Where to place a third charge so it experiences no force

In summary: If so, no, that is not what I meant.The distance from q2 is (.29 - r) . Then the force on q3 is kq2q3/(.29 - r)2The distance from q3 to q1 is r . Then the force on q3 is kq1q3/r2The problem says these two forces are equal. So you have kq2q3/(.29 - r)2 = kq1q3/r2Now you can cancel a q3 from each side and you have kq2/(.29 - r)2 = kq1/r2Now you can solve for r .
  • #1
alxkrgr
3
0

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!
 
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  • #2
Shouldn't the distance from the second charge be (.29 - r) where r is the distance from the origin?
 
  • #3
J Hann said:
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.
 
  • #4
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.
 
  • #5
J Hann said:
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)?
 
  • #6
alxkrgr said:

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 .

alxkrgr said:
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) " ?
 

1. Where should the third charge be placed to experience no force?

The third charge should be placed at the midpoint between the two existing charges, equidistant from each one. This creates a balanced and symmetrical electric field, resulting in no net force on the third charge.

2. Can the third charge be placed anywhere else to experience no force?

No, the only location where the third charge will experience no force is at the midpoint between the two existing charges. Placing it at any other location will result in a net force acting on the charge.

3. Will the magnitude of the third charge affect its placement for no force?

Yes, the magnitude of the third charge will affect its placement for no force. If the third charge has a different magnitude than the other two charges, it will need to be placed at a different distance from the two charges to experience no force.

4. What if the two existing charges have the same magnitude?

If the two existing charges have the same magnitude, the third charge should be placed at the midpoint between them for no force. This will create a symmetrical electric field and result in no net force on the third charge.

5. Will the placement of the third charge affect the electric field at other points?

Yes, the placement of the third charge will affect the electric field at other points. Placing the third charge at the midpoint between the two existing charges will create a symmetrical electric field, but placing it at a different location will result in a different electric field at other points.

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