Velocity of a point charge from work done

1. Feb 14, 2016

Jake 7174

1. The problem statement, all variables and given/known data

If a charge of +2 µC and mass 2 g is taken to (0,0,4) in the presence of an electric ﬁeld due to a ring {R : x2 + y2 = 25,z = 0} of uniform charge density ρL = +3µ C/m, and then released, ﬁnd the maximum velocity it gains.

Given in the problem:
You solve the equation 1/2 mv^2 = W for v

2. Relevant equations

W = -q * ∫E ⋅dl
E = q / (4πε0r^2)

3. The attempt at a solution

We are given 1/2 mv^2 = W (which I thought was for kinetic energy not work) and told to solve.

First I find E from the ring using variables
dq = λ r dΦ (cylindrical coordinates gives the r)
unit vector = <-r ar, z az> / sqrt (r^2 + z^2) - ar cancels
E = ∫ λ z r dΦ / [4πε0 (r^2 + z^2)^(3/2)] from Φ=0 to 2π = λ r z / [(2ε0 (r^2 + z^2)^(3/2)]

I am confused about finding work from this I have the equation W = -q * ∫E ⋅dl. I assume the q here is the point charge and dl is dz so my equation is
-q ∫ λ r z dz / [(2ε0 (r^2 + z^2)^(3/2)] I am unsure if this is correct or what my limits should be My guess is from 0 to 4.

Also, if there is a better approach I am all ears. The problem does not require me to use any specific method. Also, this is a test review with the solution given. I already know the answer.

2. Feb 14, 2016

TSny

You are right that 1/2 m v2 represents the kinetic energy, KE. The formula 1/2 mv^2 = W expresses the "work energy theorem" which states that the net work done on a particle equals the change in KE of the particle.

This approach is OK. But if you are familiar with the concept of electric potential, V, then it would be easier to find V for points on the axis of the ring. Then the work W done on the point charge is related to the change in V.

In your approach, the limits of your integral should correspond to the initial and final positions of the point charge. The final position is where the particle has maximum speed. I don't think that would be z = 0. When the charge is released, which way does it move?

3. Feb 14, 2016

Jake 7174

It will go away from the ring to some point where gravitational force equals electrical potential force. I am not sure how to find this.
How do I find V for points on the ring?

4. Feb 14, 2016

Jake 7174

dont reply yet. I think I have something. I want to see if I can get it

5. Feb 14, 2016

Jake 7174

This is what I came up with..
V = q / (4πε0r)
so if integrate a ring of radius r at a height of z i get
VA = ∫ q r dΦ / (4πε0sqrt(r^2+z^2)) from Φ= 0 to 2π and since voltage is a scalar I dont need to worry about components so my result is
VA= q r/ (2ε0sqrt(r^2+z^2))

What I dont know is where to calculate VB. If I know that I know W = qVAB

6. Feb 14, 2016

TSny

Good.

Can you see where the particle will have maximum speed? (I don't believe you are supposed to worry about gravity in the problem. )

7. Feb 14, 2016

Jake 7174

It should be at max speed just after it is released, and gravity is not mentioned in the solution. The solution shows W = Qq / (4πε0sqrt(r^2+z^2)) where Q is the ring and q is the charge.

8. Feb 14, 2016

Jake 7174

Dont reply yet. I have most of it let me type it in

9. Feb 14, 2016

Jake 7174

If max velocity is reached when the charge is released then I assume VA ≈ VB
If this assumption is good then I can say W = q V = qring Qcharge r / (2ε0sqrt(r^2+z^2)) = 1/2 mv^2
so v = sqrt (qring Qcharge r / (m ε0sqrt(r^2+z^2)))
if I sub in given numerical values I get v = 16.3 m/s

This agrees with the solution which is great but I am curious as to where W = Qq / (4πε0sqrt(r^2+z^2)) comes from.

10. Feb 14, 2016

TSny

At the point of release, the velocity is zero. The max velocity occurs somewhere else.

Think about the direction of the acceleration of the point charge once it's released. Does this direction ever change?

11. Feb 14, 2016

Jake 7174

Once released the direction of acceleration would be +z. I think the charge will accelerate in this direction until some point away from the ring. The voltage should decrease over distance causing the direction of acceleration to reverse

12. Feb 14, 2016

TSny

The direction of the acceleration is determined by the direction of the force. Does the force change direction?

13. Feb 14, 2016

Jake 7174

I suppose not

14. Feb 14, 2016

TSny

Right. For any finite distance from the ring, the particle will be repelled by the ring

15. Feb 14, 2016

Jake 7174

So when the charge is an infinite distance away the direction reverses. Then I need to find voltage at infinity, which is 0, and call that VB then W = q VAB.

Is this right? Seems like it would just make my answer negative.

16. Feb 14, 2016

TSny

Well it doesn't reverse. That particle is gone, it will never come back.

That's almost right. The work done by a conservative force (like the electrostatic force) is the negative of the change in potential energy. Think of lifting a book. The potential energy increases, ΔU > 0. But the work done by the force of gravity is negative, W < 0.

What's the sign of ΔU for your problem? So, what's the sign of W?

17. Feb 14, 2016

Jake 7174

It is the opposite of the book example. Electrical potential goes down as the charge moves away but the work is done by the froce so it will be positive. I am still confused on how to bring this home.

18. Feb 14, 2016

TSny

Yes
Can you articulate what is confusing you at this point? You have W = ΔKE and W = -ΔU.

19. Feb 14, 2016

Jake 7174

I think I am seeing it, but I am screwing up somewhere This is headed for conservation of energy.
KE0 + U0 = KEf + Uf
0 + Q q / (4 π ε0 r) = 1/2 mv^2 + 0 ; r = sqrt(z^2 + r^2) where r = radius
so that is where the W in the solution comes from
then v = sqrt[ Q q / (2 m π ε0 sqrt(z^2 + r^2)) ] if I plug in numbers given r = 5, z =4, q = 2μC, Q = 3μC; m = 2 g = .002 Kg i get v = 2.904 m/s
but I know this isnt right. Where is my error?

20. Feb 14, 2016

TSny

Yes! Most people would choose to use conservation of energy from the start. But you stated that the problem suggested using W = KE, so we followed that path.
OK, if r is interpreted properly on the left side.
Is the r on the left the same as the r on the right?
I think you are confusing the charge, Q, on the ring with the linear charge density, λ, of the ring.