Question

Three electrons form an equilateral triangle with 1.0 nm on each side. A proton is at the center of the triangle. What is the potential energy of this group of charges?

Short answer

The potential energy of this group of charges is$-1.01·{10}^{-18}\mathrm{J}$

Step-by-step solution

Step 1. Given information

Three electrons form an equilateral triangle with 1.0 nm on each side. A proton is at the center of the triangle.

Step 2. Simplify

The total energy will be the energy of the three electrons, plus the energy of the proton.

By symmetry, we only need to find the energy of one electron, and it will be the same as the three others.

An equilateral triangle whose side a is $\frac{a\sqrt{3}}{3}$away from the vertexes.

This will allow us to calculate the potential as Calculate the potential as

$V=-k\left(\frac{e}{a}+\frac{e}{a}-\frac{e}{a\sqrt{3}/3}\right)$

Step 3. Simplify

Which we can simplify to

$V=\frac{-ke}{a}·(2-\sqrt{3})$

The energy of one electron will therefore be

$E_e=\frac{k{e}^2}{a}(2-\sqrt{3})$

Due to symmetry, the potential at the position of the proton will be thrice the potential coming from one electron, which in turn is

$V=\frac{-ke}{a\sqrt{3}/3}=\frac{-ke\sqrt{3}}{a}$

The energy of the proton will therefore be

$E_p=3eV=\frac{-3k{e}^2\sqrt{3}}{a}$

Step 4. Getting the answer

Summing up the two energies we get the total energy to be

$E=3E_e+E_p=\frac{3k{e}^2}{a}(2-\sqrt{3})+\frac{-3k{e}^2\sqrt{3}}{a}=\frac{k{e}^2}{a}·6(1-\sqrt{3})$

Numerically, this value will be

$E=\frac{6·9·{10}^9·{\left(1.6·{10}^{-19}\right)}^2}{1·{10}^{-9}}(1-\sqrt{3})=-1.01·{10}^{-18}\mathrm{J}$