Question

7%) Problem 3: separated by x 5 Two electrons in a vacuum exert force ofF-4.44E-08 N on each other. They are then moved so that they are 5.2 times their original distance. 50% Part (a) What is the force, in newtons, that the electrons experience at the new separation distance? sinO cos0tan7 8 9 cotan0 asinOacos0 atan0 acotan sinh0 cosh0 tanhcotanh0 4 5 6 Degrees Radians vol ACKSPACE I give up Submit Hint Hints:0 deduction per hint. Hints remaining Feedback:--deduction per feedback. 50% Part (b) How far apart, in meters, were the electrons originally?

Answer 1

$Suppose\,\,q\,\,represents\,\,charge\,\,of\,\,each\,\,electron\,\,and\,\,r\,\,represents\,\,distance\,\,between\,\,two\,\,electrons$
$Now\,\,by\,\,Coulobm's \,\,law\,\,force\,\,F=\frac{kq^2}{r^2}\\\\ According \,\,to\,\, given \,\,condition\,\,\\ .\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4.44\times10^{-8} \,\,N=\frac{kq^2}{r^2}\,\,\,\,\,\,\,\,\,\,......(1)\\\\ When\,\,they\,\,are\,\,separated\,\,by\,\,x=5.2 r\\ .\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,F_n=\frac{kq^2}{(5.2r)^2}\,\,\,\,\,\,\,\,\,\,......(2)\\$

$Dividing\,\,eq.(2)\,\,by\,\,eq.(1)\\\\ .\,\,\,\,\,\,\,\, \frac{F_n}{4.44\times10^{-8}\,\,N}=\frac{\,\,\frac{kq^2}{(5.2r)^2\,\,\,}}{\frac{kq^2}{r^2}}\\\\ \Rightarrow F_n=\frac{1}{(5.2)^2}\times4.44\times10^{-8}\,\,N\\\\ \Rightarrow F_n=1.64\times10^{-9}\,\,N$

$.\\\\\\\ [For\,\,any\,\,doubt\,\,please\,\,leave\,\,a\,\,comment]$