Suppose that two point charges, each with a charge of 1 Coulomb are separated by a distance of 1 meter. Determine the magnitude of the electrical force of repulsion between them.
Answers
Answer:
COULOMB’S LAW
F
=
k
∣
q
1
q
2
∣
r
2
F=k∣q1q2∣r2
Coulomb’s law calculates the magnitude of the force F between two point charges, q1 and q2, separated by a distance r. In SI units, the constant k is equal to
k
=
8.988
×
10
9
N
⋅
m
2
C
2
≈
8.99
×
10
9
N
⋅
m
2
C
2
k=8.988×109N⋅m2C2≈8.99×109N⋅m2C2
The electrostatic force is a vector quantity and is expressed in units of newtons. The force is understood to be along the line joining the two charges. (See Figure 2).
Although the formula for Coulomb’s law is simple, it was no mean task to prove it. The experiments Coulomb did, with the primitive equipment then available, were difficult. Modern experiments have verified Coulomb’s law to great precision. For example, it has been shown that the force is inversely proportional to distance between two objects squared
(
F
∝
1
r
2
)
(F∝1r2) to an accuracy of 1 part in 1016. No exceptions have ever been found, even at the small distances within the atom.