An electron with an initial speed of 7x10 6 m /s isprojected along the axis midw

Question

An electron with an initial speed of 7x106m/s isprojected along the axis midway between the deection plates of acathode-ray tube. The uniform electric field between the plates hasa magnitude of 1000 V/m and is upward. F =1.6x10-16N and acceleration a =1.76x1014m/s2 (a) How far below the axis hasthe electron moved when it reaches the end of theplates?
(b)At whatangle with the axis is it moving as it leaves theplates? (c) How far below the axis willit strike the fluorescent screen S? (c) How far below the axis willit strike the fluorescent screen S?

Explanation / Answer

An electron with an initial speed u = 7x106 m / s
Theuniform electric field between the plates E =1000 V/m
Upward force F = 1.6x10-16N
acceleration a = 1.76x1014m/s^ 2 Horizontal direction: -------------------------- distance S= 6 cm=0.06 m time taken to reach end of theplate t = S / u                                                                  = 8.57*10^-9 s In vertical direction: -------------------------- Initial velocity U =0 Accleration a = 1.76 * 10^14 m / s^ 2 time t= 8.57 * 10 ^ -9s distance moved at the end of theplates in vertical direction y = ? from the relation y = Ut + ( 1/2) at ^ 2                                 = 0 + 6.465* 10 ^ -3                                = 6.465 * 10 ^ -3 m (b). required angle = tan -1 ( y / S )                                     = 6.15 degrees (c). tan = Y / ( S + S ') where   S ' = 12 cm=0.12 m from this Y = ( S+ S ' )tan                     = 0.01939 m

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