Find Solution for Electrostatic Force Neutralization

  • Thread starter Thread starter popo902
  • Start date Start date
  • Tags Tags
    Electrostatic
AI Thread Summary
To achieve electrostatic force neutralization for particle 3, it must be positioned such that the forces from particles 1 and 2 balance out. The initial approach involves setting the forces equal by using the equation F31 = F32, leading to the relationship (r^2)/(0.12 + r)^2 = Q2/Q1. The discussion highlights the challenge of solving for r, as the user is unsure how to proceed after simplifying the equation. A suggested method involves equating the forces directly and eliminating q3 from the equation, ultimately leading to a solvable quadratic equation. The conversation emphasizes the importance of correctly positioning particle 3 to achieve the desired net force of zero.
popo902
Messages
59
Reaction score
0
Electrostatic Forcce :(

Homework Statement


In Fig. 21-26, particle 1 of charge q1 = +1.4 µC and particle 2 of charge q2 = -4.2 µC, are held at separation L = 12 cm on an x axis. If particle 3 of unknown charge q3 is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the (a) x and (b) y coordinates of particle 3?

Homework Equations


F = (Q1 * Q2)K/ r^2

The Attempt at a Solution



At first i thought i could go about this problem setting the particle to the left of the two
i figured that the charge on Q2 > Q1 so its pull would be compensated by Q1's push
(i made my particle three positive)
then i set F31 = F32 so that their forces would balance out to 0
that would make my y placemnt 0 as well
but now...im stuck
i did som canceling until i finally go to this:

(r^2) Q2
------------ = ---------
(0.12 +r)^2 Q1

or
(r^2) / (0.12 +r)^2 = Q2/ Q1

and i don't know wehre to go from there
so...
is there a way out or was i supposed to envision particle three somewhere else??
 
Last edited:
Physics news on Phys.org


If you put your particle on left or right side of q1 and q2 so that it is closest to the one that pulls, you can set up two equations:

Fpull = q3*1.4/d^2
Fpush = q3*4.2/(d+12)^2

when the push and pull forces are equal...
 


Thats wuht i did
and i simplified it to get this


(r^2) / (0.12 +r)^2 = Q2/ Q1

i dint plug in the chrge values for Q1 and 2 yet
but i hope u see the jist
i pretty much started with waht u gave me, (except ur d was my r of course)
but from the equation above i can't figure out how to solve for r
so i was wondering if there was any other way to solve it?? :S
 


just equate them and solve the equation... here is a start:

q3*1.4/r^2 = q3*4.2/(r+12)^2 (--divide both sides by q3)
1.4/r^2 = 4.2/(r+12)^2 (--evaluate right side)
1.4/r^2 = 4.2/(r^2+24r+144) (--multiply both side by (r^2)(r^2+24r+144) )
1.4(r^2+24r+144) = 4.2r^2
etc...
 
Last edited:
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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