HELP Parallel plate Capacitors

[NotaPhysicsMan]

Here's the question.

The drawing (attached) shows an electron entering the lower left side of a parallel plate capacitor and exiting at the upper right side. The inital speed of the electron is 7.00x 10^6m/s. The capacitor is 2.00cm long, and its plates are separated by 0.150cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.

Here's what I got.

Seems simple enough. I use E=q/(A Eo), where (permittivity of space) Eo=8.85x10^-12C^2/N.

Ok so I plug in the charge of an electron divided by the area of the rectangle and Eo. So 1.60x10^-19/((2.00cm/100 x 0.150cm/100)(8.85x10^-12)). I get something like 6.03x10^-8 N/C.

See, I didn't use the velocity they gave me, that's where I think the problem is, but I don't see where I could use it! Maybe to find acceleration using kinematics, nope not that...

Any ideas?

Thanks

 

[Doc Al]

Seems simple enough. I use E=q/(A Eo), where (permittivity of space) Eo=8.85x10^-12C^2/N.


Ok so I plug in the charge of an electron divided by the area of the rectangle and Eo. So 1.60x10^-19/((2.00cm/100 x 0.150cm/100)(8.85x10^-12)). I get something like 6.03x10^-8 N/C.

This formula allows you to calculate the field between the plates given the charge on the capacitor and the area of the plates. But that won't help you here, since you have neither the charge nor the area! (1) The charge on the plates does not equal the charge of the electron! (2) You used the area of a sideways cross-section, not the area of the capacitor plates.

See, I didn't use the velocity they gave me, that's where I think the problem is, but I don't see where I could use it! Maybe to find acceleration using kinematics, nope not that...

That's exactly what you want to do: find the acceleration, then the force, then the field.

 

[NotaPhysicsMan]

That's odd, then what are the length's and width of the cross-section used for? In terms of acceleration, I don't have time, or distance, but the initial velocity, so I can't use kinematics.
So wait, let's see. Only the horizontal distance is important. And so the d traveled in that velocity is 2.00cm.

so t= 0.02m/7.00x10^6m/s = 2.857x 10^-9s(wow that's small)

Ok, so I have time, I will plug into x=Vt+1/2at^2.

solve for a= 2(x-volt)/t^2.
a= 2^-6/(2.857x10^-9s)
a=700m/s^2.

OK sounds good so far.

Now I use F=ma to get force.

9.11x10^-31(700m/s^2)=6.377x10^-28 N

and Now I use F=qE, solve for E=F/q.

6.377x10^-28N/(1.60x10^-19)

E=1.02x10^-46 N/C... Huh, this doesn't make sense.

 

[NotaPhysicsMan]

What the, now I get E=F/q and E=3.985x10^-9 N/C...AHHHH

 

[NotaPhysicsMan]

Any ideas? Please...

 

[NotaPhysicsMan]

No one eh..

 

[Doc Al]

Only the horizontal distance is important. And so the d traveled in that velocity is 2.00cm.

so t= 0.02m/7.00x10^6m/s = 2.857x 10^-9s(wow that's small)

Right!

Ok, so I have time, I will plug into x=Vt+1/2at^2.

solve for a= 2(x-volt)/t^2.
a= 2^-6/(2.857x10^-9s)
a=700m/s^2.

Redo this calculation: $a = 2 d / t^2$. (d = 0.0015m)

 

[NotaPhysicsMan]

Ok,

a=2(0.0015m)/(2.857x10^-9)^2
a=3.675x10^14


F=ma
=9.11x10^-31kg x 3.675x10^14
=3.345x10^-16 N.

E=F/q
=3.345x10^-16N/1.60x10^-19C
E=2093 N/C or 2.09x10^3 N/C.

NOW that looks like a more sensible answer!

Thanks.