Question

Two point charges lie on the 10.6cm axis. A charge of 9.2\mu C is at the origin, and a charge of -4.1\mu C is at x = 10.6cm .
Part A
At what position x would a third charge q_3 be in equilibrium?

Answer 1

Let me think! I'll have to do some research if I'm going to be able to answer this. Sorry it took so long - I had trouble opening the link!Although, the picture has jogged my memory!

Answer 2

its ok. Thank you for helping. Please go on explaining

Answer 3

So I'm guessing the situation is this:
We have those two charges, and they're going to exert a force on each other.
But we want to put a charge in there to cancel out the forces?

Answer 4

we want to know were to put the test charge so it will be at equilibrium

Answer 5

Can we name the charges like this, so they have names?

Answer 6

sure

Answer 7

OH! Test charge, so throw a q_3 in there so that the electrostatic force on it is 0?

Answer 8

I think that's what it's going for...

Answer 9

Would you agree?

Answer 10

Then this isn't so bad.

Answer 11

ok. I agree.

Answer 12

Okay! So\[F=k\frac{q_aq_b}{d^2}\text{ where }q_a,q_b\text{ are some charges, right?}\]

Answer 13

yes

Answer 14

Equilibrium will mean that the two forces on the test charge (one from q1, the other from q2) add up to zero! So we can start the math (algebra, I guess)!

Answer 15

\[F_{q_2,q_3}=k\frac{q_2q_3}{(\Delta x)^2}\]
\[F_{q_1,q_3}=k\frac{q_1q_3}{(\Delta x)^2}\]

Answer 16

I should have specified the different \[\Delta x\text{'s}\]sorry

Answer 17

\[F_{q_2,q_3}=k\frac{q_2q_3}{(x_{q_3}-x_{q_2})^2}\]
\[F_{q_1,q_3}=k\frac{q_1q_3}{(x_{q_3}-x_{q_1})^2}\]

Answer 18

Do you agree with the "add up to zero" part?
Then\[F_{q_1,q_2} + F_{q_3,q_2} = 0\]

Answer 19

yes. But I'm a little confused on what to put in for q3 and x

Answer 20

x is what we will find! And q3 could be anything, I think. The force on q3 from both q1 and q2 will increase proportionally, since the force is changing by a factor of .. just how much q3 changed. Look at the force equations to verify this. The change made to each force (q1 to q3 and q3 to q2) is identical (in terms of absolute value magnitude).

Answer 21

Don't worry, I think q3 should cancel out!

Answer 22

so do I make the two equation equal to each other and solve for x

Answer 23

\[F_{q_1,q_2} + F_{q_3,q_2} = 0\]\[\Downarrow\]\[F_{q_1,q_2} =- F_{q_3,q_2}\]just by subtracting one force from both sides.

Answer 24

Yep! Solve for \[x_{q_3}\]!

Answer 25

what will I put for xq1 and xq2 tho

Answer 26

They are the positions of q1 and q2 along the x-axis. It is how we keep track of the positions. The x-axis is really important because it gives us a reference to see where they all are compared to one another. :) Se the pictures to help!

Answer 27

ok, I will do it and post the answer

Answer 28

so what I rearrange is kq2q3/(xq3-xq2)^2=kq1q3/(xq3-xq1)^2

Answer 29

I just don't know how to isolate xq3

Answer 30

because when I do it, the cancel out

Answer 31

I'm about to work it out on paper as well. I hope this method isn't wrong! I would have wasted your time!

Answer 32

its due in a hour and I have another question that I can't get, I tried it, but its not working out

Answer 33

can you tell me if it works out if you get it

Answer 34

Yeah! And it's possible that there's an easier method that I don't remember.

Answer 35

oh ok.

Answer 36

thank tho.

Answer 37

You aren't given the answer, by chance, are you?

Answer 38

My equation looks incorrect.. But I have to go, sorry! E-mail others looking at physics, so maybe they can help!

Answer 39

no, I just have to sub it in. Do you have the answer tho, it might be right

Answer 40

Well, I got 31.887cm...

Answer 41

its right. Thank you so much

Answer 42

I got 62%, but I just mad stupid mistakes on some. Thank you so much for your help.

Answer 43

I'm sorry I couldn't help more!! But I guess that's over now.

Answer 44

its ok. It takes of marks if I round numbers and they don't match its numbers. Thank you for your help tho. Most of the time I don't get help at all.

Answer 45

Not as many people go into the physics section - not as much as math! I had a school program that did the same thing. It was wrong sometimes too!

Answer 46

But I know those mistakes, easy to make. If you need any help, feel free to e-mail me again.

Answer 47

I might need help in a few days, since I have another assignment on the program.

Answer 48

thank you

Answer 49

Alright, I'll try to log on every once in a while!

Answer 50

thank you so much. Physics isn't my best subject

Answer 51

I like it a lot, but I agree it's tough!

Answer 52

Good luck!

Answer 53

And take care!