Question

(19) Three 4μC point charges are placed along a straight line as shown in Fig. 19.18.
Calculate the net force on each charge.

q1 is at (0,0) q2 is 3m from the origing and q3 is 6m from the origin.

Answer 1

Given three point charges are

   q = 4*10^-6 C =q1=q2=q3

so that they are at   

   q1---3m------q2--------3m----q3

   <--------------6m-------------------->

we knwo that from Coulonm's law F = kq1*q2/r^2

here the force on q1 due to q2,q3 is

       F1 = k(q1*q2/r12^2+q1*q3/r13^2)

       F1 = kq^2(1/r12^2+1/r13^2)

       F1 =9*10^9*(4*10^-6)^2)(1/3^2+1/6^2) N = 20.0*10^-3 N = 20 mN

force on q2 due to other charges is F2

   F2 = k(q1*q2/r12^2+q2*q3/r23^2)

       F1 = kq^2(1/r12^2+1/r23^2)

   F1 =9*10^9*(4*10^-6)^2)(1/3^2+1/31^2) N = 32.0*10^-3 N = 32 mN

and force on q3 due to q1,q2 is same magnitude as that of F1, because the charge is same and are at same distance
so F3 = F1

   F3= k(q3*q2/r32^2+q1*q3/r13^2)

       F3 = kq^2(1/r32^2+1/r13^2)

       F3 =9*10^9*(4*10^-6)^2)(1/3^2+1/6^2) N = 20.0*10^-3 N = 20 mN