Question

Question: The ammonia molecule NH₃ has a permanent electric dipole moment equal to,1.47 D where,$\mathbf{1}\mathit{D}\mathit{e}\mathit{b}\mathit{y}\mathit{e}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{34}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{30}}\mathbf{\text{C.m}}$. Calculate the electric potential due to an ammonia molecule at a point 52.0nmaway along the axis of the dipole. (Set v = 0at infinity.)

Short answer

Answer:

The electric potential due to an ammonia molecule is$163\mu V$.

Step-by-step solution

The given data

1. The dipole moment of the ammonia molecule, p = 1.47D where$1D=3.34\times {10}^{-30}\text{C.m}$
2. The distance of the ammonia molecule, r = 52nm
3. The electric potential at infinity is V = 0.

Understanding the concept of the electric potential

Using the given concept of the electric potential, we can get the value of the potential using the dipole moment of the given ammonia molecule.

Formula:

The electric potential of a body due to the dipole, $V=\frac{1}{4\pi {\epsilon }_0}\frac{p}{r^2}$ (i)

Calculation of the electric potential

Now the electric potential at the distance is given using the given data in equation (i) as follows:

$$\begin{gathered} V=8.99\times {10}^9\frac{1.47\times 3.34\times {10}^{-30}}{{(52.0\times {10}^{-9})}^2} \\ =16.3\mu \text{V} \end{gathered}$$

Hence, the value of the potential is $16.3\mu \text{V}$.