Point charges and force of repulsion

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The discussion centers on calculating the force of repulsion between two point charges when their separation is reduced to 0.280 times the original distance. The relevant equation is Coulomb's law, F = (k * q1 * q2) / r^2, which indicates that the force scales inversely with the square of the distance (r). As the distance decreases, the force increases significantly; specifically, if r is reduced to 0.280, the force becomes approximately 12.8 times the original force. The mention of a negative sign is clarified as unnecessary in this context, as the force is inherently a repulsive force. Understanding the inverse square relationship is crucial for solving such problems.
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Two point charges are separated by a distance r and repel each other with a force F. If their separation is reduced to 0.280 times the original value, what is the magnitude of the force of repulsion between them?


I don't really understand this question and don't know where to start.
Would I start with the equation
E = ((k)(Q)/r^2

what follows next?
 
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What is the equation for the repulsive force between two point charges, and how does it scale with r?
 
JeffKoch said:
What is the equation for the repulsive force between two point charges, and how does it scale with r?

It would be Coulomb's law of

F = ((k)(q1)(q2))/(r^2)

So it would be -0.280 times the original r value?

The final answer is 12.8 F but I still don't understand.
 
Where do you get the minus sign?

O.K., you know that F scales as 1/r^2. When r gets smaller, F must get larger. When r is half it's original value, how much larger does F get?
 

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