Question

The distance between $H^{+}$ and $Cl^{-}$ ions in $HCl$ molecules is $1.38 \, \mathring{A}$. The potential due to this dipole at a distance of $10 \, \mathring{A}$ on the axis of dipole is

• A. $2.1 \, V$
• B. $1.8 \, V$
• C. $0.2 \, V$
• D. $1.2 \, V$

Answer 1

Answer: (C)

Here, $2a=1.38\times10^{-10} \, m, r=10 \times10^{-10}\, m $
charge, $q=1.6\times10^{-19}\, C $
As potential, $ V=\frac{P}{4\pi\varepsilon_{0} r^{2}}=\frac{q\left(2a\right)}{4\pi\varepsilon_{0} r^{2}}$
$=\frac{9\times10^{9}\times1.6\times10^{-19}\times1.38\times10^{-10}}{\left(10\times10^{-10}\right)^{2}}$
$=0.2 \, V$