91影视

The dipoles are described as the pair of opposite and equal charged magnetized poles which are separated by a distance. Moreover, the dipoles are the pair of the opposite and equal electric charges whose centers are not coincident.

The diagram of the dipoles and their distances is provided below,

The equation of the magnitude of the location of the proton is expressed as:

$r_1=\sqrt{x_1^2{y}_1^2+z_1^2}$

Here, $x_1$is the location of the proton at the x axis, $y_1$is the location of the proton at the y axis, and $z_1$is the location of the proton at the z axis.

Substitute the values in the above equation.

$$\begin{gathered} r_1=\sqrt{{\left(0\right)}^2+{\left(3脳{10}^{-8}\right)}^2+{\left(0\right)}^2}m \\ =3脳{10}^{-8}m \end{gathered}$$

The equation of the magnitude of the electric field of the proton is expressed as:

$E=k\frac{2p}{r_1\sqrt{r_1^2+s^2}}$ 鈥�(颈)

Here, k is the electric field constant that has the value $9脳{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2$, p is the dipole moment, $r_1$is the proton鈥檚 location, and s is the distance between the charged particles.

The location of the proton is much bigger than the distance between the charged particles as the proton lies in the charged axis of the dipole. Hence, $r_1鈮�s$.

Then the above equation can be reduced as:

$E=k\frac{2p}{r_1^3}$

The equation of the dipole moment is expressed as:

p=$q_{1}s$

Here, $q_1$is the charge of the charged particles which is closer to the proton and is the distance between the charged particles.

Substitute the above value in the equation (i).

$E=K\frac{2q_{1}s}{r_1^3}$

Substitute all the values in the above equation.

$$\begin{gathered} E=9脳{10}^{9}N.m^2/C^2脳\frac{2脳2脳1.6脳{10}^{-19}C脳7脳{10}^{-10}m}{{\left(3脳{10}^{-8}m\right)}^3} \\ =9脳{10}^{9}N.m^2/C^2脳\frac{4.48脳{10}^{-28}C.m}{2.7脳{10}^{-23}C/m^2} \\ =9脳{10}^{9}N.m^2/C^2脳1.65脳{10}^{-5}C/m^2 \\ =1.49脳{10}^{5}N/C \end{gathered}$$

The equation of the force on the proton due to the dipole is expressed as:

$F=q_{1}E$

Here, E is the magnitude of the electric field of the proton, and $q_1$is the charge of the charged particles which is closer to the proton.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{F}=2脳1.6脳{10}^{-19}\mathrm{C}脳1.49脳{10}^5\mathrm{N}/\mathrm{C} \\ =3.2脳{10}^{-19}\mathrm{C}脳1.49脳{10}^5\mathrm{N}/\mathrm{C} \\ =4.76脳{10}^{-14}\mathrm{N} \end{gathered}$$

As the proton is closer to the negative charge, the electric field is downward, and it has a negative component on the y axis and the direction of the force is negative.

Thus, the force on the proton due to the dipole is${\mathrm{-4.76脳10}}^{-14}\mathrm{N}$.

The equation of the magnitude of the location of the electron is expressed as:

$r_2=\sqrt{x_2^2{y}_2^2+z_2^2}$

Here, data-custom-editor="chemistry" ${\mathrm{x}}_2$is the location of the electron at the x axis, $y_2$is the location of the electron at the y axis and $z_2$is the electron鈥檚 location at the z axis.

Substitute the values in the above equation.

$$\begin{gathered} r_2=\sqrt{{\left(-3脳{10}^{-8}\right)}^2+{\left(0\right)}^2+{\left(0\right)}^2}m \\ =3脳{10}^{-8}m \end{gathered}$$

The equation of the magnitude of the electric field of the electron is expressed as:

$E=k\frac{2p}{r_2\sqrt{r_2^2+s^2}}$ 鈥�(颈颈)

Here, k is the electric field constant that has the value $9脳{10}^{9}N.m^2/C^2$, p is the dipole moment, $r_1$is the electron鈥檚 location, and is the distance between the charged particles.

The electron鈥檚 location is much bigger than the distance between the charged particles as the electron lies in the charged axis of the dipole. Hence,role="math" localid="1656931275428" $r_2鈮�s$.

Then the above equation can be reduced as:

role="math" localid="1656931306828" $E=K\frac{2p}{r_2^3}$

The equation of the dipole moment is expressed as:

role="math" localid="1656931322000" $p=q_{2}s$

Here, role="math" localid="1656931335009" $q_2$is the charge of the charged particles which is closer to the electron and s is the distance between the charged particles.

Substitute the value in the equation (ii).

role="math" localid="1656931357406" $E=K\frac{2q_{2}s}{r_2^3}$

Substitute all the values in the above equation.

$$\begin{gathered} E=9脳{10}^{9}N.m^2/C^2脳\frac{2脳2脳-1.6脳{10}^{-19}C脳7脳{10}^{-10}m}{{\left(3脳{10}^{-8}m\right)}^3} \\ =9脳{10}^{9}N.m^2/C^2脳\frac{-4.48脳{10}^{-28}C.m}{2.7脳{10}^{-23}{m}^3} \\ =9脳{10}^{9}N.m^2/C^2脳1.65脳{10}^{-5}C/m^2 \\ =1.49脳{10}^{5}N/C \end{gathered}$$

The equation of the force on the electron due to the dipole is expressed as:

$F=q_{2}E$

Here, E is the magnitude of the electric field of the electron, and $q_2$is the charge of the charged particles closer to the electron.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{F}=2脳1.6脳0^{-19}\mathrm{C}脳-1.49脳{10}^5\mathrm{N}/\mathrm{C} \\ =3.2脳{10}^{-19}\mathrm{C}脳1.49脳{10}^5\mathrm{N}/\mathrm{C} \\ =-4.76脳{10}^{-14}\mathrm{N} \end{gathered}$$

As the electron is closer to the positive charge, the electric field is upward and it has a positive component in the y axis and the direction of the force is positive.

Thus, the force on the electron due to the dipole is $4.76脳{10}^{-14}\mathrm{N}$