Attraction of two charged particles

[Gunthi]

Homework Statement

Two charged particles q1 and q2, both of m=10g:

q1-----O----------q2
|-5cm-||----10cm--|

q1=4 microC
q2=-2 microC

Find the velocity of q1 when it passes O.

Homework Equations

(for some reason latex isn't working on my browser)

a=dv/dt=(dv/dx).(dx
/t)=v.(dv/dx)

Both charges 'feel' the force:

m.a=K.(1/x)

K=q1q2/(4pi(e0))

x being the distance between the particles.

The Attempt at a Solution

v.dv=a.dx

integrating on both sides from the initial to the final states of the respective du's

(v^2)/2=-K(1/xf-1/xi)

In the same dt both travel the same dx, so if q1 is at 0 q2 must be at 5 cm from 0 so xf=5cm and xi=15cm.

Doing the math I get vf=13.8 m/s and I should be getting 4sqrt(6) m/s (aprox 9.8 m/s).

What am I doing wrong? Can anyone offer me new insight into this problem?

Cheers.

 

[gneill]

Have you considered approaching the problem from a conservation of energy point of view?

 

[Gunthi]

Have you considered approaching the problem from a conservation of energy point of view?

Yes.

Ec1 + Ec2 + V1 + V2 = Ec1'+ Ec2' + V1' + V2'

Ec1=Ec2=0 and V1=q1/(4.pi.e0.r12)

V1+V2-(V1'+V2') = 2Ec1' because same force -> same acceleration so pf1=pf2 (E=p^2/2m)

I get vf=sqrt([(1/r12-1/r12')(q1+q2)/(m.4.pi.(e0)]) with r12 the distance between the particles in the beginning and r12' the distance between them when q1 is at O.

Because r12>r12' vf becomes imaginary...

 

[gneill]

PE = k*q1*q2/r

Find the initial PE and final PE. The magnitude of the change in PE will equal the change in KE. This KE will be equally shared by both particles.

 

[Gunthi]

PE = k*q1*q2/r

Find the initial PE and final PE. The magnitude of the change in PE will equal the change in KE. This KE will be equally shared by both particles.

So KE=k.q1.q2(1/r12'-1/r12) and dKE1 = (p1'^2)/(2m).

But dKE1=KE/2 which means p1'^2 = k.q1.q2.m(1/r12'-1/r12) which is the same expression I got in the first post...

Maybe the solutions are wrong... If you have any other alternative I'm all ears, if not thanks for taking the time ;)

 

[gneill]

Initial PE: PE0 = k*q1*q2/15cm = -0.479 J

Final PE: PE1 = k*q1*q2/5cm = -1.438 J

|ΔPE| = 0.959 J

So KE is 0.959 J

There are two equally massed particles, so

(1/2)*m*v² + (1/2)*m*v² = 0.959 J

or

m*v² = 0.959 J

 

[Gunthi]

Got it. Thanks ;)