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 $\dfrac{r}{3}$ form 2q
B. At a distance of $\dfrac{2r}{3}$ from 2q
C. At a distance of $\dfrac{r}{16}$ from 2q
D. None of the above

Answer 1

The electrical potential energy of the system is the sum of potential energies of all the charge pairs in the system. To find that state we can find the expression of electrical potential energy by putting the -q charge at some arbitrary distance and then finding the value of that distance at which it will be minimum.
Formula used:
Electrical potential energy
$\dfrac{1}{4\pi {{\varepsilon }_{o}}}\dfrac{{{q}_{1}}{{q}_{2}}}{r}$

Complete step by step answer:
Let us assume that the -q charge is placed at a distance x from the 2q charge.

We will find the potential energies of all the charge pairs in the system starting with the 2q and the 8q charge pair.
$\dfrac{1}{4\pi {{\varepsilon }_{o}}}\dfrac{2q\times 8q}{r}=\dfrac{1}{4\pi {{\varepsilon }_{o}}}\dfrac{16{{q}^{2}}}{r}$
Now we will take the 2q and the -q charge pair.
$\dfrac{1}{4\pi {{\varepsilon }_{o}}}\dfrac{2q\times (-q)}{x}=-\dfrac{1}{4\pi {{\varepsilon }_{o}}}\dfrac{2{{q}^{2}}}{x}$
And finally, for the 8q and the -q charge pair. The distance between them will be r-x
$\dfrac{1}{4\pi {{\varepsilon }_{o}}}\dfrac{(-q)\times 8q}{r-x}=-\dfrac{1}{4\pi {{\varepsilon }_{o}}}\dfrac{8{{q}^{2}}}{r-x}$
Now we will take their sum and solve for the minimum value of potential energy.
Total potential energy of the system = $\dfrac{1}{4\pi {{\varepsilon }_{o}}}\dfrac{16{{q}^{2}}}{r}-\dfrac{1}{4\pi {{\varepsilon }_{o}}}\dfrac{2{{q}^{2}}}{x}-\dfrac{1}{4\pi {{\varepsilon }_{o}}}\dfrac{2{{q}^{2}}}{x}=\dfrac{2{{q}^{2}}}{4\pi {{\varepsilon }_{o}}}(\dfrac{8}{r}-\dfrac{1}{x}-\dfrac{4}{r-x})$

The extremes of this equation will exist at values where the derivative of this expression will be equal to zero. We can ignore the value outside the bracket as it is a constant and only consider the expression inside the bracket. Its derivative will be

$\dfrac{d}{dx}(\dfrac{8}{r}-\dfrac{1}{x}-\dfrac{4}{r-x})=0-(-\dfrac{1}{{{x}^{2}}})-(\dfrac{-4}{{{(r-x)}^{2}}}(-1))=\dfrac{1}{{{x}^{2}}}-\dfrac{4}{{{(r-x)}^{2}}}$
Solving for x when this expression is zero

&\dfrac{1}{{{x}^{2}}}-\dfrac{4}{{{(r-x)}^{2}}}=0\\\&\dfrac{{{(r-x)}^{2}}-4{{x}^{2}}}{{{x}^{2}}{{(r-x)}^{2}}}=\dfrac{{{r}^{2}}+{{x}^{2}}-2rx-4{{x}^{2}}}{{{x}^{2}}{{(r-x)}^{2}}}=\dfrac{{{r}^{2}}-2rx-3{{x}^{2}}}{{{x}^{2}}{{(r-x)}^{2}}}=0 \\\ \end{aligned}$$ Solving for numerator only as the expression will be zero when the numerator is zero $$\begin{aligned} & {{r}^{2}}-2rx-3{{x}^{2}}=0\Rightarrow 3{{x}^{2}}+2rx-{{r}^{2}}=0 \\\ & x=\dfrac{-2r\pm \sqrt{4{{r}^{2}}-4(3)(-{{r}^{2}})}}{2(3)}=\dfrac{-2r\pm \sqrt{4{{r}^{2}}+12{{r}^{2}}}}{6}=\dfrac{-2r\pm \sqrt{16{{r}^{2}}}}{6}=\dfrac{-2r\pm 4r}{6} \\\ & x=\dfrac{2r}{6}=\dfrac{r}{3} \\\ & x=\dfrac{-6r}{6}=-r \\\ \end{aligned}$$ There can be two values of x we will take the positive value and ignore the negative values as the negative values means the -q charge will be at a distance of r from the 2q charge along the line joining the 2q and the 8q charges but in the other direction from the 8q charge. But in the question, it is given that the charge is placed somewhere in between both the charges. **So, the correct answer is “Option A”.** **Note:** For a shortcut method we can ignore the potential energy due to the 2q and 8q charge pair as that remains constant and as can be seen in the solution, its expression also turns to zero when we take the derivative. An alternative way to solve the question is that we can find the place where the -q charge will be in equilibrium as when the system is in stable equilibrium, it has the least potential energy.