Question

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

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

Answer 1

The correct option is C $0.2V$

Here $2a=1.38\times {10}^{-10}m,r=10\times {10}^{-10}mq=1.6\times {10}^{-19}C,\theta =0^{\circ }$
As potential,

$V=\frac{Pcos\theta }{4\pi {\epsilon }_0{r}^2}$

$=\frac{q\left(2a\right)}{4\pi {\epsilon }_0{r}^2}$

$=\frac{9\times {10}^9\times 1.6\times {10}^{-19}\times 1.38\times {10}^{-10}}{{(10\times {10}^{-10})}^2}$

$V=0.2V$