Coloumbs Law when have 3 charges at unknown distances

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SUMMARY

The discussion centers on determining the position where a proton can be placed to experience zero net force due to two fixed charges: q1 = +8e at the origin and q2 = -2e located at x = L. The relevant formula used is Coulomb's Law, F = (kq1q2)/r^2, where k = 8.988 x 10^9 Nm²/C². The solution involves recognizing that placing the proton between the two charges results in a net force, while placing it outside the segment joining the charges can yield two points of equilibrium.

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  • Coulomb's Law and its application in electrostatics
  • Understanding of electric charge units (e.g., elementary charge)
  • Basic principles of force diagrams in physics
  • Concept of net force and equilibrium in electrostatic systems
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garypinkerton
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Homework Statement


I have a question that I can't figure out.

Question is
"Two charges are fixed in location: charge q1 = +8e is located at the origin and charge q2 = -2e is located on the x-axis at x = L. At what point (other than infinitely away) can a proton (a unit positive charge e) be placed so that it has net zero force acting on it"

Homework Equations


F = (kq1q2)/r^2

The Attempt at a Solution


I have got that q1 is (8e x 1.602 x 10^-19C) = 1.282 x 10^-18C
and q 2 is (-2e x 1.602 x 10^-19C) = 3.204 x 10^-19C
k = 8.988 x 10^9Nm^2C^-2
Am assuming that a unit positive charge is 1C (although did originally think it could possibly be 1.602 x 10^-19C, the size of one proton, would that be correct?)

When pumping in these values into the formula, and rearranging the formula so F = 0N, this will obviously give a distance of 0 which would be incorrect, also not sure how the added proton fits in anywhere. Have spent a few hours trying to find another way but am a bit stuck. Thanks in advance for any help =)
 
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garypinkerton said:

Homework Statement


I have a question that I can't figure out.

Question is
"Two charges are fixed in location: charge q1 = +8e is located at the origin and charge q2 = -2e is located on the x-axis at x = L. At what point (other than infinitely away) can a proton (a unit positive charge e) be placed so that it has net zero force acting on it"

Homework Equations


F = (kq1q2)/r^2

The Attempt at a Solution


I have got that q1 is (8e x 1.602 x 10^-19C) = 1.282 x 10^-18C
and q 2 is (-2e x 1.602 x 10^-19C) = 3.204 x 10^-19C
k = 8.988 x 10^9Nm^2C^-2
Am assuming that a unit positive charge is 1C (although did originally think it could possibly be 1.602 x 10^-19C, the size of one proton, would that be correct?)

When pumping in these values into the formula, and rearranging the formula so F = 0N, this will obviously give a distance of 0 which would be incorrect, also not sure how the added proton fits in anywhere. Have spent a few hours trying to find another way but am a bit stuck. Thanks in advance for any help =)
one should make out a force diagram and see /calculate the resultant of the two forces and make it zero.
 
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drvrm said:
one should make out a force diagram and see /calculate the resultant of the two forces and make it zero.
Thanks for your reply. How would I go about that if I don't have distances?
 
Try and feel it :-
if you put a +ve unit charge IN BETWEEN 8e and -2e, then it the 8e would cause repulsion and -2e would cause attraction, plot it and you would get that this is not possible.

if you put a +ve unit charge on the line joining the two charges ( but not in b/w them ) then it is possible that you find two points on which the net force =0
 
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Sahil Kukreja said:
Try and feel it :-
if you put a +ve unit charge IN BETWEEN 8e and -2e, then it the 8e would cause repulsion and -2e would cause attraction, plot it and you would get that this is not possible.

if you put a +ve unit charge on the line joining the two charges ( but not in b/w them ) then it is possible that you find two points on which the net force =0
Arrrr yes! That would work perfectly, thanks for your time in helping =)
 

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