Velocity of electron moving between charged plates

[jawad khan]

whts the relation of velocity V and electric field E?
i only know about E=F/e(relation of electric field E and Force F)
any one like to answer?

 

[berkeman]

whts the relation of velocity V and electric field E?
i only know about E=F/e(relation of electric field E and Force F)
any one like to answer?

Use the kinematic equations of motion. The force causes an acceleration, which is a change in velocity. Are you familiar with those equations?

 

[sophiecentaur]

It's accelerating all the time so I guess you want the speed on impact.
The initial Potential Energy will be eV, where e is the electron charge and V is the Potential difference between the plates.
All this PE will end up as Kinetic Energy (= half m (v squared)), where v is the velocity and m is electron mass.
Look up the constants.
Then equate the two energy values to get an equation which you can solve for v.

 

[arunma]

This is actually just a mechanics problem if you use the fact that \vec{F} = q \vec{E}.

You know that the velocity is,

v^2 = v_0^2 +2a(x-x_0)

You can simplify this somewhat by assuming that the electron starts from rest at one plate, and you can place the origin there.

v^2 = 2ax

You further know that,

ma = qE

So,

a = \dfrac{q}{m}E

Now just substitute this into the kinematic equation of motion,

v^2 = \dfrac{2q}{m}Ex

And now you can easily solve for the velocity,

v = \sqrt{\dfrac{2q}{m}Ex}

As you can see, the speed of an charged particle in an electric field does not depend only on the strength of the field, but also on the distance that it has travelled.

 

[Born2bwire]

This is actually just a mechanics problem if you use the fact that \vec{F} = q \vec{E}.

You know that the velocity is,

v^2 = v_0^2 +2a(x-x_0)

You can simplify this somewhat by assuming that the electron starts from rest at one plate, and you can place the origin there.

v^2 = 2ax

You further know that,

ma = qE

So,

a = \dfrac{q}{m}E

Now just substitute this into the kinematic equation of motion,

v^2 = \dfrac{2q}{m}Ex

And now you can easily solve for the velocity,

v = \sqrt{\dfrac{2q}{m}Ex}

As you can see, the speed of an charged particle in an electric field does not depend only on the strength of the field, but also on the distance that it has travelled.

Just to note, this is exactly equivalent to sophie's description. The voltage is Ex and so the potential energy is qEx. This requires that the electric field to be constant, which is true for infinite parallel plates.

 

[sophiecentaur]

Also, by doing it the 'Energy' way, you need not discuss the distance between the plates or consider how long it takes.
It's a one-liner.

 

[jawad khan]

dear.if the x-component and y-component of velocity is given i.e Vx &Vy.and E (electric field) is also given. then how to find the acceleration of the electron.
i only know about E=F/e(relation of electric field and force)
F=ma. can you help me in finding acceleration if Vx ,Vy and E is given??
anyone like to answer?

 

[arunma]

Just to note, this is exactly equivalent to sophie's description. The voltage is Ex and so the potential energy is qEx. This requires that the electric field to be constant, which is true for infinite parallel plates.

You're quite right, Sophie's treatment of the problem is absolutely correct. I was assuming, however, that he wanted the speed of the particle at every point in space. This is more easily done by setting the kinetic energy equal to the work done by the electric field, but I thought the kinematic description might make more intuitive sense.

dear.if the x-component and y-component of velocity is given i.e Vx &Vy.and E (electric field) is also given. then how to find the acceleration of the electron.
i only know about E=F/e(relation of electric field and force)
F=ma. can you help me in finding acceleration if Vx ,Vy and E is given??
anyone like to answer?

The acceleratio is always \dfrac{q}{m}E, assuming the electric field is constant.

 

[sophiecentaur]

So why make it straightforward when you can elaborate ad infinitum.

 

[berkeman]

So why make it straightforward when you can elaborate ad infinitum.

Now, now. There are different ways to approach this problem, depending on what you are looking for. I think that's all they are saying.

 

[arunma]

So why make it straightforward when you can elaborate ad infinitum.

Uh, I think I'll stay out of the discussion on which solution is intrinsically superior.

 

[sophiecentaur]

I sometimes like to take the pretty route, too.