Question

The ammonia molecule (NH3) has a dipole moment of 5.0×10−30C⋅m. Ammonia molecules in the gas phase are placed in a uniform electric field E⃗ with magnitude 1.9×106 N/C . Part A What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to E⃗ from parallel to perpendicular?

Answer 1

Answer:

 8 x 10^-24 J  

Explanation:

Given :

E = 1.9 x 10^6 N/m

p= 5.0 x 10^-30 C.m

Required :  

a) ΔU  

a) The change in electric potential between initial and final position so the initial energy is given by  

U_i =  - pEcos(Фi) = -pEcos(0)

      =  - 1.9 x 10^6 x 5.0 x 10^-30  

      =  - 8 x 10^-24 J  

Since the final angle between p and E is zero then the potential U is given by  

U_f =0

So the change in potential is given by  

ΔU = U_f - U_i  =  8 x 10^-24 J