Question

Three 140 nC charged objects are equally spaced on a straight line. The separation of each object from its neighbor is 0.30 m.

A) Find the force exerted on the center object if all charges are positive. B) Find the force exerted on the center object if all charges are negative. C) Find the force exerted on the center object if the rightmost charge is negative and the other two are positive.

Answer 1

let the three charges be q1. q2, q3 and its given that q1=q2=q3

so now we have

force due to q1 and q2 , from coulomb's law

F12=q1q2/r²

F12=8.99*10⁹* (140*10^-9)²/0.3²=1.96*10^-3N

and F23=1.96*10^-3N

now part a) if all the charges are positive that it means all of them will repel each other

F=F12-F23=1.96*10^-3-1.96*10^-3=0 N

so the answer is 0 N

part B) if all of them are negative , then again every charge will repel each other

F=F12-F23= 0 N

so the answer is 0N

part C) if q1 q2 are positive which are the left ones then F12 will be repulsive and if q3 is negative( the rightmost)

then F23 will be attractive

so we have,

F=F12+F23=(1.96+1.96)*10^-3N=3.92*10^-3N

so the answer is 3.92*10^-3N or 3.9 *10^-3N or 4.0 *10^-3N