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Electric Charges And Fields

Question
CBSEENPH12037178

Suppose that the particle in Question 1.33 is an electron projected with velocity vx = 2.0 x 106 m s–1. If E between the plates separated by 0.5 cm is 9.1 x 102 N/C, where will the electron strike the upper plate? (|e| = 1.6 x 10–19 C, me = 9.1 x 10–33 kg.)

Solution

Given,
An electron is projected with velocity vx= 2.0×106  m/s
Distance between the plates = 0.5 cm
Electric field between plates -E = 9.1 × 
Acceleration, 

                         a=qEm= 1.6 × 10-19×9.1×1029.1×10-31 = 1.6×1014 m/s2

Using formula     y=ut+12at2,

We get,     0.005 = 0+12×1.6×1014×t2

Simplifying for value of t, we get

                      t = 8×10-9s

The electron covers a vertical distance which is given by 
                      y = vxt    = 2.0 × 106 × 8 × 10-9    = 1.6 × 10-2m    = 1.6 cm