Calculating Net Force on an Electron using E(r) and mx

[CBA]

Homework Statement

An electron is shot at an initial velocity V0=V0ex into an inhomogeneous electric field, that has a width L. Electric field strength is E(r)=E0*sin(2pi*x/L)*ey(ex and ey being vectors), the force acting on the electron is given by: F=-e*E(r). Calculate the net force acting on the electron, the v(t) ,r(t) function in the domain [0, L/V0]and its vertical displacement.

Relevant Equations

F=m*a

I just insterted E(r) into F to get the net force. After that I wrote mx''=F_net and I don't know how to proceed (supposing that's the right way).

 

[BvU]

Hello @CBA , !

I just insterted E(r) into F to get the net force.

Which way is it pointing ?

A drawing might be helpful ...

 

[CBA]

Hello @CBA , !

Which way is it pointing ?

A drawing might be helpful ...

 

[Frabjous]

While the electric field does not vary with time, the particle experiences the changing magnitude of the electric field at different times as it moves through it. This introduces a time dependence in the force equation. Try to derive the time dependant force on the particle.

 

[BvU]

Good start. However:

• Where is $x=0$ and where is $x=L$
• You draw all $|\vec E|\;$ equal
• You don't answer which way $\vec F_{\text{net}}\;$ is pointing

 

[CBA]

I think that the "X" in E(r) is the displacement in the x-direction, which is simply v₀*t and I just instert it into the formula for F,which is =mx'' and it's simply math from there.

 

[BvU]

the formula for F,which is =mx''

Almost, but not quite -- or better: definitely not.

1. The formula is $\vec F = m\vec a$ : the acceleration is in the direction of the force

2. The force acting on the electron is given by: $\vec F=-e\;\vec E(\vec r)$ : the force is in opposite direction wrt the field

3. E(r)=E0*sin(2pi*x/L)*ey should be read as $$\vec E(\vec r)=E_0\,\sin\left ({2\pi\,x\over L}\right ) e_y$$with $e_y = \hat y$ , the unit vector in the y-direction !

As you see, a lot can go wrong if you don't pay attention to the vector character !

How about an update of the drawing ?

$\$

 

[CBA]

Oh yes I forgot that it's in the Y-direction...

 

[BvU]

Bertter than nothing, but a lot of room for improvement.
First of all

• Where is ${\bf x}$ $=0$ and where is ${\bf x}$ $= L\$

(you seem to think $L$ is in the y-direction)

And I would show only one of the two, not both. On a continuous y-axis

And my neck hurts from leaning over by $\pi/2$

 

[BvU]

So how are we doing with this nice exercise ?