Where to place a third charge so it experiences no force

[alxkrgr]

Homework Statement

A point charge of 5.6 microcoulombs is placed at the origin (x=0) of a coordinate system, and another charge of -1.9 microcouloumbs is placed on the x-axis at .29m. Where on the axis can a third charge be placed so that it experiences no charge?

q1= 5.6 E-6
q2= -1.9 E-6
d= .29m

Homework Equations

F= k*(q1*q2)/r^2

The Attempt at a Solution

Fnet = F1,3 + F2,3
Fnet = 0
F1,3 = -(F2,3)
k*(q1*q3)/r^2 =k*(q2*q3)/(r-distance)^2
i know the k and q3 variables cancel out so i get
q1/r^2 = q2/(r-distance)^2

after i get this far I can't seem to get a correct answer. any help would be greatly appreciated!

 

[J Hann]

Shouldn't the distance from the second charge be (.29 - r) where r is the distance from the origin?

 

[alxkrgr]

Shouldn't the distance from the second charge be (.29 - r) where r is the distance from the origin?

Yes, I accidentally reversed it in my post. Thanks for pointing that out.

 

[J Hann]

Also, remember that the third (test) charge cannot be between the first two charges because one charge will
attract the test charge and one charge will repel the test charge.

 

[alxkrgr]

Also, remember that the third (test) charge cannot be between the first two charges because one charge will
attract the test charge and one charge will repel the test charge.

so should the charge on q2 be q2/(.29 PLUS r)?

 

[SammyS]

Homework Statement

A point charge of 5.6 microcoulombs is placed at the origin (x=0) of a coordinate system, and another charge of -1.9 microcouloumbs is placed on the x-axis at .29m. Where on the axis can a third charge be placed so that it experiences no charge force?

q1= 5.6 E-6
q2= -1.9 E-6
d= .29m

Homework Equations

F= k*(q1*q2)/r^2

The Attempt at a Solution

Fnet = F1,3 + F2,3
Fnet = 0
F1,3 = -(F2,3)
k*(q1*q3)/r^2 =k*(q2*q3)/(r-distance)^2
i know the k and q3 variables cancel out so i get
q1/r^2 = q2/(r-distance)^2

after i get this far I can't seem to get a correct answer. any help would be greatly appreciated!

You have typo which I have indicated above.

Your equation, q₁/r² = q₂/(r - d)² is missing a negative sign. (Remember, q₂ is itself negative.)

That makes it q₁/r² = -q₂/(r - d)² .

Take the square root of both sides remembering to use a ± on one side or the other. Solve for r .

so should the charge on q2 be q2/(.29 PLUS r)?

That does not make sense. Did you mean "Should q₃ be q₂/(.29 + r) " ?