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Question: The distance between \(H^{+}\) and \(Cl^{-}\)ions in HCl molecules is 1.38 \(\overset{o}{A}\). The p...

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

A

2.1 V

B

1.8 V

C

0.2 V

D

1.2 V

Answer

0.2 V

Explanation

Solution

: Here, 2a=1.38×10−10m,r=10×10−10m2a = 1.38 \times 10^{- 10}m,r = 10 \times 10^{- 10}mcharge, q=1.6×10−19Cq = 1.6 \times 10^{- 19}C

As potential, V=P4πε0r2=q(2a)4πε0r2V = \frac{P}{4\pi\varepsilon_{0}r^{2}} = \frac{q(2a)}{4\pi\varepsilon_{0}r^{2}}

=9×109×1.6×10−19×1.38×10−10(10×10−10)2=0.2V= \frac{9 \times 10^{9} \times 1.6 \times 10^{- 19} \times 1.38 \times 10^{- 10}}{(10 \times 10^{- 10})^{2}} = 0.2V