CK-12-Physics - Intermediate

15.2. Coulomb’s Law http://www.ck12.org

Answer:

F=G

meme

r^2

=

(

6. 67 × 10 −^11

N·m^2

kg^2

)

( 9. 11 × 10 −^31 kg)^2

( 1. 00 × 10 −^2 m)^2

= 5. 54 × 10 −^67 N

c. What is the ratio of the electrostatic force to the gravitational force for the electron?

Fe

Fg

=

2. 30 × 10 −^24 N

5. 54 × 10 −^67 N

= 4. 15 × 1042

This is an enormous ratio! All things being equal, the electrical force is some 40 orders of magnitude greater than
the gravitational force!

Illustrative Example 2

A− 5. 0 μCpoint charge is placed between a positive point charge of+ 10. 0 μCand a negative point charge of
− 6. 0 μC, along a horizontal line, as shown inFigure15.14. What is the net force on the− 5. 0 μCpoint charge?

The prefixμstands for “micro” and represents one-millionth, thereforeμ= 10 −^6.

FIGURE 15.14

Answer:

Let us first draw a Free-Body-Diagram (FBD) for the− 5. 0 μCcharge.

Here,F 1 represents the force of the+ 10. 0 μCpoint charge on the− 5. 0 μCpoint charge. It is an attractive force
toward the left.F 2 represents the force of the− 6. 0 μCpoint charge on the− 5. 0 μCpoint charge. It is a repulsive
force toward the left. The net force acting on the− 5. 0 μCcharge is the vector sum of the two forces. It is best
not to include the sign of the charges in the calculation, since the resulting sign indicates only whether the force is