Initially two electrons are fixed in place with a separation of $\mathbf{\text{2.00μm}}$. How much work must we do to bring the third electron in from infinity to complete an equilateral triangle?

Distance between two fixed electrons,$\mathrm{d}=\text{2μm}=2\times {10}^{-6}\text{m}$.

The shape of the system is, Equilateral triangle.

Electric potential, the amount of work required to move a unit charge from a reference point to a specific point against an electric field.

Potential due to system of charges.

Formulae:

The work done is define by,

$\mathrm{W}=(-\mathrm{e}){\mathrm{V}}_{\mathrm{p}}$

The electric potential is defined by,

${\mathrm{V}}_{\mathrm{P}}=\frac{\mathrm{k}(-\mathrm{e})}{\mathrm{d}}+\frac{\mathrm{k}(-\mathrm{e})}{\mathrm{d}}$

Where, $\mathrm{W}$ is the work done, $\mathrm{V}$ is potential energy, $\mathrm{k}$ is the Coulomb’s constant having a value $9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$, and $\mathrm{e}$ is the electric charge having value $1.6\times {10}^{-19}\text{C}$.

The net potential at point P (the place where we are to place the third electron) due to the fixed charges is computed using,

$\begin{array}{c}{\mathrm{V}}_{\mathrm{P}}=\frac{\mathrm{k}(-\mathrm{e})}{\mathrm{d}}+\frac{\mathrm{k}(-\mathrm{e})}{\mathrm{d}}\\ =\frac{2\mathrm{k}(-\mathrm{e})}{\mathrm{d}}\\ =\frac{2(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2)(1.6\times {10}^{-19}\text{C})}{\text{2}\times {\text{10}}^{-6}\text{m}}\\ =1.44\times {10}^{-3}\text{V}\end{array}$

Then the required “applied” work is,

$\begin{array}{c}\mathrm{W}=(-\mathrm{e}){\mathrm{V}}_{\mathrm{p}}\\ =(-1.6\times {10}^{-19}\text{C})(1.44\times {10}^{-3}\text{V})\\ =2.30\times {10}^{-22}\text{J}\end{array}$

Hence, the required work is $2.30\times {10}^{-22}\text{J}$.