# Where to place a third charge so it experiences no force

#### alxkrgr

1. The problem statement, all variables and given/known data
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

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

3. 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!

Related Introductory Physics Homework News on Phys.org

#### 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

Staff Emeritus
Science Advisor
Homework Helper
Gold Member
1. The problem statement, all variables and given/known data
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

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

3. 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) " ?

### Want to reply to this thread?

"Where to place a third charge so it experiences no force"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving