Calculating Electric Field Strength in a Charged Parallel-Plate Capacitor

[agnieszka2807]

Electric field
Hi, maybe there I will find somebody much clever than me, so please help me if You can:


An electron begins to move from the surface of the negative plate in a charged parallel-plate
capacitor. What is the strength of the electric field E which exists between the plates if
electron reaches the positively-charged plate after time t=2 ms? Take the distance between
the plates as L=0.2 m, the electron's charge is q=1.6·10-31 C and its mass is m=9.1·10-31 kg.

I need it very fast... Thanx a lot :*

 

[ideasrule]

What are the relevant equations, and what have you tried already?

If you don't know how to start, find the acceleration of the electron.

 

[agnieszka2807]

I know all equations for condensator and electric field, but I don't know which of them are needed here. for example this acceleration? to find it I can use equation: a= 2L/t^2

 

[AEM]

The acceleration of the electron is caused by the electric field in the capacitor. What is the equation giving the force on an electron in an electric field? What is the relationship of the electric field to the potential difference between the two parallel plates? Answering these questions will get you started.

 

[agnieszka2807]

Equation for force of electron in electric field is: F=k x q^2/L^2
and the relation between electric field and potential is: V=V1 + V2 = k x q/r^2 + k x q/r^2

and I stil don't know what should I make next... should I use somevhere equation F= m x a? because I have here mass also

 

[AEM]

Equation for force of electron in electric field is: F=k x q^2/L^2
and the relation between electric field and potential is: V=V1 + V2 = k x q/r^2 + k x q/r^2

and I stil don't know what should I make next... should I use somevhere equation F= m x a? because I have here mass also

Okay, you've confused the force between two point charges with the force on a charge in an electric field. The force on a charge in an electric field is

$$F = qE$$

You can combine this with $F = ma$

That gives you $qE = ma$.

This last equation allows you to solve for the acceleration a in terms of the electron's mass and charge and the unknown electric field. Use that acceleration in the familiar equation

$$x = x_0 + v_0 t + \frac{1}{2} a t^2$$

Take your coordinate system's origin where the electron starts, assume the electron starts from rest, and you should be able to compute the magnitude of the electric field. The last equation works because the acceleration is constant. The equation $F = qE$ is used because to a reasonable approximation the electric field in the parallel plate capacitor is uniform and directed from the positive plate to the negative plate. I was thinking of another problem when I asked you what is the relationship between the electric field and the potential. You didn't need that here.

 

[RoyalCat]

Just on a side note, so you get the correct numerical answer at the end, the charge of an electron is $$1.6\times 10^{-19} C$$

 

[agnieszka2807]

I don't know what You mean, but I had made it like this:
L= at^2/2 => a= 2L/ t^2
F = ma => F = m x 2L/t^2
E = F/q => E = m x 2L/q x t^2 - and this is my fianal answer, I'm not sure if this is ok

 

[AEM]

I don't know what You mean, but I had made it like this:
L= at^2/2 => a= 2L/ t^2
F = ma => F = m x 2L/t^2
E = F/q => E = m x 2L/q x t^2 - and this is my fianal answer, I'm not sure if this is ok

Yes, $E = \frac{2 m L}{ q t^2}$.

Now substitute in the numerical values of the time, L, m, q, and you'll have your answer.

 

[agnieszka2807]

thank You very much for Your help :)...

 

[AEM]

thank You very much for Your help :)...

You're welcome!

 

[player1_1_1]

electric field i kondensator?:D