Question

An ammonia molecule$N{H}_3$ has a permanent electric dipole moment$5.0\times {10}^{30}Cm$. A proton is $2.0nm$ from the molecule in the plane that bisects the dipole. What is the electric force of the molecule on the proton?

Short answer

The molecule proton on electric force is$E=5.625\times {10}^6\mathrm{N}/\mathrm{C}.$

Step-by-step solution

Step: 1 Electric field:

The energy generated by a given positive ion put at a location in the field is related to the intensity of the electric field at that point.

The fundamental expression is as follows:

$E=\frac{F}{q}$

Step: 2 Electric dipole:

The electric dipole in field as

$E=\frac{p}{4\pi {\epsilon }_0{r}^3}$

The dipole moment is $p$, the distance between the dipole and the charge is $r$, and the permittivity of empty space is ${\epsilon }_0$.

Step: 3 Substituting:

Putting values in above equation,we get

$E=\frac{p}{4\pi {\epsilon }_0{r}^3}$

role="math" localid="1651406439589" $E=\left(9\times {10}^9{\mathrm{C}}^2/\mathrm{N}\cdot {\mathrm{m}}^2\right)\frac{5.0\times {10}^{-30}\mathrm{Cm}}{{\left(2.0\times {10}^{-9}\mathrm{m}\right)}^3}$

$E=5.625\times {10}^6\mathrm{N}/\mathrm{C}.$

Step: 4 Finding force:

The electric force of a molecular on a proton is computed using the induced electromagnetic and force relationship.

The equation as

$E=\frac{F}{q}$

$\begin{array}{l}F=Eq\\ F=\left(5.625\times {10}^6\mathrm{N}/\mathrm{C}\right)\left(1.6\times {10}^{-19}\mathrm{C}\right)\\ F=9\times {10}^{-13}\mathrm{N}\end{array}$