Question

Three charges $Q$, $+q$ and $+q$ are placed at the vertices of a right angled isosceles triangle as shown in the figure. The net electrostatic energy of the configuration is zero if $Q$ is equal to :

• A. $\frac{-q}{1+\sqrt{2}}$
• B. $\frac{-\sqrt{2}q}{1+\sqrt{2}}$
• C. $-2q$
• D. $+q$

Answer 1

The correct option is B $\frac{-\sqrt{2}q}{1+\sqrt{2}}$

We know P.E between two charges $=\frac{k{q}_1{q}_2}{r.}$

Now total P. E$=P\cdot E_{AB}+P\cdot E_{AC}+P\cdot E_{BC}.$

$P\cdot E_{ta}=\frac{kQq}{\sqrt{2}a}+\frac{kQq}{a}+\frac{k{q}^2}{a}$

$0=\frac{kq}{a}(\frac{Q}{\sqrt{2}}+Q+q)$

$\Rightarrow 0=Q(1+\frac{1}{\sqrt{2}})+q$

$\Rightarrow -q=Q\left(\frac{\sqrt{2}+1}{\sqrt{2}}\right)$

$\Rightarrow Q=\frac{-\sqrt{2}q}{\sqrt{2}+1}$