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

Answer 1

`V = (1)/(4\pi* 8.85x10^(-12)) (4.9098x10^(-30) *1 )/((53.1x10^(-9) m)^2)=1.566 x10^(-5) V`

And we can express the last expression like this:

`1.566 x10^(-5) V *(1 \mu V)/(10^(-6) V)= 15.66 \mu V`

Step-by-step explanation:

For this case we know the dipole moment given by 1.47 D. We know also that
`1D = 1 debyte = 3.34 x10^(-30) Cm`

And we need to calculate the electric potential due to an ammonia molecule at a point 53.1 nm away along the axis of the dipole

We can convert the dipole moment like this:

`p = 1.47 D *(3.34x10^(-30) Cm)/(1 D)= 4.9098x10^(-30) Cm`

And the potentital at the desired point can be calculated with the following formula:

`V = (1)/(4 \pi \epsilon) (p cos \theta)/(r^2)`

Where
`\epsilon = 8.85x10^(-12)` is a constant,
`r = 53.1 x10^(-9)m`, and since we are assuming V=0 we can conclude that
`cos \theta \approx 1`

If we replace the info given we got:

`V = (1)/(4\pi* 8.85x10^(-12)) (4.9098x10^(-30) Cm *1 )/((53.1x10^(-9) m)^2)=1.566 x10^(-5) V`

And we can express the last expression like this:

`1.566 x10^(-5) V *(1 \mu V)/(10^(-6) V)= 15.66 \mu V`

And this would be our final answer for this case.