Considering the Coulomb's law, if the distance between the charges is doubled, the new electrical force between the charges will be [tex]\frac{1}{4}[/tex] of the initial value.
Between two or more charges, a force called an electric force appears. That is, charged bodies suffer a force of attraction or repulsion when approaching.
The electric force with which two point charges are attracted or repelled at rest is directly proportional to their product, inversely proportional to the square of the distance that separates them and acts in the direction of the line that joins them.
That is, the modulus of the electric force depends on the value of the charges and the distance that separates them, while its sign depends on the sign of each charge. Charges of the same sign repel each other, while charges of a different sign attract.
The previously expressed is Coulomb's law and it is expressed mathematically as:
[tex]F=K\frac{q1q2}{d^{2} }[/tex]
where:
In this case, an electrical force of 8.0×10⁻⁵ N exists between two point charges, q1 and q2. The force is expressed as:
[tex]F1=K\frac{q1q2}{d1^{2} }[/tex]
If the distance between the charges is doubled, q1 and q2 don't change, and considering that the force is inversely proportional to the square of the separation between the particles, then you have: d2=2×d1. So:
[tex]F2=K\frac{q1q2}{d2^{2} }[/tex]
[tex]F2=K\frac{q1q2}{(2d1)^{2} }[/tex]
[tex]F2=K\frac{q1q2}{4d1^{2} }[/tex]
[tex]F2=\frac{1}{4}F1[/tex]
In summary, if the distance between the charges is doubled, the new electrical force between the charges will be [tex]\frac{1}{4}[/tex] of the initial value.
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