## Physics A: Problem Set 2: Electrosatic Force

### recommended reading

Barron's Let's Review: | 8.6 Coulomb's Law |

physics.info: | Coulomb's Law |

Wikipedia: | Coulomb's law, Coulomb |

HyperPhysics: | Coulomb's law |

Khan Academy: | Electrostatics (part 1) |

### practice

- Given three charges in a standard coordinate system…
- +10 μC at (+0 m, +0 m)
- −40 μC at (+0 m, +9 m)
- +20 μC at (−3 m, +0 m)

- Sketch the arrangement of the charges.
- Calculate the magnitude of the force of charge 2 on charge 1.
- In what direction does the force you just calculated point?
- Calculate the magnitude of the force of charge 3 on charge 1.
- In what direction does the force you just calculated point?
- What is the magnitude of the resultant of the two forces you just calculated?
- What is the direction of the resultant of the two forces you just calculated relative to the +
*x*axis?

Sketch it. I sketched it with the information computed as well as the information given.

Use Coulomb's law.

*F*_{21}=*kq*_{1}*q*_{2}*r*^{2}*F*_{21}=(9.0 × 10 ^{9}Nm^{2}/C^{2})(10 × 10^{−6}C)(40 × 10^{−6}C)(9 m) ^{2}Opposite charges attract. Charge 2 pulls charge 1 in the +

*y*direction (toward the top of the screen).Use Coulomb's law again.

*F*_{31}=*kq*_{1}*q*_{2}*r*^{2}*F*_{31}=(9.0 × 10 ^{9}Nm^{2}/C^{2})(10 × 10^{−6}C)(20 × 10^{−6}C)(3 m) ^{2}Like charges repel. Charge 3 pushes charge 1 in the the +

*x*direction (to the right on the screen).Use pythagorean theorem to compute the magnitude of the resultant.

*F*=√( *F*_{x}^{2}+*F*_{y}^{2})*F*=√[(0.200 N) ^{2}+ (0.0444 N)^{2}]Pick a trig function to compute the angle above the +

*x*axis. I recommend tangent, but sine and cosine also work. There are slight variations in the answers computed below due to rounding issues.tan θ = *F*_{y}*F*_{x}tan θ = 0.0444 N 0.200 N sin θ = *F*_{y}*F*sin θ = 0.0444 N 0.205 N cos θ = *F*_{x}*F*cos θ = 0.200 N 0.205 N

### homework

- Compare the magnitude of the electrostatic and gravitational forces between…
- an electron and proton in a hydrogen atom (the radius of the electron's orbit is about 0.053 nm)
- two protons in a helium nucleus (the separation between them is about 1.2 fm)
- the earth and the moon (the separation between them can be found in many references)

electrostatic

force (*F*)_{E}gravitational

force (*F*)_{g}order of magnitude

comparison (*F*/_{E}*F*)_{g}e ^{−}and p^{+}in

hydrogen atomp ^{+}and p^{+}in

helium nucleusearth and moon

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

Coulomb's law for the hydrogen atom…

*F*=_{E}*kq*_{1}*q*_{2}*r*^{2}*F*=_{E}(9.0 × 10 ^{9}Nm^{2}/C^{2})(1.60 × 10^{−19}C)^{2}(0.053 × 10 ^{−9}m)^{2}*F*= 8.2 × 10_{E}^{−8}NNewton's law for the hydrogen atom…

*F*=_{g}*Gm*_{1}*m*_{2}*r*^{2}*F*=_{g}(6.67 × 10 ^{−11}Nm^{2}/kg^{2})(1.67 × 10 ^{−27}kg)(9.11 × 10^{−31}kg)(0.053 × 10 ^{−9}m)^{2}*F*= 3.6 × 10_{g}^{−47}NThe 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…

*F*_{E}= 8.2 × 10 ^{−8}N3.6 × 10 ^{−47}N*F*_{g}*F*_{E}≈ 10 ^{39}or 39 orders of magnitude*F*_{g}Gravity is a weak force on the atomic scale.

Coulomb's law for the helium nucleus…

*F*=_{E}*kq*_{1}*q*_{2}*r*^{2}*F*=_{E}(9.0 × 10 ^{9}Nm^{2}/C^{2})(1.60 × 10^{−19}C)^{2}(1.2 × 10 ^{−15}m)^{2}*F*= 160 N_{E}Newton's law for the helium nucleus…

*F*=_{g}*Gm*_{1}*m*_{2}*r*^{2}*F*=_{g}(6.67 × 10 ^{−11}Nm^{2}/kg^{2})(1.67 × 10 ^{−27}kg)^{2}(1.2 × 10 ^{−15}m)^{2}*F*= 1.3 × 10_{g}^{−34}NThe electric force is astonishingly large. 102 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…

*F*_{E}= 102 N 1.3 × 10 ^{−34}N*F*_{g}*F*_{E}≈ 10 ^{36}or 36 orders of magnitude*F*_{g}Gravity is a weak force on the nuclear scale.

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

*F*=_{E}*kq*_{1}*q*_{2}*r*^{2}*F*=_{E}(9.0 × 10 ^{9}Nm^{2}/C^{2})(0 C)^{2}(3.84 × 10 ^{8}m)^{2}*F*= 0 N_{E}Newton's law for the Earth-moon system is not a joke. Both objects are massive (and reasonably near to one another).

*F*=_{g}*Gm*_{1}*m*_{2}*r*^{2}*F*=_{g}(6.67 × 10 ^{−11}Nm^{2}/kg^{2})(5.97 × 10 ^{24}kg)(7.35 × 10^{22}kg)(3.84 × 10 ^{8}m)^{2}*F*= 2.0 × 10_{g}^{20}NThis 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 Seventeenth Century English equivalents).

Compare, if you dare…

*F*_{E}= 0 N *F*_{g}2.0 × 10 ^{20}N*F*_{E}= 0 = n/a *F*_{g}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 (*F*)_{E}gravitational

force (*F*)_{g}order of magnitude

comparison (*F*/_{E}*F*)_{g}e ^{−}and p^{+}in

hydrogen atom8.2 × 10 ^{−8}N3.6 × 10 ^{−47}N10 ^{39}p ^{+}and p^{+}in

helium nucleus160 N 1.3 × 10 ^{−34}N10 ^{36}earth and moon

in space0 N 2.0 × 10 ^{20}Nn/a - If electric forces are so much more powerful than gravitational forces, why do we feel the Earth's gravitational force and not its electric force?
Electrostatics forces can be both attractive and repulsive. The usual state of matter around us is a neutral mix of positive and negative charges. This means that the electrostatic force between objects on the earth and the earth itself is the sum of many, many many small attractions and repulsions. Since neither sign of charge dominates on the earth, or on most objects near the earth, the sum of all these forces is zero. Many small things add up to nothing when they work against one another.

Gravity, on the other hand, is only attractive. The force between objects on the earth and the earth itself is the sum of many, many, many small attractive forces. Although these forces are weak, they add up to something large enough to be significant when the entire mass of the earth is considered. Many small things add up to something when they work together.