Three identical charges, each with a value of 1.0 × 10^ -8 C , are...

Due to the symmetry of the equilateral triangle, the electric field vectors at the centroid due to the three identical charges are equal in magnitude and separated by an angle of $120^{{}^{{}^{\circ}}}$. Their vector sum is zero. Thus, the net electric field at the centre is zero.

The distance $r$ from any corner to the centroid of an equilateral triangle of side $a$ is given by $r = \dfrac{a}{\sqrt{3}}$.

Given $a = 20 \text{ cm} = 0.2 \text{ m}$, we have $r = \dfrac{0.2}{\sqrt{3}} \text{ m}$.

The electric potential is a scalar quantity. The net potential at the centre is the sum of the potentials due to the three charges:

$V = 3 \times \dfrac{1}{4\pi\varepsilon_0} \dfrac{q}{r}$

Substituting the given values:

$V = 3 \times \dfrac{9 \times 10^9 \times 1.0 \times 10^{-8}}{\dfrac{0.2}{\sqrt{3}}}$

$V = \dfrac{270 \sqrt{3}}{0.2} = 1350 \sqrt{3}$

Using $\sqrt{3} \approx 1.732$:

$V \approx 1350 \times 1.732 \approx 2338 \text{ V} \approx 2.3 \times 10^3 \text{ V}$