A dipole is located at the origin and is composed of charged particles with charge $\mathbf{+}\mathbf{2}\mathbf{e}\mathbf{}\mathbf{and}\mathbf{}\mathbf{-}\mathbf{2}\mathbf{e}$, separated by a distance $\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{m}$along the y axis. The $\mathbf{+}\mathbf{2}\mathbf{e}$charge is on the $+\mathrm{y}$axis. Calculate the force on a proton at a location $\left(0,0,3\pm {10}^{-8}\right)\mathbf{m}$due to this dipole.

The given data is listed below,

• Separation between the charged particles,$\mathrm{s}=2\times {10}^{-10}\mathrm{m}$
• Location of charged particles,$\mathrm{r}=\left(0,0,3\times {10}^{-8}\right)\mathrm{m}$

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 p in the above expression.

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

Here, k is Coulomb’s constant with value $9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2$, s 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 $$\begin{gathered} 9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\mathrm{for}\mathrm{k},2\times {10}^{-10}\mathrm{m}\mathrm{for}\mathrm{s},3\times {10}^{-8}\mathrm{m}\mathrm{for}\mathrm{r},2\times 1.6\times {10}^{-19}\mathrm{C} \\ \mathrm{for}\mathrm{q}. \end{gathered}$$

$$\begin{gathered} \mathrm{E}=\left(9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(\frac{4\times 1.6\times {10}^{-19}\mathrm{C}\times 2\times {10}^{-10}\mathrm{m}}{{\left(3\times {10}^{-8}\mathrm{m}\right)}^3}\right) \\ =\left(9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(\frac{1.28\times {10}^{-28}\mathrm{C}.\mathrm{m}}{\left(2.7\times {10}^{-23}{\mathrm{m}}^3\right)}\right) \\ =\left(9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\times 4.74\times {10}^{-6}\mathrm{C}/{\mathrm{m}}^2 \\ =42660\mathrm{N}/\mathrm{C} \end{gathered}$$

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 and E is the magnitude of the electric field.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{F}=1.6\times {10}^{-19}\mathrm{C}\times 42660\mathrm{N}/\mathrm{C} \\ =6.82\times {10}^{-15}\mathrm{N} \end{gathered}$$

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

Thus, the force of the proton is .