Question

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

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

Answer 1

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

As triangle is isosceles right angle, so $AB=BC=a$ and $AC=\sqrt{2}a$

The net electrostatic energy of the system is $U=U_{AB}+U_{BC}+U_{AC}=0$

or $\frac{1}{4\pi {ϵ}_0}(\frac{Qq}{a}+\frac{q^2}{a}+\frac{Qq}{\sqrt{2}a})=0$

or $Q(1+\frac{1}{\sqrt{2}})=-q$

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