Q. A neutral water molecule (H$_{2}$O) in its vapour state has an electric dipole moment of $6 \times 10^{- 30}$ Cm. If the molecule is placed in an electric field of $1.5 \times 10^{4}$ $NC^{- 1}$, the maximum torque that the field can exert on it is nearly
A) $4.5 \times 10^{- 26\ }$N-m
B) $4 \times 10^{- 34\ }$N-m
C) $9 \times 10^{- 26\ }$N-m
D) $6 \times 10^{- 26\ }$N-m
Explanation
(c) Here, $\ p = 6 \times 10^{- 30}\ $C-m
$E = 1.5 \times 10^{4}$ N/C
$\tau_{\max} = pE\sin 90^{0}\ $
$= 6 \times 10^{- 30} \times 1.5 \times 10^{4} \times 1 = 9 \times 10^{- 26}$ Nm
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.