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Electric Field
Example 3.4
In the above example suppose that a negative charged particle is
projected horizontally into the uniform field with an initial velocity v o
as shown in figure 3.9.
L
Y
V o
Figure 3.9
E
(0,0)
(X,Y)
X
V
Solution
Since the direction of electric field E r in the y direction, and the charge is
negative, then the acceleration of charge is in the direction of -y.
a = -qE/m
The motion of the charge is in two dimension with constant acceleration,
with v xo = v o & v yo = 0
The components of velocity after time t are given by
v x = v o =constant
v y = at = - (qE/m) t
The coordinate of the charge after time t are given by
x = v o t
y = ½ at 2 = - 1/2 (qE/m) t 2
Eliminating t we get
qE
x
2
y = (3.8)
2
2mv0
we see that y is proportional to x 2 . Hence, the trajectory is parabola.
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