Question

Three point charges $Q$ , $+q$ and $+q$ are placed at the vertices of a right-angled isosceles triangle as shown in the figure. If the net electrostatic energy of the configuration is zero, then what is the value of $10\times \left|\frac{Q}{q}\right|$ ? $[$ Take $\sqrt{2}=1.4]$

Answer 1

Answer: 6

Net electrostatic energy of the configuration will be
$U=K\left[\frac{q\cdot q}{a}+\frac{Q\cdot q}{\sqrt{2}a}+\frac{Q\cdot q}{a}\right]$ Here, $K=\frac{1}{4\pi {\epsilon }_0}$
Putting $U=0$ , we get, $\frac{Q}{q}=\frac{-2}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}=\frac{4-2\sqrt{2}}{2}$
$=-\left(2-\sqrt{2}\right)=-0.6$
$\Rightarrow 10\left|\frac{Q}{q}\right|=6$