Please show steps, and the equations. :) Please label the final answers. ORION i
ID: 1572302 • Letter: P
Question
Please show steps, and the equations. :)Please label the final answers. ORION ignment Chapter 21, Problem o30 Z Your answer is partially correct. Try again. In the fig particles 1 and 2 are fixed in place on an x axis, at a separation of L-7.7o am. Their charges are 01 az -27e, partide 3 with darge eo ese ie to be placed on the between particles 1 and 2, so that they te and that produce a net electrostatic force F3net on e. (a) At what coordinate should partide 3 be placed to minimize the magntude of that foree (b) wht minimum magnitude? (a) Number 1,835 (b) Nut he web and Windows D LL
Explanation / Answer
let particle 1 is at x=0 and particle 2 is at x=0.077 m
let particle 3 be at x=d meters
then force on particle 3 due to particle 1 :
as both charges are positive, force is repulsive in nature.
so force will be towards +ve x axis.
magnitude=k*q1*q3/d^2
=k*e*5*e/d^2
=5*k*e^2/d^2
where k is coloumb's constant =9*10^9
force on particle 3 due to particle 2 :
as charges are of different sign, force is attractive in nature and directed along +ve x axis
force magnitude=k*27*e*5*e/(l-d)^2=135*k*e^2/(l-d)^2
so total force is along +ve x axis.
force magnitude=k*e^2*((5/d^2)+(135/(l-d)^2))
force to be minimum, derivative of total force magnitude w.r.t. d should be 0.
hence (-10/d^3)+(270/(l-d)^3)=0
==>10/d^3=270/(l-d)^3
==>27*d^3=(l-d)^3
taking cube root on both sides:
3*d=l-d
==>d=l/4=0.077/4=0.01925 m=1.925 cm
force magnitude=k*e^2*((5/d^2)+(135/(l-d)^2))
=9*10^9*(1.6*10^(-19))^2*((5/0.01925)^2+(135/(0.077-0.01925)^2))=2.487*10^(-23) N
final answers are :
distance=1.925 cm
force magntiude=2.487*10^(-23) N