- #1

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Thx

- Thread starter leon1127
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- #1

- 486

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Thx

- #2

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show some work.

how do you think it should be solved?

how do you think it should be solved?

- #3

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[tex]\frac{4.05* q_3}{(-0.34-x)^2} =\frac{-2.8 * q_3}{(0.195+x)^2} [\tex]

I dont know why it doesnt work

I dont know why it doesnt work

Last edited:

- #4

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may anyone?

- #5

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|------x2-----------------------x----------------x1----------|

if x is some value, then to find the distance, you have to use the distances as (x-x2) and (x1-x)

- #6

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how do you know if x will land in the interval between [tex] q_1 and q_2 [\tex]

the problem i get is when i assume the q3 will at equalibrium, x become imaginary when i take the square root

2.8/x^2 == -4.05 / (0.535+x)^2

i have shifted the system to the right. Then the negative sign became PAIN under the square root

the problem i get is when i assume the q3 will at equalibrium, x become imaginary when i take the square root

2.8/x^2 == -4.05 / (0.535+x)^2

i have shifted the system to the right. Then the negative sign became PAIN under the square root

Last edited:

- #7

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i'm sorry, i didn't read the question carefully enough. as far as i can tell, it has to be either left of q_2 or right of q_1. just from eyeballing the figure, it looks like it will be to the left of q_2 because the smaller repelling charge needs to be closer to cancel out the larger attracting charge.

- #8

- 486

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nvm i got it.

Thank you everyone

Thank you everyone

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