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

Answer 1

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

Explanation:

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

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

            V = 1 / (4π∈о)  $\times$  (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 $\times$ 3.14 $\times$ 8.85 $\times$ 10^-12)  $\times$ (4.90 $\times$ 10^-30 $\times$ 1) / (55.3 $\times$ 10^-9)^2

        V  = 1.44 $\times$ 10^-5 V.