Question

The permanent electric dipole moment of the water molecule $\left({\mathrm{H}}_2\mathrm{O}\right)$ is $6.2\times {10}^{-30}\mathrm{Cm}$. What is the maximum possible torque on a water molecule in a $5.0\times {10}^8\mathrm{N}/\mathrm{C}$electric field?

Short answer

Therefore, the value of the maximum possible torque $\left(\tau \right)$acting on the water molecule when it is placed in the electric field is $3.1\times {10}^{-21}\mathrm{N}·\mathrm{m}$.

Step-by-step solution

 Introduction

On a dipole, the electric field produces a torque as follows:

$\tau =pE\sin \theta$

The torque is $r$,

The dipole moment is $p$,

The electric field is $E$,

The angle between the dipole and electric field direction is $\theta$.

Step2  Explanation

Calculate the highest possible torque $\left(\tau \right)$operating on the water molecule when it is placed in an electric field using the following formula:

$\tau =pE\sin \theta$

When $\overrightarrow{p}$is perpendicular to $\overrightarrow{E}$, the torque is greatest, which means the angle between the dipole moment $\left(p\right)$and the electric field strength $\left(E\right)$is $90.0°$

Substitute $6.2\times {10}^{-30}\mathrm{C}·\mathrm{m}$for $p,5.0\times {10}^8\mathrm{N}/\mathrm{C}$for $E$, and $90.0°$for $\theta$in the above equation.

$\tau =\left(6.2\times {10}^{-30}\mathrm{C}·\mathrm{m}\right)\left(5.0\times {10}^8\mathrm{N}/\mathrm{C}\right)\sin 90°$

$=3.1\times {10}^{-21}\mathrm{N}·\mathrm{m}$