- #1

jawad khan

- 5

- 0

i only know about E=F/e(relation of electric field E and Force F)

any one like to answer?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

In summary, the relation between velocity V and electric field E is that the speed of the electron is determined by the strength of the field and the distance that it has travelled.

- #1

jawad khan

- 5

- 0

i only know about E=F/e(relation of electric field E and Force F)

any one like to answer?

Physics news on Phys.org

- #2

berkeman

Mentor

- 68,262

- 21,906

jawad khan said:

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?

- #3

sophiecentaur

Science Advisor

Gold Member

- 29,494

- 7,137

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.

- #4

arunma

- 927

- 4

You know that the velocity is,

[tex]v^2 = v_0^2 +2a(x-x_0)[/tex]

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

[tex]v^2 = 2ax[/tex]

You further know that,

[tex]ma = qE[/tex]

So,

[tex]a = \dfrac{q}{m}E[/tex]

Now just substitute this into the kinematic equation of motion,

[tex]v^2 = \dfrac{2q}{m}Ex[/tex]

And now you can easily solve for the velocity,

[tex]v = \sqrt{\dfrac{2q}{m}Ex}[/tex]

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.

- #5

Born2bwire

Science Advisor

Gold Member

- 1,780

- 24

arunma said:

You know that the velocity is,

[tex]v^2 = v_0^2 +2a(x-x_0)[/tex]

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

[tex]v^2 = 2ax[/tex]

You further know that,

[tex]ma = qE[/tex]

So,

[tex]a = \dfrac{q}{m}E[/tex]

Now just substitute this into the kinematic equation of motion,

[tex]v^2 = \dfrac{2q}{m}Ex[/tex]

And now you can easily solve for the velocity,

[tex]v = \sqrt{\dfrac{2q}{m}Ex}[/tex]

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.

- #6

sophiecentaur

Science Advisor

Gold Member

- 29,494

- 7,137

It's a one-liner.

- #7

jawad khan

- 5

- 0

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?

- #8

arunma

- 927

- 4

Born2bwire said: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.

jawad khan said:

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

- #9

sophiecentaur

Science Advisor

Gold Member

- 29,494

- 7,137

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

- #10

berkeman

Mentor

- 68,262

- 21,906

sophiecentaur said: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.

- #11

arunma

- 927

- 4

sophiecentaur said: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.

- #12

sophiecentaur

Science Advisor

Gold Member

- 29,494

- 7,137

I sometimes like to take the pretty route, too.

The velocity of an electron moving between charged plates depends on the magnitude of the electric field between the plates, as well as the charge and mass of the electron. It can be calculated using the equation v = E/m, where v is the velocity, E is the electric field, and m is the mass of the electron.

The distance between the plates does not have a direct effect on the velocity of the electron. However, it does affect the strength of the electric field between the plates, which in turn affects the velocity of the electron.

The velocity of the electron is directly proportional to the charge on the plates. This means that as the charge on the plates increases, the velocity of the electron also increases.

No, according to the theory of relativity, the speed of light is the maximum velocity at which any object can travel. Therefore, the velocity of an electron moving between charged plates cannot exceed the speed of light.

The velocity of the electron can be measured using various techniques such as the Millikan oil-drop experiment or the Thomson's cathode ray tube experiment. These experiments involve measuring the acceleration of the electron and using the equations of motion to calculate its velocity.

- Replies
- 1

- Views
- 507

- Replies
- 6

- Views
- 1K

- Replies
- 14

- Views
- 2K

- Replies
- 1

- Views
- 2K

- Replies
- 2

- Views
- 2K

- Replies
- 2

- Views
- 880

- Replies
- 8

- Views
- 1K

- Replies
- 21

- Views
- 2K

- Replies
- 11

- Views
- 2K

- Replies
- 2

- Views
- 2K

Share: