Q. The distance between $H^{+}$ and $Cl^{-}$ ions in $HCl$ molecule is $1.28\ \mathring{\mathrm{A}}.$ What will be the potential due to this dipole at a distance of $12\ \mathring{\mathrm{A}}$ on the axis of dipole
A) $0.13\ V$
B) $1.3\ V$
C) $13\ V$
D) $130\ V$
Explanation
(a) $V = 9 \times 10^{9}.\frac{p}{r^{2}}$
$= 9 \times 10^{9} \times \frac{(1.6 \times 10^{- 19}) \times 1.28 \times 10^{- 10}}{\left( 12 \times 10^{- 10} \right)^{2}} = 0.13V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.