# Uniform Electric Field

## Homework Statement

A positively charged bead having a mass of 1g falls from rest in a vaccum from a height of 5m in a uniform vertical electric field with a magnitude of 1 x 10^4 N/C. The bead hits the ground at a speed of 21 m/s. Determine a) the direction of the electric field (upward or downward), and b) the charge on the bead.

E=kq/r^2
F=kq1q2/r^2

## The Attempt at a Solution

First of all, I want to make sure that my diagram is correctly drawn. I'm sorta of confused on how the electric field and vacuum are positioned.

http://img125.imageshack.us/my.php?image=52diagramhd1.png

Thanks!

...in a uniform vertical electric field with a magnitude of...

The field in your drawing is horizontal.

Oh my, lol. How stupid of me. Well if the electric field is vertical, wouldn't the electric field be pointed downwards? or is there a trick to this question. I guess it wouldn't be that easy right? My guess is that I will need to use Newton's second law and forces?

In order to figure out the direction of the electric field, you first need to see at what speed would the bead hit the ground if there is no electric field involved.

In order to figure out the direction of the electric field, you first need to see at what speed would the bead hit the ground if there is no electric field involved.

Here's my new diagramhttp://img407.imageshack.us/my.php?image=52diagramac0.png

Btw, are the forces that act on the bead (when the field is off) its weight and the drag force? How do you calculate its drag force? I'm just confused about that. There would still be a drag force when the electric field is turned on right?

How come you need to know the speed with no electric field? Can't you just pick one of the electric fields and test it?

Btw, are the forces that act on the bead (when the field is off) its weight and the drag force? How do you calculate its drag force? I'm just confused about that. There would still be a drag force when the electric field is turned on right?

There's no drag force in vacuum.

How come you need to know the speed with no electric field? Can't you just pick one of the electric fields and test it?

Yes you can. If you consider a random direction and in the end the result says that the charge has a negative sign, than it means that you chose the wrong direction. But I would make a quick calculation to see the right direction from the beginning.

There's no drag force in vacuum.

There's no drag force in vacuum.

Yes you can. If you consider a random direction and in the end the result says that the charge has a negative sign, than it means that you chose the wrong direction. But I would make a quick calculation to see the right direction from the beginning.

So if there's no drag for in a vacuum, then the only two forces acting in the mass is gravity and the electric force caused by the electric field (F=qE) correct?

So how do you even calculate the direction? Are you suppose to use F=ma to find the acceleration? Do you need to use any of the kinematics equation?

Right now I have -W-F=ma so, -mg-qE=ma

however, I don't know the charge of the bead. What am I doing wrong/missing here?

There's no drag force in vacuum.

The http://www.grc.nasa.gov/WWW/K-12/airplane/drag1.html" [Broken] would be generated by the interaction of the bead with a fluid. But since the bead falls in a vacuum and in the vacuum there's no fluid, there's no drag force either. Correct me if I'm wrong.

Last edited by a moderator:
So if there's no drag for in a vacuum, then the only two forces acting in the mass is gravity and the electric force caused by the electric field (F=qE) correct?

So how do you even calculate the direction? Are you suppose to use F=ma to find the acceleration? Do you need to use any of the kinematics equation?

Right now I have -W-F=ma so, -mg-qE=ma

however, I don't know the charge of the bead. What am I doing wrong/missing here?

Use this equation to find the acceleration of the bead:

$$V^2={V_0}^2+2ad$$

where $$V$$ is the speed at the final point, $$V_0$$ is the speed at the starting point, $$d$$ is the distance between the starting point and the final point, and $$a$$ is the acceleration.

Also using this equation you could find the speed of the bead at the final point assuming no electric field. If the speed without the E field is smaller than the speed with the E field, what does it tell you regarding the direction of the E field?

Last edited:
Use this equation to find the acceleration of the bead:

$$V^2={V_0}^2+2ad$$

where $$V$$ is the speed at the final point, $$V_0$$ is the speed at the starting point, $$d$$ is the distance between the starting point and the final point, and $$a$$ is the acceleration.

Also using this equation you could find the speed of the bead at the final point assuming no electric field. If the speed without the E field is smaller than the speed with the E field, what does it tell you regarding the direction of the E field?

Ok, so I found the final speed of the bead without the electrical field by using the equation you gave me:

$$v=\sqrt{2*9.8*5}$$ and the final velocity would be 9.9m/s which is less than the final velocity with the electrical field so that must mean the electrical field points downwards right? in the same direction as gravity? because with the electrical field the final velocity is faster than without.

To find the charge of the bead, do I plug it into this equation?:

$$-W-F_{e}=ma$$ Where F_e=qE and then solve for (q)?? Thanks!

To find the charge of the bead, do I plug it into this equation?:

$$-W-F_{e}=ma$$ Where F_e=qE and then solve for (q)?? Thanks!

Why use $$-$$ ? Why not $$W+F_{e}=ma$$ ?

Why use $$-$$ ? Why not $$W+F_{e}=ma$$ ?

I dunno. I have always made forces that pointed downwards negative and the upward forces positive. Which is correct?? Why do you make them positive? I always have trouble deciding this by the way.

Why use $$-$$ ? Why not $$W+F_{e}=ma$$ ?

I guess I should make W and F_e positive if I make the acceleration positive? Is that how it works?

Wait, how do I know whether the acceleration is upwards or downwards? Haha, I'm getting confused again. =/

I dunno. I have always made forces that pointed downwards negative and the upward forces positive. Which is correct?? Why do you make them positive? I always have trouble deciding this by the way.

It doesn't matter if it points upwards or downwards. The two vectors representing the two sides of the equality must point in the same direction. In your case, $$W+F_e$$ points downwards, and so does the acceleration, so you need not a minus sign.

If you would've had an equality between two vectors that point in opposite directions, then you would've used a minus.

Wait, how do I know whether the acceleration is upwards or downwards? Haha, I'm getting confused again. =/

The acceleration as found from the formula I gave you, points (by definition) from the point in which the speed is $$V_0$$ (i.e the initial point) towards the point in which the speed is $$V$$ (the final point).
If the so-found acceleration would have been negative than it would have point in the opposite direction.

The acceleration as found from the formula I gave you, points (by definition) from the point in which the speed is $$V_0$$ (i.e the initial point) towards the point in which the speed is $$V$$ (the final point).
If the so-found acceleration would have been negative than it would have point in the opposite direction.

I see, ok thanks for all your help!