Download Motion of charges in uniform fields

Motion of charges in uniform fields.
[V/m] = [N/C]
q1
charged sphere or
point charge
parallel charged plates
E
K q1/r2
V
Kq1/r
ΔV/d
F = q2 E
ΔV x/d
P.E. = q2 V
d = do + vot + ½ at2
e = 1.6 x 10-19 C
me = 9.11 x 10-31 kg
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[1] Two plates are separated by 10 cm. and have a potential difference of
200 V between them, positive plate above
the negative. An electron enters the field
5 cm above the bottom plate and moving at
10 m/s horizontally (parallel to the plates).
Write expressions for the position of the
electron in the field. You may neglect
gravity here.
[2] A 10-6g oil drop with charge -6e is stationary in the electric field
between 2 plates separated by 1cm. Determine the potential difference ΔV
between the plates needed to keep the drop stationary.
[3] An electron is accelerated from rest at the
negative plate through the electric field and
through a hole in the positive plate. If the plates
have a potential difference of 10 Volts, calculate
the kinetic energy of the electron as it leaves the
hole in the positive plate.
Solutions:
[1] Force on electron is upward, the electric force F = qE . The electron has
horizontal velocity but no horizontal force. This is exactly like our old
trajectory problems with an electric field in place of the gravity field.
For x(t), we start at xo=0 with vx = 10 m/s and ax = 0 (no force)
x(t) = 10 [m/s] t + 0
For y(t), we need to determine the acceleration.
Fy = q E upward = e ΔV/d
= (1.6E-19 C) (200 V)/(0.1m)
= 3.2E-16 VC/m
= 3.2E-16 N
a = F/m = (3.2E-16 N)/(9.1E-31 kg) = 3.52E14 m/s2
Initial values are yo = 5 cm = .05m and vy0 = o
so, y(t) = 0.05m + ½ 3.52E14 t2
[2] Two forces act on the oil drop; electric and gravity. They are equal and
opposite and must balance to keep the drop stationary.
So, q E = 6e ΔV/d = mg
Solving for ΔV gives ΔV = mg d /6e
ΔV = (1E-9 kg) (9.8 m/s2)(0.1m)/(6 x 1.6E-19C)
= 9.8E-10 J / 1E-18 C = 9.8E8 V
[3] This is most easily solved using conservation of energy. The electron
begins at rest and is accelerated by the field to give kinetic energy and a
lower potential energy.
So,
PEi = PEf + KE or ΔPE = KE
ΔPE = q ΔV = (1.6E-19 C)(10 V) = 1.6E-18 J = KE.
Another energy unit is commonly used in electric force problems, the
electron volt or eV. In this problem, one electron is accelerated through 10
Volts, giving 10 eV.
This means that one eV = 1.6E-19 J.