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An electron is in motion at 4.0 × 10^6 m/s horizontally when it enters a region of space between two parallel plates, as shown, starting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 x 10^2 N/C and separation between the charged plates is 2.0 cm.


Determine the horizontal distance travelled by the electron when it hits the plate.

Respuesta :

Histrionicus

Answer:

Given that

speed u=4*10^6 m/s

electric field E=4*10^3 N/c

distance b/w the plates d=2 cm

basing on the concept of the electrostatices

now we find the acceleration b/w the plates  to find the horizontal distance traveled by the electron when it hits the plate.

acceleration a=qE/m=[tex]1.6*10^{-19}*4*10^3/9.1*10^{-31} =0.7*10^{15}[/tex]=[tex]7*10^{14}[/tex] m/s

now we find the horizontal distance traveled by electrons hit the plates

horizontal distance

[tex]X=u[2y/a]^{1/2}[/tex]

=[tex]4*10^6[2*2*10^{-2}/7*10^{14}]^{1/2}[/tex]

=[tex]3*10^{-2}[/tex]= 3 cm

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