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goformit100 Group TitleBest ResponseYou've already chosen the best response.0
Wher is it ?
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
I probably shouldn't cancel out the k's
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
the k's do cancel out. I tried it another way \[\frac1k(V_1r_1V_2r_2)=0\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
q1 is in the inside?
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
q_1 is on the inside and it's positive q_2 is on the outside and it's negative
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
I can set q_1 equal to q_2 correct? since they are equal but opposite in charge?
 one year ago

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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
elaborate.... how did we add the electric field into the equation?
 one year ago

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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
yeah no prob :)
 one year ago

TuringTest Group TitleBest 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
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
oh I see....
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
it's just q, but you are using a different formula
 one year ago

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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
Does that mean that \(q=k\Delta Vr\) ?
 one year ago

TuringTest Group TitleBest 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
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
how about this: \[\Delta V=\frac Er\] \[\frac{kq}{r}=\frac E r\] \[q=\frac{E} k\]
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
LOL I don't know ...haha
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
I think they are int he same direction since dw:1361640455529:dw correct?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
in this case, yes
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
oh yay! So what would we do with that relationship? \[\Delta V=\frac{Er}q\]
 one year ago

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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
do we assume that that is the path?
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
It probably wouldn't be parallel
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yeah, 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
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
Oh I see...
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
dw:1361641026090:dw
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
we 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...
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
dw:1361641482262:dwbetter prediction of the path of the charge in such a field^
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
what's the question?
 one year ago

JenniferSmart1 Group TitleBest 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!
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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\[\Delta V=kE\Delta r=\frac Q{4\pi\epsilon_0}\left(\frac1{r_2^2}\frac1{r_1^2}\right)\]
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
How did you get the \(kE\Delta r\) part?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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!
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
I'll go eat....I wanna walk away from this problem for a little bit....
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
It will probably make more sense then LOL.
 one year ago

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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
see ya Turing!
 one year ago
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