How to Find the Acceleration of a Charged Particle in a Uniform Electric Field?

[Clutch Cargo]

Homework Statement

What is the acceleration of a charged particle in a uniform electric field? Assume the particle moves along a straight line parallel to the electric field. Show that a particle starting from rest at x=0 and t=0 the speed and position are given by the following formulas.

Homework Equations

Vx=(eE/m)(t/Sqrt(1+(eEt/mc)^2)

and

x=(mc^2/eE)Sqrt(1+(eEt/mc)^2-1)

The Attempt at a Solution

I don't know where to start...

 

[malawi_glenn]

what is the force?

 

[Clutch Cargo]

I don't know. It is just the "uniform electric field" of unspecified potential. This question is written exactly as it was given to me.

 

[malawi_glenn]

But what course is this?

$$\vec{F} = q\vec{E}$$, where q is the charge and E the electric field

That formula you must know if someone wake you up in the middle of the night:P

EDIT:
Or w8 a minute, what is the relation between acceleration and veolcity? This must be a course in Newtonian dynamics ;)

(this is not advanced physics, I can't believe someone has this as upper level undergraduate physics..)

 

[Clutch Cargo]

Yep, it's advanced physics. And this type of question is not even in the textbook! Another thing that bugs me is the little e. The ONLY mention of the little e in the text is for the rest energy of an electron as being e=.511MeV but looking at the relevant equations I'm given we have mass, time, Energy, velocity, and position (x) and the little e is all that remains to express the electric field which I would expect to be volts per meter. I would say E were the electric field but it is a scalar quantity and the conventions used in my textbook are that the scalar E is always energy.

 

[malawi_glenn]

No the electric field is a vector quantity.

And is this correct?

x=(mc^2/eE)Sqrt(1+(eEt/mc)^2-1) ??

It would reduce to

x=(mc^2/eE)Sqrt((eEt/mc)^2) =(mc^2/eE)(eEt/mc)

 

[malawi_glenn]

This also seems to be relativistic, have a look here:

Lorentz force in special relativity
http://en.wikipedia.org/wiki/Lorentz_force

 

[Clutch Cargo]

q is the electric charge of the particle (in coulombs)
and
E is the electric field (in volts per meter)and it equals...capital E as a scalar!

Problem is there is still the question of what the little e is...

 

[malawi_glenn]

charge of the electron, the elementary charge...

sometimes they use e instead of q..

And you should bewere that since the alphabet is very short in comparison with the number of physical quantities, you must look at the context. An E in an electro magnetism problem is in 99% of the cases the electric field.

 

[Rainbow Child]

This is a special relativity question. From
$$\vec{F}=\frac{d\,\vec{p}}{d\,t}\quad (1)$$
with
$$\vec{F}=e\,\vec{E}$$
and
$$\vec{p}=\gamma\,m\,\vec{u},\gamma=\frac{1}{\sqrt{1-(\frac{u}{c})^2}}$$

Since this is a 1-dimension problem, integrate (1) to get your result.

 

[malawi_glenn]

This is a special relativity question. From
$$\vec{F}=\frac{d\,\vec{p}}{d\,t}\quad (1)$$
with
$$\vec{F}=e\,\vec{E}$$
and
$$\vec{p}=\gamma\,m\,\vec{u},\gamma=\frac{1}{\sqrt{1-(\frac{u}{c})^2}}$$

Since this is a 1-dimension problem, integrate (1) to get your result.

why not wait til OP show some work? I already pointed out that he should look at the relation between velocity and acceleration. He must also show how to get to V_x and x

 

[Rainbow Child]

@ malawi_glenn

Is there a general rule on how someone must give instructions? Is it too muh trouble to remind someone the relations that he must apply?

 

[malawi_glenn]

@ malawi_glenn

Is there a general rule on how someone must give instructions? Is it too muh trouble to remind someone the relations that he must apply?

Depends on how you see it, the OP MUST show attempt to solution before any help can be recived. And since I already have kicked him in the correct direction by saying that he should look at the relation between veolcity and acceleration and that he should look at relativistic formula. So in fact, I myself broke the forum rules too.