91Ó°ÊÓ

1. Electric field of magnitude,$E=20\text{N/C}$
2. And the above graph,$U_s=1.00×{10}^{−28}\text{J}$

The potential energy of the electric dipole placed in an electric field depends on its orientation relative to the electric field.The magnitude of the electric dipole moment is$p=qd$, where q is the magnitude of the charge, and d is the separation between the two charges.

Formula:

When placed in an electric field, the potential energy of the dipole,

(i)

$\begin{array}{c}U\left(\mathrm{θ}\right)=−\stackrel{→}{p}.\stackrel{→}{E}\\ =−pE\cos \mathrm{θ}\end{array}$

Examining the lowest value on the graph, we have the magnitude of the dipole moment using equation (i) as:

$\begin{array}{c}U=−\stackrel{→}{p}.\stackrel{→}{E}\\ −1.00×{10}^{−28}\text{J}=−20\text{N/C}\left(\stackrel{→}{p}\right)\\ \stackrel{→}{p}=5.00×{10}^{−28}\text{C.m}\end{array}$

Hence, the value of the dipole moment is.$5.00×{10}^{−28}\text{C.m}$