## JenniferSmart1 Group Title Question one year ago one year ago

1. goformit100

Wher is it ?

2. JenniferSmart1

|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

3. JenniferSmart1

I probably shouldn't cancel out the k's

4. JenniferSmart1

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

5. TuringTest

q1 is in the inside?

6. JenniferSmart1

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

7. JenniferSmart1

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

8. TuringTest

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

9. JenniferSmart1

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

10. TuringTest

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

11. JenniferSmart1

yeah no prob :)

12. 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

13. JenniferSmart1

14. JenniferSmart1

oh I see....

15. TuringTest

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

16. TuringTest

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

17. JenniferSmart1

Does that mean that $$q=k\Delta Vr$$ ?

18. TuringTest

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

19. JenniferSmart1

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

20. JenniferSmart1

LOL I don't know ...haha

21. 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

22. JenniferSmart1

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

23. TuringTest

in this case, yes

24. JenniferSmart1

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

25. TuringTest

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

26. JenniferSmart1

do we assume that that is the path?

27. JenniferSmart1

It probably wouldn't be parallel

28. 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

29. JenniferSmart1

Oh I see...

30. TuringTest

|dw:1361641026090:dw|

31. 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...

32. TuringTest

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

33. JenniferSmart1

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?

34. 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

35. TuringTest

what's the question?

36. JenniferSmart1

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

37. 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$\Delta V=-kE\Delta r=-\frac Q{4\pi\epsilon_0}\left(\frac1{r_2^2}-\frac1{r_1^2}\right)$

38. JenniferSmart1

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

39. 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!

40. JenniferSmart1

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

41. JenniferSmart1

It will probably make more sense then LOL.

42. TuringTest

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

43. JenniferSmart1

see ya Turing!