Question

St su

Answer 1

Now if 1 consider first -q image charge then for (1) -q image charge force on q = q^2/4 pi t_0 4 (d/2)^2 along x and (2) image charge force on q is = q^2/4 pi t middot 4 (d/2)^2 (-x) So they cancel Now for positive image charge (3) force on q q^2/4 pi t_0 4 (3d/2)^2 they repeat Similarly for (4) q^2/4 pi t_0 middot 1/4 (3 d/2)^2 they also repeat so net force zero this process goes on for other image charge so force on q identicalto F = 1/4 pi t_0 q^2/4 {[1/(d/2)^2 + 1/(3 d/2)^2 + 1/(5 d/2)^2 + ellipsis] - [1/(d/2)^2 + 1/(3 d/2)^2 + 1/(5 d/2)^2 + ellipsis]} = 0