Asked by Ann
An electron is projected into a uniform electric field E = (1000 N/C)i with an initial velocity (2.00* 10^6m/s)i in the direction of the field. How far the electron travel before it is brought momentarily to rest?
Answers
Answered by
Ann
the field is horizontal
Answered by
Damon
retarding force = E e = m a where m is electron mass and e is electron charge so
a = Ee/m
constant Force --- > constant de acceleration, just like gravity
if you throw something up with speed Vi
F = -m g = - m a
v = Vi - g t, when v = 0, t = Vi/g
x = Vi t - (1/2) g t^2
x = Vi^2/g - (1/2) g Vi^2/g^2 = (1/2) Vi^2/g
here instead of g we have for acceleration E e/m
so
x = (1/2) Vi^2 [ m / Ee ]
a = Ee/m
constant Force --- > constant de acceleration, just like gravity
if you throw something up with speed Vi
F = -m g = - m a
v = Vi - g t, when v = 0, t = Vi/g
x = Vi t - (1/2) g t^2
x = Vi^2/g - (1/2) g Vi^2/g^2 = (1/2) Vi^2/g
here instead of g we have for acceleration E e/m
so
x = (1/2) Vi^2 [ m / Ee ]
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