The distance between ${H^ + }$ and $C{l^ - }$ ions in $HCl$ molecule is $1.28\,\mathop A\limits^o $. What will be the potential due to this dipole at a distance of $12\,\mathop A\limits^o $ on the axis of dipole......$V$
Easy
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(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}}}}{{{{(12 \times {{10}^{ - 10}})}^2}}}$ $= 0.13\,V$
$ = 9 \times {10^9} \times \frac{{(1.6 \times {{10}^{ - 19}}) \times 1.28 \times {{10}^{ - 10}}}}{{{{(12 \times {{10}^{ - 10}})}^2}}}$ $= 0.13\,V$
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