Asked by Sam

Two particles with positive charges q1 and q2 are separated by a distance s.

I got up to here:

(kq1)/x^2 = (kq2)/(s-x)^2

I now have to solve for x by cross multiplying and using the quadratic equation but I can't seem to isolate x without it being canceled out. =S

Answers

Answered by bobpursley
Well, then, you have a serious math error.
Post it, and we should be able to find it.Start with this:

q1(s-x)^2=q2*x^2

Answered by Sam
Is it

I keep getting

Kq1x^2 = Kq2x^2 - Kq1s^2
Answered by drwls
You seem to have left out part of the question. How is x defined? Is it the distance from q1 at which the forces on a third charge cancel?

If so, all you can solve for is x/s (or s/x), not x itself.

q2/q1 = (s-x)^2/x^2 = [(s/x)-1]^2
(s/x) - 1 = sqrt (q2/q1)
s/x = 1 + sqrt (q2/q1)
Answered by Sam
It says:
Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges zero?
Answered by Sam
Can you also help me simply this equation. I'm having so much difficult simplifying questions with mostly variables in it.

1/2*mv^2 = k[(-3q^2/(d/sqrt2))+(5q^2/(d/sqrt2))+(2q^2/(d/sqrt2))- (-3q^2)/d+ 2q^2/d + 5q^2/{(root2)d}]
Answered by Sam
sorry in that above one i have to isolate for v
Answered by drwls
I seem to have interpreted the foirst part of your question correctly, so the solution I derived should work..

As for your other question (which should have been posted separately so that more teachers read it), you can simplify by factoring out q^2/d. That will leave you with a bunch of constants to add up inside the parentheses.
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