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 minimum?

• A. At a distance of $r/3$ from $2q$
• B. At a distance of $2r/3$ from $2q$
• C. At a distance of $r/16$ from $2q$
• D. None of the above

Answer 1

Answer: (D)

None of the above

Let the third charge $-q$ is placed at a distance of $x$ from the charge $2q$. Then, the potential energy of the system is

$U=k[\frac{2q\left(8q\right)}{r}-\frac{\left(2q\right)\left(q\right)}{x}-\frac{\left(8q\right)\left(q\right)}{r-x}]$

Here, $k=\frac{1}{4\pi {\epsilon }_0}$

$U$ 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, $\frac{dy}{dx}=0$

$-\frac{2}{x^2}+\frac{8}{{(r-x)}^2}=0$

$\Rightarrow \frac{x}{r-x}=\sqrt{\frac{2}{8}}=\frac{1}{2}\Rightarrow x=\frac{r}{3}$

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 cannot be any point between them where the potential energy is minimum.