Electric force between point charges—two examples
We saw that Coulomb’s Law said that
qq
FE = (9 x 109 N m2 /C2 ) 1 2 2 .
R
Let’s consider two force examples. We will take for the first example two pith balls, each
carrying charge 0.05 C and having mass 25 g, while separated by 0.50 m. In the first case,
we have
FE = (9 x 109 N m2 /C2 )
[0.05 C][0.05 C]
[0.50 m]2
= (9 x 109 N m2 /C2 ) x [0.0025 C2]/0.50 m2 = 9.0 x 107 N = 90 MN.
This is clearly a very large force, and the amount of charge 0.05 C must be a very large
amount. In many cases, microcoulombs or nanocoulombs of charge are involved. In other
words, a coulomb is a lot of charge—it represents the charge on 6.2 x 1018 protons (6.2
billion billion!).
Using Newton’s Second Law, F = ma, we may find the acceleration experienced by either
pith ball as
a = F/m = 9.0 x 107 N/25 g = 9.0 x 107 N/2.5 x 10-2 kg = 3.6 x 109 N/kg
= 3.6 x 109 m/s2.
In the second example, we will find the magnitude of the force exerted between the proton
and the electron in the hydrogen atom. The force between these tiny charges, each one of
size 1.6 x 10-19 C (but opposite in sign, which we ignore here) separated by a distance of
5.29 x 10-11 m is
FE = (9 x 109 N m2 /C2 )
[1.6 x 10-19 C][1.6 x 10-19 C]
[5.29 x 10-11 m]2
= (9 x 109 N m2 /C2 ) x [2.56 x 10-38 C2]/[5.29 x 10-11 m]2
= 8.23 x 10-8 N = 82.3 nN.
Energy, Ch. 4, extension 1 Electric force between two point charges
Clearly, the pith balls have a much greater-sized force acting between them than the
proton and electron in the hydrogen atom.
What about the acceleration? The electron mass is 9.11 x 10-31 kg, while the proton’s
mass is 1.67 x 10-27 kg, almost 2000 times greater. The electron’s acceleration (assuming
the much more massive proton to be roughly stationary) is given by Newton’s Second
Law as
a = F/m = 8.23 x 10-8 N/9.11 x 10-31 kg = 9.04 x 1022 N/kg = 9.04 x 1022 m/s2.
This is much, much greater than the acceleration of the pith ball.
In E03.3, we found that we could write a = v 2/R. A typical atomic electron, then, with R
= 5.29 x 10-11 m will have a speed of
———————————————
——
v = √ aR = √(9.04 x 1022 m/s2) x (5.29 x 10-11 m) = 2.2 x 106 m/s.
Atomic motions may occur at high speed, but the atomic forces are very small compared
to those we deal with every day of our lives.
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