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anonymous
 3 years ago
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anonymous
 3 years ago
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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw: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
 3 years ago
Best ResponseYou've already chosen the best response.0I probably shouldn't cancel out the k's

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the k's do cancel out. I tried it another way \[\frac1k(V_1r_1V_2r_2)=0\]

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1q1 is in the inside?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0q_1 is on the inside and it's positive q_2 is on the outside and it's negative

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I can set q_1 equal to q_2 correct? since they are equal but opposite in charge?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1yes, so you can factor it out by superposition\[\sum_i\vec E_i=\frac q{4\pi\epsilon_0r_i}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0elaborate.... how did we add the electric field into the equation?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1I'm trying to do this from scratch, but I guess I'm a little rusty, hold on please

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1\[\Delta V=\int\limits_C\vec E\cdot d\vec\ell=\frac q{4\pi\epsilon_0}\int_a^b\frac{dr}{r^2}\]or is it 2q... I should find something to doublecheck, but the rest is right

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1it's just q, but you are using a different formula

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1your formula is the same, of course. You are probably just highlighting the connection to potential energy and voltage.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Does that mean that \(q=k\Delta Vr\) ?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1\[q=k\Delta Vr=\frac1{4\pi\epsilon_0}\cdot\frac q{4\pi\epsilon_0 r^2}\cdot r\]so no :p

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how about this: \[\Delta V=\frac Er\] \[\frac{kq}{r}=\frac E r\] \[q=\frac{E} k\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0LOL I don't know ...haha

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1for 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
 3 years ago
Best ResponseYou've already chosen the best response.0I think they are int he same direction since dw:1361640455529:dw correct?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh yay! So what would we do with that relationship? \[\Delta V=\frac{Er}q\]

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1yes, as long as the path we are checking voltages between is parallel to the electric field lines

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do we assume that that is the path?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It probably wouldn't be parallel

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1yeah, I mean the displacement really, we could go in the pathdw:1361640896393:dwbut the *displacement* is all that matters for calculating voltage if the field lines are parallel to it

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1361641026090:dw

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1we need the more general line integral for for things where the Efield is not parallel to the displacement at all times, likedw:1361641180439:dwwhere a charge is going like...

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1361641482262:dwbetter prediction of the path of the charge in such a field^

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since 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
 3 years ago
Best ResponseYou've already chosen the best response.1if 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
 3 years ago
Best ResponseYou've already chosen the best response.1what's the question?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{kq}{r}=E\cdot r\] \[q=\frac{E\cdot r}{k}\] I'm soo stuck....LOL! But it seems sooo simple!

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1in 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
 3 years ago
Best ResponseYou've already chosen the best response.0How did you get the \(kE\Delta r\) part?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1so 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
 3 years ago
Best ResponseYou've already chosen the best response.0I'll go eat....I wanna walk away from this problem for a little bit....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It will probably make more sense then LOL.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1yes, it is quite simple and you are overthining it see you later, I am learning a little statistics
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