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 55.3 nm away along the axis of the dipole. (Set V = 0 at infinity.)

The electric potential due to ammonia at a point away along the axis of a dipole is 1.44 10^-5 V.

Explanation:

Given that 1 D = 1 debye unit = 3.34 × 10-30 C-m.

Given p = 1.47 D = 1.47 3.34 10^-30 = 4.90 10^-30.

            V = 1 / (4π∈о)    (p cos(θ)) / (r^2)

where p is a permanent electric dipole,

           ∈ο is permittivity,

            r is the radius from the axis of a dipole,

            V is the electric potential.

        V = 1 / (4 3.14 8.85 10^-12)   (4.90 10^-30 1) / (55.3 10^-9)^2

        V  = 1.44 10^-5 V.

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