Question

Question

Answer 1

Wher is it ?

Answer 2

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

Answer 3

I probably shouldn't cancel out the k's

Answer 4

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

Answer 5

q1 is in the inside?

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

yeah no prob :)

Answer 12

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

Answer 13

oh I see....

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

LOL I don't know ...haha

Answer 20

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

Answer 21

I think they are int he same direction since correct?

Answer 22

in this case, yes

Answer 23

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

Answer 24

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

Answer 25

do we assume that that is the path?

Answer 26

It probably wouldn't be parallel

Answer 27

yeah, I mean the displacement really, we could go in the pathbut the *displacement* is all that matters for calculating voltage if the field lines are parallel to it

Answer 28

Oh I see...