Question

2. A parallel plate capacitor is formed with two plates separated by 5.00 mm as shown in Figure 1.Each plate is a 10.0 cm X 10.0 cm square. We do not know the charge on the plates. An electron beam isshot in from one edge of the capacitor. It enters the capacitor very close (call it 0 mm) from the top plateand travelling parallel to the plates. The electrons in the beam are moving at 1.50x 107 m/s as they enterthe capacitor (this may seem extremely fast, but it is easy to make electrons go this fast). They exit theother side of the capacitor just missing the edge of the bottomplate. (a) Which plate is the positive one?Explain. (b) From the information given determine the acceleration of theelectrons (c) Find the strength of the electric field inside the capacitor. (d) Determine how much charge is on each plate of the capacitor

Answer 1

Concept :- we first find the time taken by the electron to cross the plate. Then we use that to find the acceleration using kinematics equation, Then we use Newton's law of motion and force on charged particle in an electric field to find E as follows, ***************************************************************************************************
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