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Wher is it ?

I probably shouldn't cancel out the k's

the k's do cancel out. I tried it another way
\[\frac1k(V_1r_1-V_2r_2)=0\]

q1 is in the inside?

q_1 is on the inside and it's positive
q_2 is on the outside and it's negative

I can set q_1 equal to q_2 correct? since they are equal but opposite in charge?

yes, so you can factor it out
by superposition\[\sum_i\vec E_i=\frac q{4\pi\epsilon_0r_i}\]

elaborate....
how did we add the electric field into the equation?

I'm trying to do this from scratch, but I guess I'm a little rusty, hold on please

yeah no prob :)

oh I see....

it's just q, but you are using a different formula

Does that mean that \(q=k\Delta Vr\) ?

\[q=k\Delta Vr=\frac1{4\pi\epsilon_0}\cdot\frac q{4\pi\epsilon_0 r^2}\cdot r\]so no :p

how about this:
\[\Delta V=-\frac Er\]
\[\frac{kq}{r}=-\frac E r\]
\[q=\frac{-E} k\]

LOL I don't know ...haha

I think they are int he same direction since |dw:1361640455529:dw| correct?

in this case, yes

oh yay!
So what would we do with that relationship?
\[\Delta V=-\frac{Er}q\]

yes, as long as the path we are checking voltages between is parallel to the electric field lines

do we assume that that is the path?

It probably wouldn't be parallel

Oh I see...

|dw:1361641026090:dw|

|dw:1361641482262:dw|better prediction of the path of the charge in such a field^

what's the question?

\[\frac{kq}{r}=E\cdot r\]
\[q=\frac{E\cdot r}{k}\]
I'm soo stuck....LOL! But it seems sooo simple!

How did you get the \(-kE\Delta r\) part?

I'll go eat....I wanna walk away from this problem for a little bit....

It will probably make more sense then LOL.

yes, it is quite simple and you are over-thining it
see you later, I am learning a little statistics

see ya Turing!