Question

The distance between $H ^{+}$ and $Cl^-$ ions in $HCl$ molecule is $1.28\, \mathring{A}$. What will be the potential due to this dipole at a distance of $12\, \mathring{A}$ from one end and on the axis of dipole?

• A. 0.13 V
• B. 1.3 V
• C. 13 V
• D. 130 V

Answer 1

Answer: (A)

$V =9 \times 10^{9} \cdot \frac{p}{r^{2}}$
$=9 \times 10^{9} \times \frac{\left(1.6 \times 10^{-19}\right) \times 1.28 \times 10^{-10}}{\left(12 \times 10^{-10}\right)^{2}}$
$=0.13\, V$