Electron in a parallel plate capacitor - finding plate's magnitude

In summary, the conversation discusses the motion of an electron through a parallel plate capacitor. The initial speed of the electron, the length and separation of the plates, and the assumption of a uniform electric field are given. The focus is on finding the magnitude of the electric field by using the second law of dynamics and kinematics to determine the acceleration and force acting on the electron.
  • #1
vaxopy
26
0
The drawing shows an electron entering the lower left side of a parallel plate capacitor and exiting at the upper right side. The initial speed of the electron is 7.00×106 m/s. The capacitor is 2.00 cm long, and its plates are separated by 0.150 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.

See the picture attached..

I don't really know how to start this.. (at all).

would i try to find the centripital force of the electron, and let that equal to the plates strength?
 

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  • #2
Write the second law of dynamics.There's no centripetal force involved.

Daniel.
 
  • #3
but how do you find its acceleration?
 
  • #4
By identifying which forces act on the particle...??And then applying the second law of dynamics for the moving particle..??

Daniel.
 
  • #5
michaelw said:
but how do you find its acceleration?
Using kinematics. There is a uniform acceleration in the vertical direction; constant speed in the horizontal. Find the acceleration, then the force, then the field.
 

1. What is an electron in a parallel plate capacitor?

An electron in a parallel plate capacitor is an electrically charged particle that is placed between two parallel plates with opposite charges. It experiences a force due to the electric field created by the plates.

2. How do you find the magnitude of the plates in a parallel plate capacitor?

The magnitude of the plates in a parallel plate capacitor can be found by using the formula Q = CV, where Q is the charge on the plates, C is the capacitance, and V is the voltage. By rearranging the formula, we can find the magnitude of the plates (Q/C) by dividing the charge by the capacitance.

3. What factors affect the magnitude of the plates in a parallel plate capacitor?

The magnitude of the plates in a parallel plate capacitor is affected by the distance between the plates, the surface area of the plates, and the dielectric material between the plates. A larger distance between the plates and a larger surface area of the plates will result in a larger magnitude of the plates.

4. How does the magnitude of the plates in a parallel plate capacitor affect the electric field?

The magnitude of the plates in a parallel plate capacitor is directly proportional to the electric field between the plates. This means that a larger magnitude of the plates will result in a stronger electric field and vice versa.

5. Can the magnitude of the plates in a parallel plate capacitor ever be zero?

No, the magnitude of the plates in a parallel plate capacitor can never be zero. This is because there will always be some amount of charge and capacitance present, even if they are very small. Therefore, the magnitude of the plates will always be a non-zero value.

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