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Two particles with positive charges q_{1} and q_{2} are separated by a distance s.

I got up to here:

(kq_{1})/x^{2} = (kq_{2})/(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

If two positive point charges are present then at a point between them the net field is zero. This is because at that point the two fields are equal in magnitude and are in opposite directions. The electric field of the point charge can be described as

* E = (k×q) ⁄ r ^{2}*

It is it?

I keep getting

Kq1x^2 = Kq2x^2 – Kq1s^2

You have made a serious math mistake.

It is possible to post it and it should be found.

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

Your foirst question seems to have been correctly understood by me, so I believe the solution that I came up with should work.

Your other question, which should have been posted separately to allow more teachers to read it, can be simplified by simply factoring out Q2/d. This will give you a lot of constants that you can add to the parentheses.

It reads:

What distance is there from the charge Q1 to the electric field zero along the line connecting them?