a simple model of water molecule has a dipole moment of 1891

a simple model of water molecule has a dipole moment of 1.89*10^-29cm at 60 degree in the xz plane from x axis. a uniform electric field of -3i in/c is now aplied
a) determine the magnitude and direction of the tourqe acting on the dipole when the field is applied
by) determine the potential energy of the dipole in the electric field when the field is applied

Solution

a) torque acting on a dipole , t =   p x E   

                                                        pE sin(Q) , Q is the angle made by p with E

                                     p = 1.89 * 10^29 C. cm , or 1.89 * 10^27C. m (the units of dipole momentum is C .cm or C.m)

                                 3 N / C pointing negative x -axis

                         t = 1.89 * 10^27 * 3 * sin (60) N .m

                                           = 4.9103 * 10^ 27 N.m

b) potential energy of the dipole V = - p . E   (- sign is due to the opposite direction of p and E)

                                                            = - pE cos (Q)

                                                            = - 1.89 * 10^27 * 3 * cos (60)

                                                          V = - 2.835 x 10^-27 N m