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An electron is shot at an initial speed of V0 m/s, at an angle theta from the x-axis. It moves through a uniform electric field E j^ . A screen for detecting electrons is positioned parallel to the y-axis , ata a distance x'. What is the velocity of the electron when it hits the screen.?
hi, x component of velocity = V0 cos θ accleration in x direction is zer0 x component of velocity after tsec , v1x = V0 cos θ , & time taken to reach the sc reen ,t = x'/ V0 cos θ vertical component of velocity = V0 sin θ & accl. = -eE /m vertical velocity at time t , v1y = V0 sin θ - eE /m * x'/ V0 cos θ = V0 sin θ - eE x'/ mV0 cos θ velocity after time t ,v1 = √ v1x^2 + v1y ^2 = √ { V0 cos θ)^2 + ( V0 sin θ - eE x'/ mV0 cos θ )^2 }
hi,
x component of velocity = V0 cos θ
accleration in x direction is zer0
x component of velocity after tsec , v1x = V0 cos θ ,
& time taken to reach the sc reen ,t = x'/ V0 cos θ
vertical component of velocity = V0 sin θ & accl. = -eE /m
vertical velocity at time t , v1y = V0 sin θ - eE /m * x'/ V0 cos θ
= V0 sin θ - eE x'/ mV0 cos θ
velocity after time t ,v1 = √ v1x^2 + v1y ^2
= √ { V0 cos θ)^2 + ( V0 sin θ - eE x'/ mV0 cos θ )^2 }
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