Two point charges

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phunphysics2
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Greetings fellow physics adorers/ mathematicians,

I was wondering if someone could help me with my posted problem. All the necessary details from the template are included in the screenshot on the right.
 
Attachments
  • Handwritten algebra deriving a quadratic in x from q_B/x^2 = q_A/(x-d)^2, ending with the quadratic formula applied to 3q_A x^2 - 8dx q_A + 4d^2 q_A = 0, giving x = (8d ± sqrt(64d q_A^2 - 48q_A^2 d^2))/(6q_A).
    response.jpg
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  • Handwritten physics derivation on whiteboard: number line with charges qB and qA and point P, distance x, field equations E = k|q|/d^2 and E = k|q|/x^2, then sqrt(qB)/x = sqrt(qA)/(x-d) with question marks below.
    phunphysics.png
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From the second attachment (the one on the right) it looks as though you need only cross multiply and solve for x to finish.
 
Your starting formula is wrong.

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 ...
 
Coulomb's law is not used. My professor said to only use the E=k[q]/rsquared formula...
 
I know that the order of the point charges goes as followed


qb -------------------qa--------------------P
 
phunphysics2 said:
I know that the order of the point charges goes as followed


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.
 
phunphysics2 said:
I don't know how to do so mathematically...

Really? :confused:

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

You've never seen this?