The discussion revolves around the forces between two charges and how the force changes when the distance between them is altered. The original poster presents a scenario where two charges attract each other with a force of 1.5 N and questions how this force will change if the distance is reduced to one-ninth of its original value, referencing the equation for electrostatic force.
The discussion is active, with participants offering various perspectives on the problem setup and the mathematical relationships involved. Some guidance has been provided regarding the nature of the force's dependence on distance, although there is no explicit consensus on the correct approach yet.
Participants are grappling with the implications of the inverse square law in the context of electrostatic forces, and there are references to similar problems that may provide additional context. There is also mention of confusion regarding the mathematical representation of distance in the force equation.
Set it up so that 1.5 N is the force when the distance = R.aChordate said:I don't think I am setting this up right, but would I multiply 1.5N by 1/9 and by the constant k?
Doc Al said:Another way to approach this is to just ask yourself what happens to the force if the distance doubles? If the distance is cut in half? That should point you in the right direction.
Careful. The force is inversely proportional to the distance squared. So if the distance doubles, the force decreases by a factor of 4. (Multiplied by (1/2)2 = 1/4.)aChordate said:If the distance doubles, the force increases x4 (exponentially?).
No, the force would increase by a factor of 4.If the distance is halved, the force decreases.
Realize that ##(\frac{1}{R})^2 = \frac{1}{R^2}##cathal84 said:Could you explain why the squared sign is outside the last bracket. Should it not be inside the bracket as it is the distance squared.
##(\frac{1}{R/2})^2 = (\frac{2}{R})^2 = 4(\frac{1}{R})^2##cathal84 said:Also, how did you simplify it to be 4*(const), how did you calculate the 4?