The distance between H^+ and Cl^- ions in an | vedclass

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(A) The electric potential $V$ due to a dipole at a point on its axis at a distance $r$ from the center is given by $V = \frac{1}{4\pi\epsilon_0} \cdo

Vedclass · Accepted Answer

$0.13$

Vedclass · Answer

(A) The electric potential $V$ due to a dipole at a point on its axis at a distance $r$ from the center is given by $V = \frac{1}{4\pi\epsilon_0} \cdot \frac{p}{r^2}$.Here,the dipole moment $p = q 	imes d$,where $q = 1.6 	imes 10^{-19} \, C$ and $d = 1.28 	imes 10^{-10} \, m$.The distance $r = 12 	imes 10^{-10} \, m$.Substituting the values:$V = (9 	imes 10^9) 	imes \frac{(1.6 	imes 10^{-19}) 	imes (1.28 	imes 10^{-10})}{(12 	imes 10^{-10})^2}$.$V = (9 	imes 10^9) 	imes \frac{2.048 	imes 10^{-29}}{144 	imes 10^{-20}}$.$V = 9 	imes \frac{2.048}{144} 	imes 10^0 = 0.128 \, V \approx 0.13 \, V$.

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

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