Question

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 is shot in from one edge of the capacitor. It enters the capacitor very close (call it 0 mm) from the top plate and travelling parallel to the plates. The electrons in the beam are moving at 1.50 × 107 m/s as they enter the capacitor (this may seem extremely fast, but it is easy to make electrons go this fast). They exit the other side of the capacitor just missing the edge of the bottom plate.

(a) Which plate is the positive one? Explain.

(b) From the information given determine the acceleration of the electrons

(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

10-3 d 5.0 mm m - 5.0x10-3 m 1 mm 10 m 2. A (10.0 cm (10.0 cm) = 100 cm2 0.01 m 1 cm2 (a) As the electron beam are attracted towards downward, the lower plate must be positive plate which attracts the electrons that have negatively charged particles and the upper must be negative charged plate plate Therefore, the lower plate is the positive one. (b) The electron beam is shot in horizontally at top of the upper plate and hence, the electron beam has only horizontal component of the velocity, that is, u 1.5x10 m/s The vertical component of the velocity of the electron beam is Use the following equation of kinetic equation to calculate the time taken by the electron beam to cross the lower plate. Rearrange the above equation fort and substitute the values 102 10 cm 1 cm --6.67 x10 s t=_ = (1.5x10 m/s Use the following kinematic equation 1 yutat 2 =(0)t+at Rearrange the above equation for a 2y f 10-2 2 10 cm m 1 cm (6.67x109 s S = 4.49x1015 m/s Therefore, the acceleration of the electrons is |4.49x1015 m/s2

(c) The force on the electron is, F qE ma qE Rearrange the above equation for E ma E (9.1x10-31 kg )(4.49 x1015 m/s2 (1.6x10"19 C) =25536.87 N/C Therefore, the electric field inside of the capacitor is 25536.87 N'C (d) The equation for the potential difference between the plates of the capacitor is V = Ed The equation for the capacitance (1) on the capacitor is C d (2) The equation for the charge on the each of the capacitor plate is, О3 ст Substitute the equations (1) and (2) in above equation. Ed) d =(8.85x1012 F/m (0.01 m2)(25536.87 N/C) 2.26x10 1nC 109 C =2.26 nC on each of the capacitor plate is|2.26 nC| Therefore, the charge