Electrostatics
Electromagnetics
Electricity & Magnetism
© The scientific sentence. 2010

Electrostatics: Coulomb's law
1. Coulomb's law

The electric force on the two stationary charges, qa and qb,
distant by r, is given by the Coulomb's law:
The constant ε_{o} is called the permittivity of free space
or vaccum.
The forces in the figure are oriented according to
Newton's third law. Here the charges are of like sign.

2. Example: Three points charges
We want to calculate the Coulomb forces at the point c of charge  q_{c}.
The charge on the point a is positive and the charge on
the point c is negative, so the related forces are attractive.
Fac = Fca = k qa qc/n^{2}.
Fac has no components on the yaxis.
The charge on the point b is negative and the charge
on the point c is negative, so the related forces
are repulsive.
Fbc = Fcb = k qb qc/(n^{2} + m^{2})
Fbc has two components: one on the yaxis, and the
other on the zaxis.
The components of Fbc:
Fbcy = Fbc cos(π  θ) =  Fbc cos θ
Fbcz = + Fbc sin θ
Finally, the components of the
net force F on c are:
Fy =  Fbc cos θ
Fz =  Fac + Fbc sin θ
cos θ = m/[m^{2} + n^{2}]^{1/2}
sin θ = n/[m^{2} + n^{2}]^{1/2}
F = [Fy ^{2} + Fz^{2}]^{1/2}
tg (β) = Fz/Fy
qa = + 1 µC
qb =  2 µC
qc =  3 µC
m = 4.0 m
n = 3.0 m
k = 9 x 10^{9} N.m^{2} /C^{2}
Therefore
Fbc = k qb qc/(n^{2} + m^{2})
= 9 x 10^{9} (2 x 10^{6}) (3 x 10^{6}) /25
= 2.16 x 10^{3} N
Fac = k qa qc/n^{2}
= 9 x 10^{9} (1 x 10^{6}) (3 x 10^{6}) /9 =
3.0 x 10^{3} N
cos θ = 3/5
sin θ = 4/5
Fy =  2.16 x 10^{3} (3/5) =  1.30 x 10^{3} N
Fz =  Fac + Fbc sin θ =
 3.0 x 10^{3}N + 2.16 x 10^{3} (4/5) N =
 3.0 x 10^{3}N + 1.72 N =
 1.18 x 10^{3} N
F = [Fy ^{2} + Fz^{2}]^{1/2} =
[1.69 + 1.4]^{1/2} x 10^{3}
= 1. 75 x 10^{3} N
tg (β) = Fz/Fy =  1.18 /( 1.30) = + 0.91
β = 42^{o}.

