**Examples
for Coulomb's law**

Example #1

Problem:

How many coulomb's
of charge is contained in the electrons of one *kg* of
hydrogen?

Solution:

One kg of hydrogen is 1000 grams, and has therefore 1000 times Avagadro's number of electrons.

Q = 1000* *N*_{A}
* *e*

where *e* = 1.602E-19 (*C*)
, and *N*_{A} = Avogadro's number =
6.02E23

*Q*
= -4.82E7 (*C*)

Example #2

Problem:

Consider one gram of hydrogen gas which has been separated into its electrons and protons. The electrons are placed on Earth's north pole and the protons are placed on Earth's south pole. What is the magnitude of the force the electrons feel due to the presence of the protons?

Solution:

Remember that one gram of H gas has one mole (Avagadro's number) of hydrogen atoms.

The force between the two is given by Coulomb's law:

where *k* = 8.99E9
(*Nm*^{2}*/C*^{2})
, *r* = 2*6.37E6 (*m*)
, *q*_{a}* = q*_{b}*
= N*_{A}**e , e *= 1.602E-19 (*C*)
, and *N*_{A} = Avogadro's number =
6.02E23

*F*
= 5.14E5 (*N*)

Example #3

Problem:

What is the
elecric force between 2 *u*-quarks separated by 1.0E-16
meters? This is a typical separation inside a proton.

Given: The charge of an up quark is *(2/3)e*.

Solution:

The force between the two is given by Coulomb's law:

where *r*
= 1.0E-16 (*m*) , *q*_{a}*
= q*_{b}* = *(2/3)**e*

*F*
= 1.03E4 (*N*)

Example #4

Problem:

Two charges feel a
repulsive force of 96 (*N*). What is the force if the
separation, *r*, is quadrupled.

Solution:

Since Coulomb's law scales as *r*^{-2},
and *r* becomes larger by a factor of 4, the new force
should be (1/4) squared, or one sixteenth of the old force.

*F* = 96/16 (*N*)

*F*
= 6.0 (*N*)

Example #5

Problem A:

Imagine 3 charges,
separated in an equilateral triangle as shown above, with *L*
= 2.0 *cm*, *q* = 1.0 *nC*. What is the
magnitude and direction of the force felt by the upper charge?

Solution:

The Force due to the lower left charge has a magnitude:

where the factor of 2
comes from the fact that the upper charge is 2*q*. The
direction of this contribution is 30 degrees from vertical. After
adding in the contribution from the lower right charge, the net
force felt by a will be in the vertical direction, and downwards
since it is attractive. Thus multiply the magnitude above by
2cos(30^{o}), where the factor of 2 is from two charges
and the cos(30^{o}) projects out the vertical component.
(Remember cos(30^{o}) is sqrt(3/2)

=
7.79E-5 (*N*), down

Problem B:

What is the net electric potential energy of the configuration?

Solution:

The net Potential energy is the
sum of three contributions. For 3 charges *a,b,c* it is
the sum of the *ab* part, the *ac* part and the *bc*
part. The contribution from the lower two charges is *kq*^{2}*/L*,
while the contribution of the pairs which include the upper
charge is -2 time this, thus the sum is:

=1.35E-6
(*J*)

Problem C:

How much energy does it cost to remove the upper charge?

Solution:

This is just the sum of the two terms used above that include the upper charge.

=-1.80E-6
(*J*)