Question

A water molecule perpendicular to an electric field has $1.0\times {10}^{-21}\mathrm{J}$ more potential energy than a water molecule aligned with the field. The dipole moment of a water molecule is $6.2\times {10}^{-30}\mathrm{Cm}$. What is the strength of the electric field?

Short answer

The electric field's magnitude is$1.61\times {10}^8\mathrm{N}/\mathrm{C}$.

Step-by-step solution

Given Information

Given:

• The potential energy of a water molecule perpendicular to an electric field is$1.0\times {10}^{-21}\mathrm{J}$ higher than that of a water molecule aligned with the field.
• Dipole moment, $P=6.2\times {10}^{-30}\mathrm{C}-\mathrm{m}$

Determine the electric field strength

A water molecule aligned with the field has $1.0\times {10}^{-21}\mathrm{J}$more potential energy than a water molecule perpendicular to the field.

$U_{\text{perpendicular}}-U_{\text{parallel}}=1\times {10}^{-21}\mathrm{J}\dots ..\left(1\right)$

Compute the water molecule's potential energy perpendicular to the electric field.

$\begin{array}{rcl}{U}_{\text{perpendicular}}& =& -PE\cos 90\\ & =& 0\end{array}$

Estimate the water molecule's potential energy when it is aligned with the electric field.

$U_{\text{parallel}}=-PE\times \cos 0$

Plugging the values in the above equation

$U_{\text{parallel}}=-6.2\times {10}^{-30}E\times 1$

Insert the values in the above equation

$$\begin{gathered} 0-\left(-6.2\times {10}^{-30}E\times 1\right)=1\times {10}^{-21} \\ E=1.61\times {10}^8\mathrm{N}/\mathrm{C} \end{gathered}$$