#3. I figure if I get caught up on section 2 then I can dive into three. xP

I already need help on HW #2. :oops:
I don't even know where to start on #8.
The momentum changed due to a force.
I'm assuming the force is due to the repulsion and the attractive forces on either side of the field.
But I don't know how to use MOMENTUM to get a force? :sweat:

Ah gotcha!

Ok let me take a look real quick...
So a proton goes from 5.0E-23 to 1.5E-22 in 6.3microSeconds. This one?

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Let me draw a picture.

Yes, that's the one.

Ok that'll be just a minute...

Hurry up slow poke! I've got three or four other things to do tonight! xD

You hurry up! Sheesh... I was expecting you'd ask for number 7 :lol:

We did that one in class. :lol:

9 & 10 suck, too.

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Mmm, PB&J. :drool:

Oh. Oh well. I had fun with #7 anyways :lol:

9 isn't as bad as it looks and 10 is just a little more difficult.

You have a PB&J?!?! :o *makes grabby hands* GIMMIE GIMMIE!!!

Mine. *smacks hands away*

*noms sandwich and sips milk*

Okay, so lay it on me nerdly!

You're the devil... :stare:

Ok so... I'm a little stumped. Would you like me to get you started on 9?

I guess so.

Ok. So did you draw a picture?

Uhm, yes, but Bruce kept it. xD [He had us turn in 9 or 10 so he could see what needed to be discussed more].
I'll redraw it real quick.

:rofl: ok

The point way out there (Left or right of the axis of the dipole, whichever you chose) is going to experience the force of the E field.

So the force of the E field is going to be the sum of the field due to the positive charge and the field due to the negative charge.

What's gonna happen is we'll need to use Coulomb's Law to figure out the force at that point for both charges and then add them together. Since it's all on the axis of the Dipole you won't have to deal with Vector parts.

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In my picture the positive 6nC charge is on the left and the - one on the right and I put the point I'm trying to find the E field charge out to the right.

K. I got distracted. Just hold your horses for a minute.

Badadup badadup badadup.... CWEEEE CWEEE CWEEE!!! I like Chocolate milk... Can I have some Chocolate milk?

From Cheese from Foster's Home For Imaginary Friends! Together they make BlooCHEESE and Mac&CHEESE!!! :XD

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Ok and I got 8

Don't be random, you could get in trouble for that shit.

And sweet, lemme finish writing this thing up then I'll get back to the homework.
He wasn't completely happy with all parts of my lab last week so I had to rewrite. :lol:

It's not random! It's CHEESE from Fosters home for Imaginary friends!!! I should probably put it in quotes come to think of it...

Aw that sucks :(

It was only one part and I only had to add a few sentences.
Not too big of a deal.

I'm cold!

Oh ok good! :stare: Hurry up slow poke! :XD :lol:

I'm a smidge below neutral at the moment. In other news it took me 5 tries to spell neutral correctly...

*throws a fuzzy blanket at Estrella's head* That should keep you warm!

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LMAO! Haha. Way to go. :stare:

Okay, picture is drawn for number 9.
So, the magnitude of the force for the positive charge would be

F = k (6*-6)/2.12^2

and then for negative it'd be the same except divided by 2?

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I left out units and conversions 'cause I'm too lazy to type it.

ok lets see... so... No. :XD :rofl: Let me extrapolate

so you are going to use Coulomb's law except that because you are trying to find the force of the E field at a point in the field, you only need one source charge q in the equation. so it would be like this...

I hate physics. :drool:

force for positive charge is F=k(6)/(2.06)squared

force for negative charge is F=k(-6)/(1.88)squared

This way the resulting force will give us N/C instead of just N

I don't understand the 2.06. :\
Or the 1.88.
Or for that matter why you wouldn't take the force you solved for and use E = F/q.

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Nevermind, I see where the meters are coming from.
Away from THE CENTER of the dipole.
I'm an idiot.

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I was just drawing it like it was 2m away from one of the charges. :oops: