Physics I - A Review: Distance Time Mass
Physics I - A Review: Distance Time Mass
Physics I - A Review: Distance Time Mass
7 cm
+3 Coul. -5 Coul
Electric Force - example
From Coulomb’s Law, we know that there is
an electric force between any two charges:
F = kq1q2/r122 , with the direction determined
by the signs of the charges.
F = (9x109 Nt-m2/C2) * (3 C) * (5 C) / (.07 m)2 =
2.76 x 1013 Nt. Note that we ignore the sign on
any charge when calculating the magnitude.
Since the charges are opposite, the force is
attractive!
7 cm
+3 Coul. -5 Coul
Electric Force - example
F = 2.76 x 1013 Nt.
Note that this force is huge: over 27 trillion
Newtons which is equivalent to the weight
of about 6 billion tons! What this indicates
is that it is extremely hard to separate
coulombs of charges. Most of the time, we
can only separate picoCoulombs or
nanoCoulombs of charge.
Electric Field - example
The Electric Field can be found two different ways.
1. Since we know the electric force and the charge
at the field point, we can use: F = qE, or
Eat 1 = F/q1 = 2.76 x 1013 Nt / 3 C = 9.18 x 1012 Nt/C.
Since the charge at the field point is positive, the
force and field point in the same direction.
2. Since we are dealing with the field due to a point
charge (the -5 C charge), we can use:
Eat 1 = kq/r2 = (9x109Nt-m2/C2) * (5 C) / (.07m)2 =
9.18 x 1012 Nt/C; since the charge causing the field
is negative, the field points towards the charge.-5 Coul
+3 Coul. 7 cm
Another Force Example
Suppose that we have an electron orbiting a
proton such that the radius of the electron in
its circular orbit is 1 x 10-10m (this is one of
the excited states of hydrogen). How fast
will the electron be going in its orbit?
qproton = +e = 1.6 x 10-19 Coul
qelectron = -e = -1.6 x 10-19 Coul v
r = 1 x 10-10 m, p
r e
melectron = 9.1 x 10-31kg
Force Example