Question

An electric field can induce an electric dipole in a neutral atom or molecule by pushing the positive and negative charges in opposite directions. The dipole moment of an induced dipole is directly proportional to the electric field. That is, $\overrightarrow{p}=\alpha \overrightarrow{E}$, where $\alpha$is called the polarizability of the molecule. A bigger field stretches the molecule farther and causes a larger dipole moment.

a. What are the units of $\alpha$?

b. An ion with charge $q$is distance$r$from a molecule with polarizability $\alpha$. Find an expression for the force ${\overrightarrow{F}}_{ionondipole}$.

Short answer

a. The units of $\alpha$is$\frac{{\left(C\right)}^2\times {\left(s\right)}^2}{kg}$

b. The required expression is :${\overrightarrow{F}}_{ionondipole}=\frac{[2\alpha \times {\left(k\right)}^2\times {\left(q\right)}^2]}{r^5}$

Step-by-step solution

Given information and formula 

Given :

The dipole moment of an induced dipole is directly proportional to the electric field. That is :$\overrightarrow{p}=\alpha \overrightarrow{E}$

A bigger field stretches the molecule farther and causes a larger dipole moment.

Theory used :

Unit of Dipole moment is coulomb-meter

Unit of Electric field is$m\cdot kg\cdot s^{-3}\cdot A^{-1}$

Calculating units of α and expression for Force of ion on dipole

Solving $\overrightarrow{p}=\alpha \overrightarrow{E}$for units of $\alpha$:

localid="1649133782229" $$\begin{gathered} \overrightarrow{p}=\alpha \overrightarrow{E} \\ \Rightarrow \alpha =\frac{\overrightarrow{p}}{\overrightarrow{E}}=\boxed{\frac{{\mathbf{\left(}\mathbf{C}\mathbf{\right)}}^{\mathbf{2}}\mathbf{\times }{\mathbf{\left(}\mathbf{s}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{k}\mathbf{g}}} \end{gathered}$$

On a dipole, the ion's force $F$is:

localid="1649133791721" $F=\boxed{\frac{\mathbf{[}\mathbf{2}\mathbf{\alpha }\mathbf{\times }{\mathbf{\left(}\mathbf{k}\mathbf{\right)}}^{\mathbf{2}}\mathbf{\times }{\mathbf{\left(}\mathbf{q}\mathbf{\right)}}^{\mathbf{2}}\mathbf{]}}{{\mathbf{r}}^{\mathbf{5}}}}$