Electron motion 1
Question:
A particle of charge q and mass M with an initial velocity V
o enters an electric
field E at right angles to the field.
We assume E is uniform , i.e , its value is constant at all points in the region
between plates of length L, except for small variations near the edges of the plates which we
shall neglect.
Question. If the particle in the problem is an electron of kinetic energy
10
-17 J (kinetic energy = ½ mv
2) if the electric field strength is 0.06 volt/cm
and if L = 3.5 cm find,
The vector velocity as it leaves the region between the
plates
The angle for the particle as it leaves the plates
The point of intersection
between the x-axis and the direction of the particle as it leaves
the
field
Answer:
Energy = ½ mv
2 = 10
-17
Joules
Taking:
the charge on the electron to be 1.6x10
-19 C and its
mass to be 9x10
-31 kg.
Velocity in the x direction (V
o) = 4.71x10
6 ms
-
1.
Therefore time in between the plates = s/v = 3.5x10
-2/4.71x10
6 =
7.43x10
-9 s
Acceleration in the y direction at right angles top the plates is
gained by:
F = ma = Electric field (E) x electron charge (e) and so a =
Ee/m
Using the figures above gives a = 8.9x10
11 ms
-2Then using v
= u + at with u = 0 in the y direction
v in the y direction = 6x10
3 ms
-1
Therefore velocity on emerging from the plates is found by adding the squares of
the velocities in the x and y directions and then taking the square root of the result. This
gives the final velocity as just over 4.71x10
6 ms
-1 in a direction of 0.073 degrees to
the axis of the plates.
The electron will emerge 2.46x10
-5 m above the line along
which it was originally moving.
The intercept of the motion on the x-axis is 0.0193 m or
1.93 cm from the end where the electron emerges.
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