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 $\Big(\text{x}=\frac{\text{r}}{2}\Big)$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$A :$ The excess pressure inside a small liquid drop is more than that of a big drop.
$B :$ As the aeroplane moves fast on the runway the pressure is more on the upper surface of its wings and less on the bottom surface of the wings.