A 3.5 nC charge is at the origin and a -10 nC charge is at x = 2 cm

[Count]

Homework Statement

A 3.5 nC charge is at the origin and a -10 nC charge is at x = 2 cm.

At what x-coordinate could you place a proton so that it would experience no net force?

Relevant Equations

(3.5/x^2)=(10/(.02+x)^2)

Apparently the answer is not 2.9 cm or 7.4mm. I've looked at similar questions in this thread and solving the exact same way gives me the same answers. I set it up as (3.5/x^2)=(10/(.02+x)^2).
What am I doing wrong?

 

[kuruman]

We cannot tell you what you did wrong if we don't know what you did. Please post your solution.

 

[Orodruin]

Homework Statement: A 3.5 nC charge is at the origin and a -10 nC charge is at x = 2 cm.

At what x-coordinate could you place a proton so that it would experience no net force?
Relevant Equations: (3.5/x^2)=(10/(.02+x)^2)

Apparently the answer is not 2.9 cm or 7.4mm. I've looked at similar questions in this thread and solving the exact same way gives me the same answers. I set it up as (3.5/x^2)=(10/(.02+x)^2).
What am I doing wrong?

The distance squared to the -10 nC charge is not (0.02+x)^2 … even if x is in meters. (Please use appropriate units in equations)

 

[Count]

The distance squared to the -10 nC charge is not (0.02+x)^2 … even if x is in meters. (Please use appropriate units in equations)

I don't understand. If 3.5nC is at the origin, and -10nC is 2cm to the right of that, then the only place for a point charge to be placed is to the left of the origin. That unknown distance is x, so the distance from -10nC to the proton is .02m+x. Solving for x would give the results I got, and my mistake may be in not using the negative sign to show the point charge to the left of the origin, so the answer would be -2.9cm. Is this not correct? If so, why not?

 

[haruspex]

I don't understand. If 3.5nC is at the origin, and -10nC is 2cm to the right of that, then the only place for a point charge to be placed is to the left of the origin. That unknown distance is x, so the distance from -10nC to the proton is .02m+x. Solving for x would give the results I got, and my mistake may be in not using the negative sign to show the point charge to the left of the origin, so the answer would be -2.9cm. Is this not correct? If so, why not?

I agree with 2.9cm left of origin. You can rule out the other solution to the quadratic.
Taking x as distance left of the origin without saying so was a bit confusing.

 

[kuruman]

and my mistake may be in not using the negative sign to show the point charge to the left of the origin,

That is your mistake. The statement of the problem asks "At what x-coordinate could you place a proton so that it would experience no net force?" Coordinates to the left of the origin are conventionally negative.