Electrostatics
Electromagnetics
Electricity & Magnetism
© The scientific sentence. 2010

Electrostatics:
Effect of the field on a charged particle.
1. Free charged particle in a electric field
We are considering the simplest situation where the
electric field is uniform .
In an electric field E, if the only significant
force on a charged particle q is the electric force
F, then the Coulomb's law and Newton second law together
give the expression of the net force:
F = q E = m a or a = q E/m
a is the acceleration of the charged particle of
mass m.
1. A charged particle released in a uniform electric
field E will be accelerated a long the electric field.
At the origin, we place the release point and orient
the field parallel to the xaxis.
The kinematics of the particle released from rest
(v_{o} = 0), like a freefall particle, is:
a_{x} = q E/m
x = (1/2)a_{x}t^{2} = (1/2)(qE/m)t^{2}
v_{x} = a_{x}t = (qE/m)t
a_{x} = q(E/m)
x = (1/2)a_{x}t^{2} = (1/2)(qE/m)t^{2}
2. Projectile charged particle in a electric field
Now if the positive charged particle enters the field E oriented
along the yaxis, perpendicularly
with an initial velocity v_{o} at t = 0, like
a projectile, its kinematics equations are:
a_{x} = 0
a_{y} = qE/m
x = v_{o}t
v_{x} = v_{o}
v_{y} = a_{y}t = (qE/m)t
x = v_{o}t
y = (1/2) ayt^{2} = (1/2) (qE/m)t^{2} =
(1/2) (qE/m)(x/v_{o})^{2} =
(1/2) (qE/m)(1/v_{o})^{2}. x^{2}
That is the particle follows the parabolic path.
v_{x} = v_{o}
v_{y} = a_{y}t = (qE/m)t
x = v_{o}t
y = (1/2) a_{y}t^{2} = (1/2) (qE/m)t^{2} =
(1/2) (qE/m)(x/v_{o})^{2} = (1/2) (qE/m)(1/v_{o})^{2}. x^{2}

