The Physics
Hypertextbook
Opus in profectus

# Coulomb's Law

## Practice

### practice problem 1

Compare the magnitude of the electrostatic and gravitational forces between…
1. an electron and proton in a hydrogen atom (the radius of the electron's orbit is about 0.053 nm)
2. two protons in a helium nucleus (the separation between them is about 1.2 fm)
3. the Earth and the moon (the separation between them can be found in many references)
Compile your results in a table like the one below.
electro­static force (FE) gravita­tional force (Fg) order of magnitude comparison (FE/Fg)
e and p+ in a
hydrogen atom
p+ and p+ in a
helium nucleus
Earth and moon
in space

#### solution

This problem involves repeated application of Coulomb's law of electric forces and Newton's law of universal gravitation.

1. Coulomb's law for the hydrogen atom…

FE =
 kq1q2 r2
FE =
 (9.0 × 109 N m2/C2)(1.60 × 10−19 C)2 (0.053 × 10−9 m)2
FE = 8.2 × 10−8 N

Newton's law for the hydrogen atom…

Fg =
 Gm1m2 r2
Fg =
 (6.67 × 10−11 N m2/kg2)(1.67 × 10−27 kg)(9.11 × 10−31 kg) (0.053 × 10−9 m)2
Fg = 3.6 × 10−47 N

The electric force may seem like a small number, but keep in mind that the electron doesn't have much mass. The electric force is sufficient to keep the electron in "orbit" around the proton in the hydrogen atom. At 39 orders of magnitude smaller, the gravitational force might as well be zero. Gravity does not do anything to keep a hydrogen atom together.

Compare…

FE  =
 8.2 × 10−8 N 3.6 × 10−47 N
Fg
FE  ≈  1039 or 39 orders of magnitude
Fg

Gravity is a weak force on the atomic scale.

2. Coulomb's law for the helium nucleus…

FE =
 kq1q2 r2
FE =
 (9.0 × 109 N m2/C2)(1.60 × 10−19 C)2 (1.2 × 10−15 m)2
FE = 160 N

Newton's law for the helium nucleus…

Fg =
 Gm1m2 r2
Fg =
 (6.67 × 10−11 N m2/kg2)(1.67 × 10−27 kg)2 (1.2 × 10−15 m)2
Fg = 1.3 × 10−34 N

The electric force is astonishingly large. 160 newtons is something like the weight of a one year old child — a one year old child pushing on an proton! How does the nucleus stay together? Being 36 orders of magnitude weaker than electricity, gravity isn't doing much to help. What keeps the nucleus from blowing apart? The answer appears elsewhere in this book.

Compare…

FE  =
 160 N 1.3 × 10−34 N
Fg
FE  ≈  1036 or 36 orders of magnitude
Fg

Gravity is a weak force on the nuclear scale.

3. Coulomb's law for the Earth-moon system is a joke. Neither body is charged.

FE =
 kq1q2 r2
FE =
 (9.0 × 109 N m2/C2)(0 C)2 (3.84 × 108 m)2
FE = 0 N

Newton's law for the Earth-moon system is not a joke. Both objects are massive (and reasonably near to one another).

Fg =
 Gm1m2 r2
Fg =
 (6.67 × 10−11 N m2/kg2)(5.97 × 1024 kg)(7.35 × 1022 kg) (3.84 × 108 m)2
Fg = 2.0 × 1020 N

This astronomical number is big enough to explain this astronomical phenomenon. Newton verified the law of universal gravitation by comparing the acceleration of the moon in its orbit to the acceleration of an apple falling from a tree (a statement that is metaphorically true, not literally true). In essence, the equation was tested with these numbers (or their 17th century English equivalents).

Compare, if you dare…

 FE = 0 N Fg 2.0 × 1020 N FE = 0 =  n/a Fg

Zero is not a power of ten so no order of magnitude comparison can even be made. Electricity is absolutely irrelevant to this problem. Gravity holds the moon in its orbit — end of discussion.

Compile your results in a table like the one below.

electro­static force (FE) gravita­tional force (Fg) order of magnitude comparison (FE/Fg)
e and p+ in a
hydrogen atom
8.2 × 10−8 N 3.6 × 10−47 N 1039
p+ and p+ in a
helium nucleus
160 N 1.3 × 10−34 N 1036
Earth and moon
in space
0 N 2.0 × 1020 N n/a

### practice problem 2

If the moon were held in its orbit by an electrostatic rather than gravitational force…
1. What quantity of charge would be needed?
2. How many elementary charges is this?

If all of the charge came from the separation of hydrogen atoms into electrons and protons…

1. What mass of hydrogen would be required?
2. How many liters of H2 gas at STP would be required?

### practice problem 3

Given three charges in a standard coordinate system, calculate the magnitude and direction of the net electrostatic force on each.
1. +10 μC at (+0 m, +0 m)
2. −20 μC at (+0 m, +3 m)
3. +25 μC at (−4 m, +0 m)