Velocity of a point charge from work done

[Jake 7174]

Homework Statement

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

Homework Equations

W = -q * ∫E ⋅dl
E = q / (4πε₀r^2)

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.

 

[TSny]

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

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.

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.

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?

 

[Jake 7174]

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?

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?

 

[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?

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

 

[Jake 7174]

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

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
V_A = ∫ q r dΦ / (4πε₀sqrt(r^2+z^2)) from Φ= 0 to 2π and since voltage is a scalar I don't need to worry about components so my result is
V_A= q r/ (2ε₀sqrt(r^2+z^2))

What I don't know is where to calculate V_B. If I know that I know W = qV_AB

 

[TSny]

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
V_A = ∫ q r dΦ / (4πε₀sqrt(r^2+z^2)) from Φ= 0 to 2π and since voltage is a scalar I don't need to worry about components so my result is
V_A= q r/ (2ε₀sqrt(r^2+z^2))

Good.

What I don't know is where to calculate V_B. If I know that I know W = qV_AB

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

 

[Jake 7174]

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. )

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.

 

[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.

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

 

[Jake 7174]

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

If max velocity is reached when the charge is released then I assume V_A ≈ V_B
If this assumption is good then I can say W = q V = q_ring Q_charge r / (2ε₀sqrt(r^2+z^2)) = 1/2 mv^2
so v = sqrt (q_ring Q_charge 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.

 

[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?

 

[Jake 7174]

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?

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

 

[TSny]

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

 

[Jake 7174]

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

I suppose not

 

[TSny]

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

 

[Jake 7174]

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

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 V_B then W = q V_AB.

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

 

[TSny]

So when the charge is an infinite distance away the direction reverses.

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

Then I need to find voltage at infinity, which is 0, and call that V_B then W = q V_AB.
Is this right? Seems like it would just make my answer negative.

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?

 

[Jake 7174]

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?

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.

 

[TSny]

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.

Yes

I am still confused on how to bring this home.

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

 

[Jake 7174]

Yes

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

I think I am seeing it, but I am screwing up somewhere This is headed for conservation of energy.
KE₀ + U₀ = KE_f + U_f
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 isn't right. Where is my error?

 

[TSny]

I think I am seeing it, but I am screwing up somewhere This is headed for conservation of energy.

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.

KE₀ + U₀ = KE_f + U_f
0 + Q q / (4 π ε₀ r) = 1/2 mv^2 + 0

OK, if r is interpreted properly on the left side.

r = sqrt(z^2 + r^2) where r = radius

Is the r on the left the same as the r on the right?

then v = 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 isn't right. Where is my error?

I think you are confusing the charge, Q, on the ring with the linear charge density, λ, of the ring.

 

[Jake 7174]

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.

the r on the left is from the generic expression for electric potential energy. So, from some point on the ring to my charge the distance is sqrt(z^2 + r^2) I am plugging this into the general expression. I should have used different variables.

And I am confusing λ with Q
If I multiply off length (2 π r λ ; r =5) I get Q = 3 π / 100000
when I put this into the equation I get the correct answer of 16.277 m/s

Is this correct?

 

[TSny]

I believe that's correct. Good work.

 

[Jake 7174]

I believe that's correct. Good work.

Thank you. This is probably the 5th time you have helped me. I really appreciate your help.

 

[TSny]

You are welcome. Enjoy your studies.