Find net velocity of charged particle in electric field (symbols only)

[Jaccobtw]

Homework Statement

A small sphere of mass m and charge +q with zero velocity falls under the influence of gravity from height h onto an infinite uniformly charged plane with positive charge density σ.


Assuming the charge density is sufficiently low that the sphere falls, what will the velocity of the sphere be when it hits the plane? Note that the electric field above an infinite plane of charge is constant everywhere, pointing away from the plane (for a positive charged plane) and equal to $\frac{\sigma}{2\epsilon_o}$

Give an expression for the magnitude of $\vec{v}_{F}$
in terms of m ,g, q, h, σ ,and ϵ

.

Relevant Equations

F = qE
F = mg
Fnet = mg - qE
v^2 = 2ax

We know the net force on the charged particle in the uniform electric field pointing up is mg - qE.

To get acceleration, divide the net force by mass to get g - qE/m

Plug into kinematic equation and get velocity by itself and substitute$$\sqrt{h(2g - \frac{q \sigma}{\epsilon_o m})}$$

 

[Orodruin]

Did you have a question?

 

[Jaccobtw]

Did you have a question?

Yeah apparently I did it wrong. But I'm not sure what I did wrong

 

[haruspex]

Plug into kinematic equation

Try that part again, taking it in small steps. If you still don't see it, post them.

 

[Jaccobtw]

Did you have a question?

Do you know what I did wrong?

 

[Jaccobtw]

Try that part again, taking it in small steps. If you still don't see it, post them.

Did I plug it it wrong? I’m still getting the same answer

 

[Orodruin]

Yeah apparently I did it wrong. But I'm not sure what I did wrong

Why do you think so and what are you told it is supposed to be?

 

[Jaccobtw]

Why do you think so and what are you told it is supposed to be?

I don’t have the answer but the homework just says it’s incorrect

 

[hutchphd]

Look at your result. How big would the electric field need to be to make v=0? You are off by a factor of 2
The field above an infinitessimal sheet of charge is not the same as a conductor with charge.

 

[Jaccobtw]

Look at your result. How big would the electric field need to be to make v=0? You are off by a factor of 2
The field above an infinitessimal sheet of charge is not the same as a conductor with charge.

Why would v = 0 except as an initial velocity?

 

[Jaccobtw]

Look at your result. How big would the electric field need to be to make v=0? You are off by a factor of 2
The field above an infinitessimal sheet of charge is not the same as a conductor with charge.

Wouldn’t the electric field have to be equal to mg/q for v initial to be 0?

 

[hutchphd]

Wouldn’t the electric field have to be equal to mg/q for v initial to be 0?

Yes. My mistake sorry.

 

[Orodruin]

Look at your result. How big would the electric field need to be to make v=0? You are off by a factor of 2

He is not off. The electric field would have to be $mg/q$ to balance the forces.

The field above an infinitessimal sheet of charge is not the same as a conductor with charge.

He has the field correctly as $\sigma/(2\epsilon_0)$. Hence, for the forces to balance it is necessary that
$$
g = \frac{q\sigma}{2m\epsilon_0}
$$
which is when his expression for the velocity goes to zero as it should.

 

[Orodruin]

but the homework just says it’s incorrect

Is this an autocorrected homework?

 

[Steve4Physics]

$$\sqrt{h(2g - \frac{q \sigma}{\epsilon_o m})}$$

Unless I’m being silly (quite possible), I get the same answer as you.

Some thoughts:

In your Post #1, you state the answer should be in terms of $\epsilon$ (not $\epsilon_0$). So (unless this is a mistake) try entering your answer without the subscript 0.

The software which checks your answer may not be able to parse it correctly. So try a different format, e.g. $$\sqrt{2gh - \frac{q \sigma h}{\epsilon_o m}}$$
Sometimes the ‘official’ answer is wrong – it happens. One possibility is that whoever worked out the official answer made a mistake and used $E = \frac {\sigma}{\epsilon_0}$ instead of $E = \frac {\sigma}{2\epsilon_0}$. You could try entering the answer based on the assumption that this mistake has been made to see what happens.