Question

WILL GIVE BRAINLIEST! In the nucleus of an atom, two protons are separated by a distance of 1*10^-15m. What is the magnitude of the electric force between them?
A.) 115 N

B.) 720 N

C.) 142 N

D.) 230 N

Answer 1

Answer:

hi!

Explanation:

answer D.) 230 N

Answer 2

Answer:

The magnitude of the electric force between them is 230 N (option D)

Explanation:

The electric force is one that is generated between electric charges, and is described by Coulomb's law.

Coulomb's law indicates that charged bodies suffer a force of attraction or repulsion when approaching. The value of this force is proportional to the product of the value of its loads. Charges of the same sign repel each other, while charges of different signs attract. This law also indicates that the force is inversely proportional to the square of the distance that separates them.

Then:

$F=k*\frac{q1*q2}{r^{2} }$

where:

• k: a constant of proportionality, whose value is 9*10⁹ $\frac{N*m^{2} }{C^{2} }$

• q1, q2:  represent the amount of charge, in units of coulombs [C]
• r: separation distance between charges in meters [m]

In this case:

• q1=q2= 1.6*10⁻¹⁹ C, which is the value of the charge of each proton.
• r=1*10⁻¹⁵ m.

Then

$F= 9*10^{9} \frac{N*m^{2} }{C^{2} } *\frac{1.6*10^{-19} C*1.6*10^{-19}C }{(1*10^{-15} )^{2} m^{2} }$

F=230.4 N≅230 N

The magnitude of the electric force between them is 230 N (option D)