# Electron in electric field

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1. Jul 10, 2015

### Fabon

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

I'm right now trying to solve a question concerning the direction of an electron in an electric field.
There is the picture I added given and you know that there is a homogeneous electric field.
The blue line represents the movement of an electron (there is no movements up or down) which enters the field in the point A.

The question: In which direction points the electric field?

2. Relevant equations

3. The attempt at a solution

I'm aware of rules like the right-hand-rule, but I'm especially confused since there is an electric (and not magnetic) field - I have never before dealt with electric fields.

I hope you can help me.

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2. Jul 10, 2015

### ehild

It is a homogeneous electric field. What do you know about the force that acts on the electron in that field? Is the force constant during the motion?
How does a particle move if a constant force acts on it? What is its acceleration? Is the acceleration constant during the motion?

3. Jul 10, 2015

### Fabon

Yes! The force and the acceleration is constant.

Maybe I can use the right-hand-rule? Or just for magnetic fields?

4. Jul 10, 2015

### ZapperZ

Staff Emeritus
Er... why are you invoking a magnetic field when there isn't any?

You may not be aware of this, but you HAVE solved a similar type of problem already! This is no different than solving a projectile motion in a gravitational field, where the gravitational force is constant and uniform (i.e. intro kinematics in first year physics!). Just because the nature of the force is different (going from gravitational force/weight to electrostatic force), it doesn't mean that how it is treated is also different.

So unless you slept through your intro physics class, you should already have the skills to answer this.

Zz.

5. Jul 10, 2015

### Fabon

Well that's a great explanation - thank you!
Concerning your explanation, the field would point in a direction as I drew on the picture?

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6. Jul 10, 2015

### EM_Guy

This is pretty much the starting point of electric fields. The electric force acting on a charge in the presence of an electric field is:
$\newcommand{\vect}[1]{\boldsymbol{#1}}$
$\vect{F} = q\vect{E}$

where $q$ is the charge (measured in Coulombs) and $\vect{E}$ is the electric field (measured in V/m or equivalently N/C).

Is the charge of an electron negative or positive? What does this tell you about the direction of the force due to an electric field acting on an electron?

Don't confuse yourself with the right hand rule. This is all about the electric field, not the magnetic field. The magnetic field is different. The force due to the electric field is much simpler than the force due to a magnetic field.

7. Jul 10, 2015

### EM_Guy

This is not kind, nurturing, or helpful. If, on the other hand, your goal is to make people feel stupid, you're probably on to something.

Sometimes even smart people don't see what is obvious. When that happens, I think a patient approach is in order, in which you guide the individual back to what he does know and proceeding from there to the unknown. It is not appropriate in such a case to insult the guy's intelligence or to call into question his work ethic.

8. Jul 10, 2015

### ehild

The force might point in that direction, but what about the electric field? If it points upward, what is the direction of the acceleration of the electron?
And you can figure out the exact direction from the figure. You can read the coordinates of the turning point, where the x component of the velocity becomes zero. Apply the equations valid for uniformly accelerating motion.

9. Sep 3, 2015

### JeffT

Hi, I've asked myself a similar question to this problem.
Will it go in a parabolic way downwards?

Is this the right direction for the electric field?

In school, we've only discused about electrons which are arriving perpendicular or parallel to the electric field. Can someone give me a small hint, how to do this in this case?

Thanks,
Jeff

10. Sep 3, 2015

### ehild

You are right, the electric field points downward and to the right. You can find the slope from the figure:
The electron accelerates both in the x and y direction. Write up the kinematic equations of the uniformly accelerating motion.
Read the coordinates of the turning point, where the x component of the velocity becomes zero. You can assume that the smallest scale corresponds to 1 m , so the turning point is (19,3).
You get the time when the electron reaches the turning point from the equation vx=vo+axt =0
Using that t in the equations for x and y, you get expressions both for ax and ay, in terms of vo. Take the ratio, vo cancels. That ratio is the slope of the force. The slope of the field is opposite.

Last edited: Sep 4, 2015
11. Sep 3, 2015

### TSny

Suppose you rotate your figure so that the parabola is oriented like the trajectory of a ball thrown in the air. Does your black arrow point parallel (or anti-parallel) to the acceleration of gravity?

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12. Sep 12, 2015

### JeffT

Sorry for my long absence, first I thought I've solved the problem, but after thinking a little bit, it doesn't make sense.
I think parallel.

For the time t I had: $$t=- \frac{V_0}{a_x}$$
And for $$a_x=\frac{V_0^2}{2s}$$
(V0+ax* t=0 and s = 1/2 ax* t^2)
Is this right? I thought, the acceleration should be negative, because it slows the electron down?

I've took the ratio of ax and ay and calculated the angle, unfortunately it is rounded ca. 60 degrees, which doesn't sounds right.

13. Sep 12, 2015

### ehild

Your result is wrong, What is the numerical value of s? How did you get ay?

14. Sep 13, 2015

### TSny

Your black arrow does not point parallel or antiparallel to the field direction. I thought it would be helpful to consider the parabolic trajectory of a ball kicked from ground level and returning to the ground. In this case, you know how the direction of the gravitational field is oriented relative to the parabola.

15. Sep 13, 2015

### JeffT

I used s for the distance: sx=19.3 and sy=11.4
I think I found my mistake:
I used the time tx in the equation: $$s_y=\frac{a_y}{2}t^2=\frac{a_y}{2}*4*\frac{V_0^2}{a_x^2}$$
Then I replaced ax with $$a_x=\frac{V_0^2}{2s}$$
And calculated ay: $$a_y=\frac{V_0^2*s_y}{8(s_x)^2}$$
Now it make sense: $$tan^-1(\frac{a_y}{a_x})=8.4^{\circ}$$

Thanks, that makes it more easier to understand.

16. Sep 13, 2015

### ehild

What are these numbers? What are sx and sy? Why do you not use the coordinates of the turning point?

17. Sep 13, 2015

### JeffT

Your're right, it's easier to use the coordinates of the turning point S(19 | 3)(sx|sy), instead of the whole way. With these point, $$a_y=\frac{V_0^2*s_y}{2s_x^2}$$ the result is the same. Is this correct?

18. Sep 13, 2015

### ehild

Still overcomplicated a bit. For the turning point, x=ax/2 t2, and y=ay/2 t2. You need ay/ax.