Electrons are initially traveling at 2.0 x 10^6 m/s in the horizontal direction. They enter a region between two charged plates of length 2 cm and experience an acceleration of 4 x 10^14 m/s? vertically upward. Find :
(a) the vertical position as they leave the region between the plates;
(b) the angle at which they emerge from between the plates.
Answers
Answer:
An electron is a negatively charged particle, so it will be attracted by the positive plate with force F=eE. Hence, acceleration of electron along y-axis will be
a=
m
F
=
m
eE
=
md
eV
{asE=
d
V
} (i)
So, from equation of motion, v
2
=u
2
+2as along the x-axis,
v
x
=v
0
=10
6
(ms
−1
){asa
x
=0} (ii)
And along the y-axis, v
y
2
=2ay
0
∣{asu=0ands=y
0
}
Now, as y
0
=1cm (given) and is given by equation (i), the electron will hit the top plate with the velocity
v
y
=
md
2y
0
eV
=
9×10
−31
×3×10
−3
2×1.6×10
−19
×20×1×10
−3
=(4
2
/3)×10
6
=1.885×10
6
ms
−1
(iii)
So, the electron will hit the upper plate with the velocity
v
x
=10
6
(ms
−1
) and v
y
=1.885×10
6
ms
−1
Note
In this problem, time taken by the electron to hit the plate
t=
a
2y
0
=
eE
2y
0
m
=
eV
2y
0
md
=
1.6×10
−19
×30
2×(2×10
−3
)×(9×10
−31
)×(3×10
−3
)
=
2
2
3
×10
−9
s=1.06×10
−6
s
And in this time, electron will travel a horizontal distance
s=v
0
t=10
6
×1.06×10
−9
=1.06×10
−3
m