Answer:
[tex] V = \frac{1}{4\pi* 8.85x10^{-12}} \frac{4.9098x10^{-30} *1 }{(53.1x10^{-9} m)^2}=1.566 x10^{-5} V [/tex]
And we can express the last expression like this:
[tex] 1.566 x10^{-5} V *\frac{1 \mu V}{10^{-6} V}= 15.66 \mu V[/tex]
Explanation:
For this case we know the dipole moment given by 1.47 D. We know also that [tex]1D = 1 debyte = 3.34 x10^{-30} Cm [/tex]
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:
[tex] p = 1.47 D *\frac{3.34x10^{-30} Cm}{1 D}= 4.9098x10^{-30} Cm[/tex]
And the potentital at the desired point can be calculated with the following formula:
[tex] V = \frac{1}{4 \pi \epsilon} \frac{p cos \theta}{r^2}[/tex]
Where [tex] \epsilon = 8.85x10^{-12}[/tex] is a constant, [tex] r = 53.1 x10^{-9}m[/tex], and since we are assuming V=0 we can conclude that [tex] cos \theta \approx 1[/tex]
If we replace the info given we got:
[tex] V = \frac{1}{4\pi* 8.85x10^{-12}} \frac{4.9098x10^{-30} Cm *1 }{(53.1x10^{-9} m)^2}=1.566 x10^{-5} V [/tex]
And we can express the last expression like this:
[tex] 1.566 x10^{-5} V *\frac{1 \mu V}{10^{-6} V}= 15.66 \mu V[/tex]
And this would be our final answer for this case.
