In Fig. 24-67, we move a particle of charge $\mathbf{+}\mathbf{2}\mathbf{e}$in from infinity to the x-axis. How much work do we do? Distance $\mathbf{D}$ is$\mathbf{\text{4.00m}}$.

The distance, $\mathrm{D}=4.00\text{m}$

The charge, $\mathrm{q}=+2\mathrm{e}$

Potential due to the system of charge and the relation between potential and work done.

Formulae:

The work done is defined by,

$\mathrm{W}=\mathrm{q}({\mathrm{V}}_{\mathrm{net}}-{\mathrm{V}}_{\mathrm{\infty }})$

Here, $\mathrm{q}$ is the charge, ${\mathrm{V}}_{\mathrm{net}}$ is the net potential, and ${\mathrm{V}}_{\mathrm{\infty }}$ is the potential at infinity which is zero.

The potential is defined by,

${\mathrm{V}}_{\mathrm{net}}=\frac{\mathrm{kq}}{\mathrm{D}}$

Here, $\mathrm{k}$ is the Coulomb’s constant having a value $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ and $\mathrm{q}$ is the electric charge having a value $1.60\times {10}^{-19}\text{C}$.

Note that the net potential (due to the "fixed" charges) is zero at the first location ("at $\mathit{\infty }$") being considered for the movable charge $\mathrm{q}$ (where $\mathrm{q}=+2\mathrm{e}$).

The net potential on the x axis due to the other 2 charged is,

$\begin{array}{c}{\mathrm{V}}_{\mathrm{net}}=\frac{\mathrm{K}\left(2\mathrm{e}\right)}{2\mathrm{D}}+\frac{\mathrm{K}\left(\mathrm{e}\right)}{\mathrm{D}}\\ =\frac{2\mathrm{Ke}}{\mathrm{D}}\\ =\frac{\left(2\right)(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}})(1.60\times {10}^{-19}\text{C})}{4.00\text{m}}\\ =7.19\times {10}^{-10}\text{V}\end{array}$

Thus, the required work is as below.

$\begin{array}{c}\mathrm{W}=\mathrm{q}({\mathrm{V}}_{\mathrm{net}}-{\mathrm{V}}_{\mathrm{\infty }})\\ =\left(2\mathrm{e}\right)(7.19\times {10}^{-10}\text{V}-0)\\ =2.30\times {10}^{-28}\text{J}\end{array}$

Hence, the required work is$2.30\times {10}^{-28}\text{J}$ .