## Section33.7Dielectrics Bootcamp

### Subsection33.7.6Miscellaneous

An ammonia molecule has an electric dipole moment of $5.0 \times 10^{-30}$ C.m. It is placed in an external electric field of $1,000$ V/m. (a) What is the torque on it at the time the dipole is pointed at a $60^{\circ}$ angle with respect to the electric field? (b) Once aligned, how much energy will it take to flip the orientation of the dipole by $180^{\circ}\text{?}$ Write your answer in the electron volt (eV) unit. The unit $1$ eV = $1.6\times 10^{-19}$ J.

Hint

Use definitions

(a) $4.3\times 10^{-27}$ N.m, (b) $6.3\times 10^{-8}$ eV.

Solution

(a) The torque on a dipole is equal to the vector product of the dipole moment and the electric field. Therefore, the magnitude of the torque will be

\begin{equation*} \tau = p\:E\:\sin\theta = 5.0\times 10^{-30}\:\textrm{C.m}\times 1000\:\dfrac{\textrm{V}}{\textrm{m}}\:\sin 60^{\circ} = 4.33\times 10^{-27}\:\textrm{N.m}. \end{equation*}

The direction of the torque vector can be obtained by using the right-hand rule. therefore, if you place the vector $\vec p$ towards the positive $x$-axis and vector $\vec E$ in the $xy$-plane towards a direction in the space $y>0\text{,}$ then the torque will be pointed towards the positive $z$-axis.

(b) Flipping of a dipole's direction requires energy equal to $2pE\text{.}$

\begin{equation*} U = 2pE = 2\times 5.0\times 10^{-30}\:\textrm{C.m}\times 1000\:\dfrac{\textrm{V}}{\textrm{m}} = 1.0\times 10^{-26}\:\textrm{J}. \end{equation*}