91影视

1. Dipole moment,$p=3.02x{10}^{鈭�25}\text{C.m}$
2. Electric field of magnitude,$E=46.0\text{N/C}$
3. Initial angle,${胃}_i={64}^o$
4. Final angle of the dipole,${胃}_f={180}^o$

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$\stackrel{鈫�}{p}=q\stackrel{鈫�}{d}$, where,qis the magnitude of the charge, and dis the separation between the two charges. When placed in an electric field, the potential energy of the dipole is given by

$\begin{array}{c}\mathbf{U}\mathbf{\left(}\mathbf{胃}\mathbf{\right)}\mathbf{=}\mathbf{鈭�}\stackrel{\mathbf{鈫�}}{\mathbf{p}}\mathbf{.}\stackrel{\mathbf{鈫�}}{\mathbf{E}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{鈭�}\mathbf{p}\mathbf{E}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{胃}\end{array}$

Formulae:

Therefore, if the initial angle between p and E isand the final angle is$胃$, then the change in potential energy would be

$\begin{array}{c}螖U=U\left(胃\right)鈭�U_o\left(胃\right)\\ =鈭�pE(cos胃鈭�cos{胃}_o)\end{array}$ (i)

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

鈥淭orque and energy of an electric dipole in an electric field,鈥� thus we can find the amount of work can be given as the net potential difference and using equation (i) it is given as follows

$\begin{array}{c}W=U({胃}_o+蟺)鈭�U_o\left({胃}_o\right)\\ =鈭�pE\left(cos({胃}_o+蟺)鈭�cos\left({胃}_o\right)\right)\\ =2pEcos{胃}_o\\ =2(3.02脳{10}^{鈭�25}\text{C.m})\left(\frac{\text{46.0N}}{\text{C}}\right)cos{64.0}^o\\ =1.22脳{10}^{鈭�23}\text{J}\end{array}$

Hence, the value of the amount of work done for the required change in angle is.$1.22脳{10}^{鈭�23}\text{J}$