Explanation:
$\text{V}=\frac{\text{KQ}}{\text{r}}$
V → Electric Potential
$\text{V}_\text{p}=\frac{\text{KQ}}{\text{x}}+\frac{\text{KQ}}{\text{r}-\text{x}}=\frac{\text{KQ}\text{r}}{\text{x}(\text{r}-\text{x})}$
$\frac{\text{dvp}}{\text{dx}}=\frac{-\text{kQr}(\text{r}-2\text{x})}{\big(\text{r}(\text{r}-\text{x})\big)^2}=0$
$\text{r}=2\text{x},\ \text{x}=\frac{\text{r}}{2}$
$\text{V}_\text{Pmin}=\frac{\text{KQ}}{\frac{\text{r}}{2}}+\frac{\text{KQ}}{\frac{\text{r}}{2}}=\frac{4\text{KQ}}{\text{r}}$ at $\Big(\text{x}=\frac{\text{r}}{2}\Big)$
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