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- anonymous

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- Stacey Warren - Expert brainly.com

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- goformit100

Wher is it ?

- anonymous

|dw:1361637181191:dw|
Given \[q_1=-q_2\]
I have to find the potential difference.
Here is what I got so far.....
\[V_1=\frac{kq_1}{r_1}\rightarrow q_1=\frac{V_1r_1}{k}\]
\[V_2=\frac{kq_2}{r_2} \rightarrow q_2=\frac{V_2 r_2}{k}\]
\[V_1 r_1=-V_2r_2\]
\[V_1r_1+V_2r_2=0\]
but I have to get delta V

- anonymous

I probably shouldn't cancel out the k's

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## More answers

- anonymous

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

- TuringTest

q1 is in the inside?

- anonymous

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

- anonymous

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

- TuringTest

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

- anonymous

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

- TuringTest

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

- anonymous

yeah no prob :)

- TuringTest

\[\Delta V=-\int\limits_C\vec E\cdot d\vec\ell=-\frac q{4\pi\epsilon_0}\int_a^b\frac{dr}{r^2}\]or is it 2q... I should find something to double-check, but the rest is right

- anonymous

##### 1 Attachment

- anonymous

oh I see....

- TuringTest

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

- TuringTest

your formula is the same, of course. You are probably just highlighting the connection to potential energy and voltage.

- anonymous

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

- TuringTest

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

- anonymous

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

- anonymous

LOL I don't know ...haha

- TuringTest

for the other, I eant\[q\neq\Delta Vr=\frac1{4\pi\epsilon_0}\cdot\frac q{4\pi\epsilon_0 r}\cdot r=\frac q{(4\pi\epsilon_0)^2}\]about your next one\[\Delta V=-\frac{Er}q\]\(if\) E and r are in the same direction whe whole time

- anonymous

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

- TuringTest

in this case, yes

- anonymous

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

- TuringTest

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

- anonymous

do we assume that that is the path?

- anonymous

It probably wouldn't be parallel

- TuringTest

yeah, I mean the displacement really, we could go in the path|dw:1361640896393:dw|but the *displacement* is all that matters for calculating voltage if the field lines are parallel to it

- anonymous

Oh I see...

- TuringTest

|dw:1361641026090:dw|

- TuringTest

we need the more general line integral for for things where the E-field is not parallel to the displacement at all times, like|dw:1361641180439:dw|where a charge is going like...

- TuringTest

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

- anonymous

since we have a sphere, it shouldn't have a potential difference on it's own. But it has a constant potential.....so: no potential difference on a sphere? just between spheres?

- TuringTest

if there were a potential difference between one part of a conducting sphere and another, current would flow across it until the potential difference was neutralized

- TuringTest

what's the question?

- anonymous

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

- TuringTest

in that case, all it's basically asking is for you to find\[\Delta V=\frac1q\Delta U=k(E_{outside}-E_{insode})\]and since |Q1|=|Q2| in this case, we can factor it out to give\[\Delta V=-kE\Delta r=-\frac Q{4\pi\epsilon_0}\left(\frac1{r_2^2}-\frac1{r_1^2}\right)\]

- anonymous

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

- TuringTest

so many typos, my connection is terrible
getting a new computer soon
last line should be\[\Delta V=-\frac1QE\Delta r\] but really this is better seen as an integral..\[\Delta V=-\int_a^b Edr=1\frac Q{4\pi\epsilon_0}\int_a^b\frac{dr}{r^2}\]I don't know what else to say right now... go wat and think of a more definitive question to ask!

- anonymous

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

- anonymous

It will probably make more sense then LOL.

- TuringTest

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

- anonymous

see ya Turing!

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