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 $10\times \left|\frac{Q}{q}\right|$ ? $\left[\right.$ Take $\sqrt{2}=1.4\left]\right.$

Answer 1

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, $K=\frac{1}{4 \pi \varepsilon_{0}}$
Putting $U=0$ , we get, $\frac{Q}{q}=\frac{- 2}{2 + \sqrt{2}}\times \frac{2 - \sqrt{2}}{2 - \sqrt{2}}=\frac{4 - 2 \sqrt{2}}{2}$
$=-\left(2 - \sqrt{2}\right)=-0.6$
$\Rightarrow 10\left|\frac{Q}{q}\right|=6$