(II) How much work must be done to bring three electrons from a great distance apart to \({\bf{1}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 10}}}}\;{\bf{m}}\) from one another (at the corners of an equilateral triangle)?

The value of the work done can be calculated by multiplying the value of the magnitude of the charge with the difference in the potential. Its value is altered linearly to the value of the magnitude of the charge.

The distance is,\(L = 1.0 \times {10^{ - 10}}\;{\rm{m}}\).

The charge on the electron is, \(e = - 1.6 \times {10^{ - 19}}\;{\rm{C}}\)

The schematic diagram for the problem can be drawn as:

Initially, there is no other charge is present. Thus, the work done to bring the first electron from its initial position will be:

\({W_1} = 0\)

The expression for the potential difference between the electrons 1 and 2 can be written as:

\(\begin{aligned}\Delta {V_{12}} &= \frac{{k\left( { - e} \right)}}{L}\\\Delta {V_{12}} &= \frac{{ - ke}}{L}\end{aligned}\)

The expression for the work done in bringing the second electron of charge \( - e\)with a distance \(L\) to the first electron is as follows:

\(\begin{aligned}{W_2} &= \left( { - e} \right)\Delta {V_{12}}\\{W_2} &= \left( { - e} \right)\left( {\frac{{ - ke}}{L}} \right)\\{W_2} &= \frac{{k{e^2}}}{L}\end{aligned}\)

The expression for the potential difference between the electrons 1 and 3 can be written as:

\(\begin{aligned}\Delta {V_{13}} &= \frac{{k\left( { - e} \right)}}{L}\\\Delta {V_{13}} &= \frac{{ - ke}}{L}\end{aligned}\)

The expression for the potential difference between the electrons 2 and 3 can be written as:

\(\begin{aligned}\Delta {V_{23}} &= \frac{{k\left( { - e} \right)}}{L}\\\Delta {V_{23}} &= \frac{{ - ke}}{L}\end{aligned}\)

The expression for the work done in bringing the third electron to the position with the equal distance of the charge 1 and 2 is as follows:

\(\begin{aligned}{W_3} &= \left( { - e} \right)\left( {\frac{{ - ke}}{L}} \right) + \left( { - e} \right)\left( {\frac{{ - ke}}{L}} \right)\\{W_3} &= \frac{{2k{e^2}}}{L}\end{aligned}\)

The expression for the work done in bringing the third electron to the position with the equal distance of the charge 1 and 2 is as follows:

\(\begin{aligned}{W_3} &= \left( { - e} \right)\left( {\frac{{ - ke}}{L}} \right) + \left( { - e} \right)\left( {\frac{{ - ke}}{L}} \right)\\{W_3} &= \frac{{2k{e^2}}}{L}\end{aligned}\)

The total work done in bringing the three charges from a greatest distance from one another can be calculated as:

\(\begin{aligned}W &= {W_1} + {W_2} + {W_3}\\W &= 0 + \frac{{k{e^2}}}{L} + \frac{{2k{e^2}}}{L}\\W &= \frac{{3k{e^2}}}{L}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}W &= \frac{{\left( 3 \right)\left( {9 \times {{10}^9}\;{{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} \mathord{\left/{\vphantom {{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} {{{\rm{C}}^{\rm{2}}}}}} \right.} {{{\rm{C}}^{\rm{2}}}}}} \right){{\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)}^2}}}{{\left( {1.0 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}}\\W &= 6.9 \times {10^{ - 18}}\;{\rm{J}}\end{aligned}\)

Thus, the total work done in bringing the three charges from a great distance apart to \(1.0 \times {10^{ - 10}}\;{\rm{m}}\) from one another is \(6.9 \times {10^{ - 18}}\;{\rm{J}}\).