A dipole is located at the origin and is composed of charged particles with charge $\mathbf{+}\mathbf{e}$and$\mathbf{-}\mathbf{e}$ , separated by a distance$\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{\hspace{0.33em}}\mathbf{m}$ along the y axis. The $\mathbf{+}\mathbf{e}$charge on -y axis. Calculate the force on a proton due to this dipole at a location$\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\hspace{0.33em}}\mathbf{m}$ .

The given data is listed as follows,

• Separation between the charged particles,$\mathrm{s}=6\times {10}^{-6}\hspace{0.33em}\mathrm{m}$
• Location of charged particles,$\mathrm{r}=4\times {10}^{-8}\hspace{0.33em}\mathrm{m}$

The electric field’s magnitude is directly proportional to the dipole moment and inversely proportional to the cube of the distance of the particle.

The electric field’s magnitude gives the force on a proton.

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

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

Substitute qs for q in the above expression.

$\mathrm{E}=\mathrm{k}\frac{2\mathrm{qs}}{{\mathrm{r}}^3}$

Here, k is Coulomb’s constant with value role="math" localid="1656929030530" $9\times {10}^9\mathrm{N}-{\mathrm{m}}^2/{\mathrm{C}}^2$, is the separation between charged particles, q is the charge on an electron, r is the location of charged particles and E is the magnitude of the electric field.

Substitute,$9\times {10}^9\hspace{0.33em}\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2\mathrm{for}\mathrm{k},6\times {10}^{-6}\hspace{0.33em}\mathrm{m}\mathrm{for}\mathrm{s},4\times {10}^{-8}\hspace{0.33em}\mathrm{m}\mathrm{for}\mathrm{r},1.6\times {10}^{-19}\hspace{0.33em}\mathrm{C}\mathrm{for}\mathrm{q}.$

$$\begin{gathered} \mathrm{E}=\left(9\times {10}^9\mathrm{N}-{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(\frac{2\times 1.6\times {10}^{-19}\mathrm{C}\times 6\times {10}^{-6}\mathrm{m}}{{\left(4\times {10}^{-8}\mathrm{m}\right)}^3}\right) \\ =\left(9\times {10}^9\mathrm{N}-{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(\frac{1.92\times {10}^{-24}\mathrm{C}.\mathrm{m}}{\left(6.4\times {10}^{-23}{\mathrm{m}}^3\right)}\right) \\ =\left(9\times {10}^9\mathrm{N}-{\mathrm{m}}^2/{\mathrm{C}}^2\right)\times 0.03\mathrm{C}/{\mathrm{m}}^2 \\ =2.7\times {10}^8\mathrm{N}/\mathrm{C} \end{gathered}$$

As both the charges lies in the y axis, then the proton will also lie in the y axis.

The equation of the force on a proton is expressed as:

$F=qE$

Here, q is the charge on electron with value $1.6\times {10}^{-19}\mathrm{C}$and E is the magnitude of the electric field.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{F}=1.6\times {10}^{-19}\mathrm{C}\times 2.7\times {10}^8\mathrm{N}/\mathrm{C} \\ =4.32\times {10}^{-11}\mathrm{N} \end{gathered}$$

As negatively charged dipole is closer to the proton, then the proton exerts an attraction force which will move towards the negative y axis, hence the force will also be negative.

Thus, the force of the proton is .