The distance between H+ and Cl− ions in HCI molecule is 1.28 A∘ . What will be the potential due to this dipole at a distance of 12 A∘ on the axis of dipole?

The distance between $H^{+}$ and $C l^{-}$ ions in HCI molecule is $1.28 \overset{\circ}{A}$ . What will be the potential due to this dipole at a distance of $12 \overset{\circ}{A}$ on the axis of dipole?

A
0.13 V
B
13 V
C
1.3 V
D
130 V

Solution:

Here, $2 a = 1.28 \overset{\circ}{A} = 1.28 \times 10^{- 10} m$

$V = ?, r = 12 \overset{\circ}{A} = 12 \times 10^{- 10} m$

As the point is on axial line of dipole

$∴$ $V = \frac{k q (2 a)}{r^{2}}$

$= \frac{9 \times 10^{9} \times 1.6 \times 10^{- 19} (1.28 \times 10^{- 10})}{{(12 \times 10^{- 10})}^{2}}$

$= 0.128 \approx 0.13 V$

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