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Section 33.7 Dielectrics Bootcamp

Subsection 33.7.1 Electric Dipoles

Subsection 33.7.2 Force and Torque on Electric Dipoles

Subsection 33.7.3 Potential Energy of Electric Dipoles in an External Field

Subsection 33.7.4 Electric Potential of a Dipole

Subsection 33.7.5 Linear Dielectrics

Subsection 33.7.6 Miscellaneous

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.


Use definitions


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


(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*}