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

AI Thread Summary
To find the electric field magnitude in a parallel plate capacitor, start by analyzing the motion of the electron as it enters and exits the plates. The electron has an initial speed of 7.00×10^6 m/s and travels through a 2.00 cm long capacitor with a plate separation of 0.150 cm. Since the electric field is uniform, use kinematics to determine the vertical acceleration, which results from the electric force acting on the electron. Apply the second law of dynamics to relate the forces and calculate the electric field strength. This approach leads to the necessary calculations for determining the electric field between the plates.
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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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Write the second law of dynamics.There's no centripetal force involved.

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

Daniel.
 
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.
 
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