Electric Field Question

[thebigbluedeamon]

I need a little guidance in this problem...

Two point charges lie along the y-axis. A charge of q1=-9 mu*C is at y=6.0m, and a charge of q2=-8.0 mu*C is at y=-4.0 m. Locate the point (other than infinity) at which the total electric field is zero.

So,

I made the statement

E1y = -E2y

and

Ke (q1/(r-4)^2) = Ke (q2/(r+4)^2)
or
q1/(r-6)^2 = q2/(r+4)^2

But that makes it very hard to solve for r. Is that equation set up correctly? If so, what is the easiest way, algebraicly, to solve for r.

[e(ho0n3]

Ke (q1/(r-4)^2) = Ke (q2/(r+4)^2)
or
q1/(r-6)^2 = q2/(r+4)^2

But that makes it very hard to solve for r. Is that equation set up correctly? If so, what is the easiest way, algebraicly, to solve for r.
I think you have a typo in the equation above. Take the square root of both sides and simplify.

[thebigbluedeamon]

I think you have a typo in the equation above. Take the square root of both sides and simplify.

I did have a typo....It was supposed to be a 6 instead of a 4 in the first equation.

Let me try this and hopefully I can solve.

[thebigbluedeamon]

Okay...so I took the square root of both sides of the equation, but that doesn't seem to get me very far.

I end up with:

sqrt(q1/q2) = (r-6)/(r-4)

And frankly I don't know how to solve for r in this situation. I could use Maple or something, but it seems that this problem shouldn't require that. I think I might have got the initial set up wrong.

[Doc Al]

sqrt(q1/q2) = (r-6)/(r-4)
Two problems: (1) another typo, and (2) when you take square roots you better be sure your answer is positive.

Your equation should be: sqrt(q1/q2) = (6-r)/(r+4)
And frankly I don't know how to solve for r in this situation.
Start by multiplying both sides by (r+4). It's a simple linear equation.

[thebigbluedeamon]

Of course it is. I just had a mental block. I wouldn't have caught the "(6-r)" though. Thanks for your help.