91Ó°ÊÓ

a) Initial orientation of the electric dipole i,${\mathrm{θ}}_i={20.0}^0$

b) Final orientation of the electric dipole f, ${\mathrm{θ}}_f={20.0}^o$

c) Strength of the uniform external electric field $\stackrel{→}{E}=3.00x{10}^6\text{N/C}.$

d) The electric dipole moment, $p=1.60x{10}^{−27}\text{C.m}$

Using the concept of an electric dipole, we can get the potential energy of the dipole. This determines the change in the potential energy of the system.

Formula:

Potential Energy of an Electric Dipole is given by:$U=−\stackrel{→}{p}.\stackrel{→}{E}=−pEcos\mathrm{θ}$ (i)

Where,

$\stackrel{→}{p}$and $\stackrel{→}{E}$are the dipole moment and Electric field

From the given data and the figure, we know that the initial and final angles of orientation ${\mathrm{θ}}_i={70.0}^0$are${\mathrm{θ}}_f={110.0}^o$, we can get the change of the potential energy using equation (i) as follows:

$\begin{array}{c}\mathrm{Δ}U=−pE(cos70.0°−cos110°)\\ =−\left(1.60x{10}^{−27}\text{C.m}\right)\left(3.00x{10}^6\text{N/C}\right)\left(0.684\right)\\ =−3.28x{10}^{−21}\text{J}\end{array}$

Hence, the value of the potential energy is.$−3.28x{10}^{−21}\text{J}$