A water molecule is shown. The angle between the OH-bonds is theta = 106 degree

Question

A water molecule is shown. The angle between the OH-bonds is theta = 106 degree and the OH-distance is d = 9.42 times 10^-11 m. The charges on the oxygen and hydrogen are fractions of the elementary charge ["partial charge"]" q_o =-0.67 e and q_H = 0.335 e, respectively. An electron is placed at the point P = (x = 0, y = - 1.1 times 10^-10 m). a) Find the force on the electron due to the oxygen atom. b) Find the force on the electron due to the two hydrogen atoms. c) Find the net force on the electron due to the water molecule! Is the electron attracted or repelled by the water molecule?

Explanation / Answer

given

H2O molecule

the OH bond angle is theta = 106 degrees

OH distance d = 9.42*10^-11m

charges

qo = -0.67*e = -0.67*(1.6*10^-19) C = -1.072*10^-19C

qH = 0.335*(1.6*10^-19) C = 5.36*10^-20 C

electron at point P(x2,y2)= (0m,-1.1*10^-10m)

and the distance from origin to the H atom is sin theta = x/d

x = d*sin theta

x = 9.42*10^-11 m*sin 106 m

x = 9.0551*10^-11 m

so the position of the H atom is (x1,y1) (9.0551*10^-11m,0m)

the distance from H to electron is r = sqrt((x2-x1)^2+(y2-y1)^2)

r = sqrt((0-9.0551*10^-11)^2+(-1.1*10^-10-0)^2) m

r = 1.424763*10^-10 m

Now

a) force on the electron at P due to the oxygen atom is

F = kqo*e/r_op^2

let distance from origin to O is y = d cos theta = 9.42*10^-11 cos 106 m = (2.59650)*10^-11 m

so coordiantes of " O " is (0m,2.59650*10^-11m)

r_op = sqrt((0-0)^2+(-1.1*10^-10-2.59650*10^-11)^2) m = 1.35965*10^-10 m

F1 = 9*10^9(-1.072*10^-19)*(-1.6*10^-19)/(1.35965*10^-10)^2 N

F1 = (8.350)*10^-9 N

b)

force on electron due to two H atoms

F2 = kqH*e /r^2

F2 = 9*10^9(5.36*10^-20)*(-1.6*10^-19)/(1.424763*10^-10)^2N

F = (-3.802)*10^-9 N

due to two H atoms is FH = 2*(-3.802)*10^-9 N=(-7.604)*10^-9 N

c) net force is

FT = F1+F2 = (8.350)*10^-9+(-7.604)*10^-9 N

FT = 7.46*10^-10 N

the net force on electron due to water molecule is a repulsive force

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