Question

1) Water is a polar molecule. There is an effective accumulation of negative charge near the oxygen atom and an effective lack of negative charge ( positive charge) near each of the hydrogen atoms. The molecule is triangular in shape, as diagramed below, with the angle near the oxygen vertex equal to 104.5°. d is the distance from each of the hydrogen atoms to the oxygen atom +Q 5225 52.25 -20 +Q Derive a formula for the electric field produced by the molecule for points along the positive x axis beyond the hydrogen atoms, in terms of Q, d and x. The oxygen atom is at the origin and has charge -2Q. Each hydrogen atom has charge Q

Answer 1

Water is a polar molecule. There is an effective accumulation of negative charge near the oxygen. Electric filed at x E_2 - 2 Q + e_2 Q + e_0 vertical components of field = E_1 sin theta - E_2 and sin Q = 0[ Vertical-bar E_1 Vertical-bar = Vertical-bar E_2 Vertical-bar ] Let field at x = [- k (2 theta)/Psi^2 + k(theta)/l^2 cos theta + k Q/l^2 cos 0] I \r\n where cos 52.25 degree = x^2 + d^2 - l^2 /2 x d = 0.6122 rightarrow l^2 = x^2 + d^2 - 2 x d (0.6122) l^2 = x^2 + d^2 - 1.22 x d ra E = 2 k Q [cos theta/l^2 - 1/x^2] I = 2 k Q [1/l^2 [x^2 + l^2 - d^2/2 times l] - 1/x^2] I = 2 k Q [1/2 - [x^2 - d^x + x^2 - 1.22 x d + d^x/2 x l] - 1/x^2] i