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- Thread starter phunphysics2
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- #2

gneill

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- #3

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I don't know how to do so mathematically...

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Draw a picture of the two charges along a line, then equate Coulomb attraction of a unit (positive) test charge to qB to repulsion of that test charge from qA.

Hint: the zero-field point might be between qA and qB, or it might not ...

- #5

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Coulomb's law is not used. My professor said to only use the E=k[q]/rsquared formula...

- #6

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qb -------------------qa--------------------P

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qb -------------------qa--------------------P

Never mind, you called x the distance from p to qB and d the distance between qA and qB, so what you wrote is fine.

- #8

gneill

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I don't know how to do so mathematically...

Really?

If ##\frac{a}{b} = \frac{c}{d}## then ##a \cdot d = b \cdot c##

You've never seen this?

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