A positive point charge q of mass m is released from rest in a uniform electric field E directed along
the x-axis as in the figure.
(A) What is the acceleration of the positive charge q? Is it in uniform acceleration or not?
(B) What will be the final velocity of the charge after having a displacement x?
(C) What will be the kinetic energy of the charge after it has moved a distance x?
Answers
Answered by
3
Answer:
initial velocity u=0,
acceleration a=qE/m,
time t=t,
final velocity v=?
by v=u+at, we get
v=0+(
m
qE
)t
orv=
m
qEt
so the final kinetic energy is
K(t)=
2
1
m(
m
qEt
)
2
⇒K(t)=
2m
q
2
E
2
t
2
Answered by
3
Concept:
- Electric field and charges
- One-dimensional motion
Given:
- Charge = q
- mass of charge = m
- Electric field = E
Find:
- Acceleration of charge q
- Final velocity of charge after displacement x, v
- Kinetic energy of charge after moving distance x, KE
Solution:
a) F = qE
F= ma
a =F/m
a = (qE/m) m/s^2
This is uniform acceleration
b) v^2 = u^2 +2as
v^2 = 0 +2qE/m (x)
v = √(2qxE/m) m/s
c) Kinetic energy KE = 1/2mv^2
KE = 1/2m (2qxE/m)
KE = qxE J
The uniform acceleration is (qE/m) m/s^2. The final velocity is √(2qxE/m) m/s. The kinetic energy is (qxE) J.
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