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

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

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.