Two parallel, charged plates, both of length L= 15 m create an electric field be

Question

Two parallel, charged plates, both of length L= 15 m create an electric field between them as shown here. An electron is shot into a region where an electric field exists, as shown. The electron is shot in with a speed of v= 191200 m/s, at an angle of theta = 60 degree, right at the left edge of the bottom plate. The electric field has a strength of E = 43 N/C. Assuming the electron doesn't hit the upper plate, where will the electron strike the lower plate again? The electron will hit at a distance is equal to____

Explanation / Answer

Solution:

The mass of the electron = m = 9.1 x10^-31 kg

Charge on the electron = q = 1.6 x 10^-19 C

E= F/q

=> electric force = F = qE = (1.6 x 10^-19) ( 43) = 68.8 x 10^-19 = 6.88 x 10^-18 N

Acceleration = a = F/m = ( 6.88 x 10^-18)/(9.11 x 10^-31) = 7.55 x 10^12 m/s^2

Time taken to reach the bottom plate is calculated using the kinematic equation:

y = vo y - 1/2aT^2

=> T = 2 voy /a

here voy = initial Y commponent of velocity = vo sin theta = (191200) sin 60 = 165584 m/s

T = 2 (165584) / (7.55 x 10^12) = 4.386 x 10^-8 s = 43.9 nano seconds

The distance where the electron hits the bottom plate = vox * T

= horizontal component of velocity x time = (191200 cos 60)(4.38 x 10^-8) = 0.0042 m = 4.2 mm

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