 9)
known nucleus charge is e= 1.6 x 10-19.
I have been asked whether the magnitude of known nuclues 1, and unknown nuclus is attractive or repulsive and I reckon its repulsive since the product of q1 and q2 has been given as a positive (5.5 x 10-9)

The thing that has been driving me mad is how to calculate the electric charge on the unknown nucleus. The only thing I can think of is to use my rearranged equation (q = Fe X r squared, divided by Ke), and then take the answer, which I have worked out to be q = 1.528 x 10 -37 to calulate q2. Because if the sum of q1 and q2 is 1.528 x 10 -37 , then q2 must equal 1.528 x 10 -37 - 1.6 x 10-19 (known nucleus charge) = -1.6 x 10*19.

Is this right? and does this make it then an attractive force? Also is it right that both nuclei have same charge, one negative and one positive (I guess electrons and protons have the same charges...)

Sorry this is so long, but I have got myself in a complete muddle.
Any help greatly appreciated
Thank you
Jill

2. Aug 9, 2004

Locrian

I didn't understand some of your post Jill.

Yes, the force must be repulsive. Your rationale that they would attract if $$F_e$$ was negative is correct.

As for calculating $$q_2$$, your formula is slightly off, I believe. I think it should be:

$$q_2 = (F_e * r^2)/(K_e * q_1)$$

However, you may have included $$q_1$$ later in your calculations. In any case, I did not get the answer you did. You talk about a sum, but I don't understand why there is a sum involved at all.

3. Aug 9, 2004

jillmus

Hi Locrian

thanks for your reply. I'm not susprised you didn't understand some of my post I was mightily confused! I have tried again to rearrange the equation now you have steered me right to what it should be, (I did get this previously but wasn't sure if it was correct)so I shall recalculate and post my new answer. If its Ok then I don't need to confuse you further regarding 'the sum'!

Thank you again
Phew
Jill

4. Aug 9, 2004

chroot

Staff Emeritus
Jill,

In general, it is best to post the complete question you are struggling with, exactly as your teacher or professor gave it to you, along with your thoughts or failed attempts. With that information in hand, we're much better equipped to help you.

- Warren

5. Aug 9, 2004

jillmus

Right, here goes.

Rearranging coulomb's equation we get (with help from Locrian)
q2 = (Fe x r squared)/Ke x q1

Substituting known values into the equation gives:

q2 = (5.5 x 10 E-9) x (5.0 x 10 E-10)2 / (8.99 x 10E9) x (1.6 x 10E-19) =

9.6 x 10E-19 C

6. Aug 9, 2004

jillmus

Hi Chroot
Just seen your post, thank you for replying. I think I may have it now, but for clarities sake I'll write the first bit of the question I have been trying to solve.

A hydrogen nucleus and a nucleus of an 'unknown' atom are stationary, 5.0 x 10E-10 apart. The mass of the hydrogen nucleus is mh = 1.7 x 10E-27 kg and the electric charge of the hydrogen nuclus is e = 1.6 x 10E-19C. You may assume thet the gravitational constant is G = 6.7 x 10E-11.

state whether this force is attractive or repulsive.

Calculate the electric charge on the unknown nucleus that is necessary to give a force of this magnitude.

There are many more parts to this question (to do with mass of unknown nuclus) but I just needed this first bit to get started.

Sorry if this seems really simple to you guys, but for some reason I came unstuck with this.

could someone explain what an integer multiple of the charge is?

Thanks everyone

Jill

7. Aug 9, 2004

chroot

Staff Emeritus
Looks like the question is asking you to balance gravitational and electromagnetic forces. The gravitational force is always attractive, so, for the two nuclei to remain stationary, the electromagnetic force must be repulsive. That means, of course, that the unknown nucleus is also positively charged.

Are you sure this is the COMPLETE description of the initial conditions? It doesn't appear that there is enough information to say much more; you might have a massive but strongly charged unknown nucleus, or a lightweight but weakly charged one, and the forces would be balanced in either case.

- Warren

8. Aug 10, 2004

Locrian

We can't be sure we are giving you good information without the rest, but hopefully have helped.

Integer multiple of charge refers to how many $$1.6*10^-19$$ charges there are. If there are 4 protons, they will have 4 times that value; if there are 4 electrons, they will have -4 times that value.

9. Aug 10, 2004

jillmus

Thank you Locrian

You all have helped an awful lot. BTW, does my answer now match yours?

Cheers from accross the sea in England

Jill

10. Aug 10, 2004