Coloumbs Law when have 3 charges at unknown distances

AI Thread Summary
To determine where a proton can be placed to experience zero net force between two fixed charges, one positive (+8e) and one negative (-2e), it is essential to analyze the forces acting on the proton. The positive charge will repel the proton, while the negative charge will attract it. Placing the proton between the two charges results in a net force that cannot equal zero due to opposing directions of the forces. However, positioning the proton outside the segment joining the two charges may yield two points where the forces balance out. This approach allows for a solution to the problem, confirming that the proton can be placed at specific locations along the line extending beyond the two charges.
garypinkerton
Messages
3
Reaction score
0

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 =)
 
Physics news on Phys.org
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.
 
  • Like
Likes garypinkerton
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
 
  • Like
Likes garypinkerton
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 =)
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top