Question

The ammonia molecule NH₃ has a permanent electric dipole moment equal to 1.47 D, where 1 D = 1 debye unit = 3.34 × 10^-30 C-m. Calculate the electric potential in volts due to an ammonia molecule at a point 49.6 nm away along the axis of the dipole. (Set V = 0 at infinity.)

Answer 1

Electric potential due to a charge particle having charge ' q ' at a distance ' r ' is given by

Where,

Electric potential due to a dipole at a distance ' r ' on its axis :

Arrangement of two charge particles of opposite polarity at a separation is known as dipole.

Here, Electric potential due to negative charge (-q) at point P is  

  

And electric potential due to the positive charge (+q) at point P is

So, total potential at point P is

  

  

  

Where, p = 2aq is the dipole moment

If r >> a then

...............................................(1)

Now, for ammonia molecule : 0.1 nm < a < 0.4 nm

For the given problem , r = 49.6 nm

Hence, r >> a .

Therefore, we can use equation (1) in this case.

Given,

So, potential due to an ammonia molecule at a point 49.6 nm away along the axis of the dipole is

  

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