(a) what is the electric potential energy of two electrons separated by 2.00 nm? (b) If the separation increases, does the potential energy increase or decrease?

1. Separation of the two electrons,$\mathrm{d}=2.0\times {10}^{-9}\mathrm{m}$
2. Charge of the electron,$\mathrm{e}=-1.6\times {10}^{-19}\mathrm{C}$

Using the concept of the electric potential energy, we can get the value of the potential energy due to the charges present in the system. Using this formula, we can see that the separation value is inversely proportional to the potential energy of the system.

Formula:

The electric potential energy of the system due to the point charges,$\mathrm{U}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{d}}$ (i)

Using the given data in equation (i), we can get the electric potential energy of the system as follows:

$\begin{array}{rcl}\mathrm{U}& =& \left(9.0\times {10}^9\right)\times \frac{\left(1.6\times {10}^{-19}\right)\left(1.6\times {10}^{-19}\right)}{\left(2\times {10}^{-9}\right)}\\ & =& 1.152\times {10}^{-19}\mathrm{J}\end{array}$

Hence, the value of the energy is $1.152\times {10}^{-19}\mathrm{J}$.

As U>0, and from the above expression, we can get the potential energy and separation relation as follows:

$\mathrm{U}\propto \frac{1}{\mathrm{d}}$

Thus, by increasing the value of separation distance, the value of potential energy decreases.