Question

Two very small spherical metal objects, each with 1 coulomb of charge, are brought together in a vacuum so that the separation distance between their centers is 1 mm. what is the force of repulsion between the two object? (coulomb's constant is k=9.0 × 109 n.m2/c2.)

Answer 1

Answer: 9*10^15 N

Force=kqq/r^2

F=[(9*10^9)(1)(1)]/.001^2=9.0*10^15

Answer 2

Answer:

The force of repulsion between the two object is 9*10¹⁵ N

Explanation:

Coulomb's law indicates that charged bodies suffer a force of attraction or repulsion when approaching. The value of said force is proportional to the product of the value of its loads and inversely proportional to the square of the distance that separates them. This is expressed mathematically by the expression:

$F=K*\frac{Q*q}{r^{2} }$

where:

• F is the electrical force of attraction or repulsion. In the International System it is measured in Newtons (N).
• Q and q are the values ​​of the two point charges. In the International System they are measured in Culombios (C).
• r is the value of the distance that separates them. In the International System it is measured in meters (m).
• K is a constant of proportionality called constant of Coulomb's law.   It depends on the medium in which the loads are located. For vacuum K is approximately 9*10⁹ $\frac{N*m^{2} }{C^{2} }$ in the International System.

From this law it is possible to predict the electrostatic force of attraction or repulsion between two particles according to their electrical charge and the distance between them.

The force will be of attraction if the charges are of opposite sign and of repulsion if they are of the same sign.

In this case:

• Q=q= 1 C
• r=1 mm= 0.001 m

Then:

$F=9*10^{9} \frac{N*m^{2} }{C^{2} } *\frac{1 C*1C}{(0.001m)^{2} }$

So:

F=9*10¹⁵ N

The force of repulsion between the two object is 9*10¹⁵ N