Question

Two point charges $2q$ and $8q$ are placed at a distance $r$ apart. Where should a third charge $-q$ be placed between them so that the electrical potential energy of the system is a minimum

• A. at a distance of $\frac{r}{3}$ from $2q$
• B. at a distance of $\frac{2r}{3}$ from $2q$
• C. at a distance of $\frac{r}{16}$ from $2q$
• D. None of the above

Answer 1

The correct option is D None of the above

Let the third charge $-q$ is placed at a distance of $x$ from the charge $2q$. Then potential energy of the system is
$U=k[\frac{\left(2q\right)\left(8q\right)}{r}-\frac{\left(2q\right)\left(q\right)}{x}-\frac{\left(8q\right)\left(q\right)}{r-x}]$
Here, $k=\frac{1}{4\pi {ϵ}_0}$
$U$ is minimum, where
$\frac{2}{x}+\frac{8}{r-x}$ is maximum.
Let $\frac{2}{x}+\frac{8}{r-x}=y$
For $y$ to be maximum $\frac{dy}{dx}=0$
or $-\frac{2}{x^2}+\frac{8}{(r-x{)}^2}=0$
or $\frac{x}{r-x}=\sqrt{\frac{2}{8}}=\frac{1}{2}$
or $x=\frac{r}{3}$
But at $x=\frac{r}{3},\frac{d^{2}y}{d{x}^2}=$ positive
i.e., at $x=\frac{r}{3}$, $y$ is minimum or $U$ is maximum. So, there can't be any point between them where potential energy is minimum.