In Fig. 24-66, point P is at distance${\mathit{d}}_{\mathbf{1}}\mathbf{}\mathbf{=}\mathbf{4}\mathbf{.00}\mathbf{\text{m}}$ from particle 1$\mathbf{(}{\mathbf{\text{q}}}_{\mathbf{1}}\mathbf{-}\mathbf{\text{2e}}\mathbf{)}$ and distance${\mathit{d}}_{\mathbf{2}}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.00}\mathbf{\text{m}}$ from particle 2 $\mathbf{(}{\mathbf{q}}_{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{e}\mathbf{)}$, with both particles fixedin place.

(a) With $\mathit{V}\mathbf{=}\mathbf{}\mathbf{0}$at infinity, what is V at P? If we bring a particle ofcharge${\mathit{q}}_{\mathbf{3}}\mathbf{+}\mathbf{}\mathbf{2}\mathit{e}$ from infinity to P,

(b) how much work do we do and

(c) what is the potential energy of the three-particle system?

After reading the question,

From the figure:

The values of the charges are as below.

$q_1=-2e$and $q_2=+2e$

The distances are given below.

$d_1=4.0\text{m}$and $d_2=2.0\text{m}$

Potential due to the group of charges and Potential energy of the system of charges.

Formulae are as follow:

The potential is define by,

$V_{net}=\frac{1}{4\pi {\epsilon }_o}\left[\frac{q_1}{d_1}+\frac{q_2}{d_2}\right]$

The potential energy is,

$U=U_{13}+U_{23}+U_{12}$

Here, U is total potential energy, d is distance, q is charge, and V is net potential.

At infinity, the potential $V=0$.

Determine the net potential as below.

$\begin{array}{c}{V}_{net}=V_1+V_2\\ =\frac{1}{4\pi {\epsilon }_o}\left[\frac{q_1}{d_1}+\frac{q_2}{d_2}\right]\end{array}$

Substitute known values in the above equation.

$\begin{array}{c}{V}_{net}=9.0\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\left[\frac{-2e}{4.0\text{m}}+\frac{2e}{2.0\text{m}}\right]\\ =9.0\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\times \left(\frac{e}{2}\right)\\ =9.0\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\times (\frac{1}{2}\times 1.6\times {10}^{-19})\\ =7.2\times {10}^{-10}\text{Volt}\end{array}$

Hence, the net potential is,${\text{7.2×10}}^{\text{-10}}\text{Volt}$.

If the charge, $q_3=+2e$

Then the work done is,

$\begin{array}{c}W=V_{net}{q}_3\\ =V_{net}(+2e)\\ =\left({\text{7.2×10}}^{\text{-10}}\text{Volt}\right)\times (2\times 1.6\times {10}^{-19}\text{C})\\ =2.3\times {10}^{\text{-28}}\text{J}\end{array}$

Therefore, the work done is $2.3\times {10}^{-28}\text{J}$.

The total electric potential energy of the system:

$\begin{array}{c}U=U_{13}+U_{23}+U_{12}\\ =\frac{1}{4\pi {\epsilon }_o}[\frac{q_1{q}_2}{d_1}+\frac{q_1{q}_2}{d_2}+\frac{q_1{q}_2}{\sqrt{d_1^2+d_2^2}}]\end{array}$

Putting known values in the above equation, and you have

$\begin{array}{c}U=9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left[\frac{(-2e)\left(2e\right)}{4.0\text{m}}+\frac{\left(2e\right)\left(2e\right)}{2.0\text{m}}+\frac{(-2e)\left(2e\right)}{\sqrt{4^2+2^2}\text{m}}\right]\\ =(9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2)\times 4e^2\left[-0.25{\text{m}}^{-1}+0.5{\text{m}}^{-1}-0.224{\text{m}}^{-1}\right]\\ =(9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2)\times 4\times {(1.6\times {10}^{-19}\text{C})}^2\left[{\text{0.026m}}^{-1}\right]\\ =2.4\times {10}^{-29}\text{J}\end{array}$

Hence, the total electric potential energy of the system is $2.4\times {\text{10}}^{\text{-29}}\text{J}$.