Question

An ammonia molecule has permanent electric dipole moment = 1.47D, where 1D = 1 debye unit `= 3.34xx10^(-30) Cm`. Calculate electric potential due to this molecule at a point 52.0 nm away along with axis of the dipole. Assume V = 0 at infinity.

Answer 1

`P = 1.47 D = 1.47xx3.34xx10^(-30) Cm`
`V = ?, r = 52 nm = 52.0xx10^(-9) m`
On the axis of dipole ,
`V = (p)/(4pi in_(0) r^(2)) = (9xx10^(9)xx1.47xx3.34xx10^(-30))/((52.0xx10^(-9))^(2))`
`= 1.63xx10^(-5)V`