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The distance between $\mathrm{H}^{+}$ and $\mathrm{Cl}^{-}$ ions in $HCl$ molecule is $1.25 A^{\circ}$ . what will be the potential at a distance of $12 \AA$ on the axis of dipole -
Option: 1
$0.13 \mathrm{~V}$

Option: 2
$1.3 \mathrm{~V}$

Option: 3
$13 \mathrm{~V}$

Option: 4
$130 \mathrm{~V}$

Answers (1)

best_answer

Potential is given by -

$V =\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{2}}$
$=9 \times 10^{9} \times \frac{1.6 \times 10^{-19} \times 1.28 \times 10^{-10}}{\left(12 \times 10^{-10}\right)^{2}}$

$V =\frac{14.4 \times 10^{-10} \times 1.28 \times 10^{-10}}{144 \times 10^{-20}}$
$V =\frac{14.4}{144} \times 1.28$
$V =\frac{1}{10} \times 1.28=\frac{1.28}{10}$

$V =0.128=0.13$ Ans

Posted by

Divya Prakash Singh

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