Answered

An electric dipole, consisting of charges each of magnitude 2.11 nC and separated by 8.37 micrometers, is in an electric field of strength 1578 N/C. What is the magnitude of the difference in potential energy (in Joules) corresponding to dipole orientations parallel to and anti-parallel to the electric field



Answer :

Hashirriaz830

Answer:

54*10^-13 J

Explanation:

given:

charge =2.11 nC

distance=8.37*10-6

electric field intensity=1578 N/C

solution:

dipole moment magnitude=q.2a

                                            =2.11 nC*2*8.37*10-6

                                            =18*10^-15

potential energy of dipole = -P*E

                                           = -P*E*cos Ф

for parallel orientation, Ф=0

potential energy = -18*10^-15*300*cos 0

                             =-54*10^-13 J

for perpendicular orientation, Ф=90

potential energy = -18*10^-15*300*cos 90

                             =0

           difference =0-(-54*10^-13 )J

                            =54*10^-13 J