91Ó°ÊÓ

The given data can be listed below as,

• The dipole moment of a water molecule is,$\mathrm{p}=6.2×{10}^{−30}\text{ â¶Ä‹C}â‹\dots \text{m}$.
• The electric field of a water molecule is, $\mathrm{E}=4×{10}^5\text{ â¶Ä‹N/³¾}$.

The electrical dipole moment is the distance between the positively charged particle and the negatively charged particle in an electric field system. The electric dipole shows the strength of the electric field system.

The expression for the torque is given as,

$\mathrm{Ï„}=\mathrm{\pm è·¡²õ¾\pm ²Ôθ}$

Here,$p$ is the dipole moment and its value is, $6.2×{10}^{−30}\text{ â¶Ä‹C}â‹\dots \text{m}$,$\mathrm{E}$ is the electric field and its value is,$4×{10}^5\text{ â¶Ä‹N/³¾}$ , $\mathrm{θ}$is the angle for perpendicular its value is$90°$ .

Substitute, $6.2×{10}^{−30}\text{​â¶Ä‰C}â‹\dots \text{m}$for $\mathrm{p}$and $4×{10}^5\text{ â¶Ä‹N/³¾}$for$E$ and$90°$ for $\mathrm{θ}$.

$\begin{array}{l}\mathrm{Ï„}=6.2×{10}^{−30}\text{ C}â‹\dots \text{m}×4×{10}^5\text{​â¶Ä‰N/³¾}×\text{sin(90}°)\\ \mathrm{Ï„}=2.48×{10}^{−24}\text{ N}â‹\dots \text{C}\end{array}$

Hence, the value of torque is $2.48×{10}^{−24}\text{ N}â‹\dots \text{C}$.

Considering a dipole with charges$+\mathrm{q}$and $-q$separated by a distance$d$, dipole is placed in a electric field $E$and the axis of the dipole forms an angle $\mathrm{θ}$with the electric field.

The expression for the force on charges is,

$\stackrel{→}{\mathrm{F}}=\mathrm{q}\stackrel{→}{\mathrm{E}}$

The components of force perpendicular to dipole expressed as,

$\mathrm{F}=\mathrm{qE}\text{ }\mathrm{²õ¾\pm ²Ôθ}$

Since, the magnitude of the force are equal and they are separated by distance$\mathrm{d}$, then the torque on dipole is expressed as,

$\begin{array}{c}\mathrm{Torque}=\mathrm{Force}×\mathrm{Distance}\\ \mathrm{Ï„}=\mathrm{qE}\text{ s¾±²Ô}\mathrm{θ}×\mathrm{d}\end{array}$

The dipole moment is expressed by,

$\mathrm{p}=\mathrm{qd}$

Substitute the expression in the torque equation. The torque on dipole is expressed as,

$\begin{array}{l}\mathrm{Ï„}=\mathrm{\pm è·¡²õ¾\pm ²Ôθ}\text{ }\\ \mathrm{Ï„}=\stackrel{→}{\mathrm{p}}×\stackrel{→}{\mathrm{E}}\end{array}$

Hence, it is proved.

Hence, the vector torque is $(\stackrel{→}{\mathrm{p}}×\stackrel{→}{\mathrm{E}})$.