physics---electricity1

(i) The vertical forces are all balanced, hence
kQq/h^2 + k(2q)Q/h^2 + R = W
where k is the electrostatic constant (= 9 x10^9 N.m^2/C^2)
R is the normal reaction at the bearing
i.e. R = W - 3kqQ/h^2 -------------- (1)

Taking moment about the left-hand end of the rod,
[k(2q)Q/h^2].L + R(L/2) = Wx
i.e. W.x = 2kqQL/h^2 + (L/2).(W - 3kqQ/h^2)
W.x = kqQL/2h^2 + WL/2
x = kqQL/2Wh^2 + L/2 = (L/2).(kqQ/Wh^2 + 1)

(ii) From (1), when R = 0, W = 3kqQ/h^2
h = square-root[3kqQ/W]