Two charged dust particles exert a force of $4.2 \times 10^{-2} \textrm{ N}$ on each other. What will be the force if they are moved so they are only one-eighth as far apart?

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The force between these two dust particles initially we call that force one is Coulombs constant times the charge in one dust particle times the charge in the other dust particle divided by the distance between them squared. And we're told that's 4.2 times ten to the minus two Newtons and then the second case it's going to be the same charges same dust particles and same Coulombs constant of course and divide that by different distance r two squared. And we're told that r two is one eighth of r one . So we substitute that in for r two here and this becomes K Q one Q two over r one squared over 64 and then Multiply top and bottom by 64 and the 64 is cancel on the bottom leaving us with 64 times K Q one Q two over r one squared. And this business here K Q one Q two over r one squared is force one so forced two then is 64 times force one. So the force when they moved so the only one eighth the distance apart will be 64 times 4.2 times ten to the minus two Newtons which is 2.7 Newtons.