Answer: If the distance between a pair of electric charges is cut in a half, the electrostatic force changes in size by a factor of four.
Explanation:
The electrostatic force between the charges can be calculated with the Coulomb's law, that in its scalar form is given by:
[tex]F = K\frac{Q1 *Q2}{d^{2} }[/tex]
Where [tex]K[/tex] is the Coulomb's constant; [tex]Q1[/tex] and [tex]Q2[/tex] are the electric charges; and [tex]d[/tex] is the distance between the charges. From the equation above, you can establish that the force and the distance between the charges have are inversely proportional.
If the distance between the charges is cut in half, then you have:
[tex]F = K\frac{Q1 *Q2}{(\frac{d}{2})^{2} }[/tex]
Organizing and doing small calculations:
[tex]F = K\frac{Q1 *Q2}{(\frac{d^{2} }{2^{2} })}[/tex]
[tex]F = K\frac{Q1 *Q2}{(\frac{d^{2} }{4 })}[/tex]
[tex]F = 4 * K\frac{Q1 *Q2}{(\frac{d^{2} }{4 })}[/tex]
So, you can see that the resultant force changes in size by a factor of four.
