91Ó°ÊÓ

a) Electric field,$E=1100\text{N/C}$

b) Charges of magnitude,$q=1.50\text{nC}$

c) Charges separated by,$r=6.20\text{μ³¾}$

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,$\stackrel{\mathbf{→}}{\mathbf{p}}\mathbf{=}\mathit{q}\stackrel{\mathbf{→}}{\mathbf{d}}$ where $\mathit{q}$,is the magnitude of the charge, andis the separation between the two charges. When placed in an electric field, the potential energy of the dipole is given by

$\begin{array}{c}\mathbf{U}\mathbf{\left(}\mathbf{θ}\mathbf{\right)}\mathbf{=}\mathbf{−}\stackrel{\mathbf{→}}{\mathbf{p}}\mathbf{.}\stackrel{\mathbf{→}}{\mathbf{E}}\\ \mathbf{=}\mathbf{−}\mathbf{p}\mathbf{E}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{θ}\end{array}$

Formulae:

Therefore, if the initial angle between p and E is${\mathrm{θ}}_o$and the final angle is,$\mathrm{θ}$ then the change in potential energy would be

$\begin{array}{c}\mathrm{Δ}U=U\left(\mathrm{θ}\right)−U_o\left(\mathrm{θ}\right)\\ =−pE(cos\mathrm{θ}−cos{\mathrm{θ}}_o)\end{array}$ (i)

The electric dipole moment of a system, $p=qd$ (ii)

Substituting the given data in equation (ii), we find the magnitude of the dipole moment to be:

$\begin{array}{c}p=(1.50×{10}^{−6}\text{C})(6.20×{10}^{−6}\text{m})\\ =9.30×{10}^{−15}\text{C.m}\end{array}$

Hence, the value of the dipole moment is.$9.30×{10}^{−15}\text{C.m}$

The initial and the final angles are${\mathrm{θ}}_o=0^o$(parallel) and, ${\mathrm{θ}}_o=\text{18}{0}^o$so the difference between the potential energies is given using equation (i) as follows:

$\begin{array}{c}\mathrm{Δ}U=U\left({180}^o\right)−U_o\left(0\right)\\ =2pE\\ =2(9.30×{10}^{−15}\text{C.m})\left(\frac{1100\text{N}}{\text{C}}\right)\\ =2.05×{10}^{−11}\text{J}\end{array}$

The potential energy is a maximum ($U_{max}=+pE$) when the dipole is oriented antiparallel to E, and is a minimum ($U_{\min }=−pE$) when it is parallel to E.

Hence, the value of the difference is.$2.05×{10}^{−11}\text{J}$