Question

A natural water molecule (H2O) in its vapor state has an electric dipole moment of magnitude, p = 6.2 x 10-30 C.m. (a) Find the distance of the positive and negative charge centers of the molecule. Note that there are 10 electrons and 10 protons in a natural water molecule. (b) If the molecule is placed in a uniform electric field, E = 2 x 104 N/C find the maximum torque acting on the molecule. (c) How much work is needed to rotate this molecule by 180° in this field starting from the initial position, for which θ = 0? Hint: θ is the angle between the electric dipole moment and the electric field.

Answer 1

a)Dipole moment p=aq

p= 6.2 x C.m

a=dipole length=?

q=charge =10e

e=charge of one electron=1.6xC

q=10e=10x1.6x=1.6xC

p=aq

a=p/q= 6.2 x /1.6x=3.875xmeter

b)maximum torque =pE

E=electric field=2x

maximum torque =pE=6.2 x x2x=1.24xNm

c)Work to rotate W=PE(cos-cos)

=0

=180

W=PE(cos-cos)=1.24x(cos0-cos180)=1.24xx[1-(-1)]=1.24xx2=2.48xjoule