# Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if Q is equal to  Option 1) $\frac{-q}{1 +\sqrt2}$ Option 2) $\frac{-\sqrt2 q}{1 +\sqrt2}$ Option 3) -2q Option 4) +q

V Vakul

As we have learnt,

Potential Energy Of System Of two Charge -

$\dpi{100} U=\frac{kQ_{1}Q_{2}}{r}$  $\dpi{100} \left ( S.I \right )$

$\dpi{100} U= \frac{Q_{1}Q_{2}}{r}$  $\dpi{100} \left ( C.G.S \right )$

- wherein

$\dpi{100} K=\frac{1}{4\pi \epsilon _{0}}$

Potential energy of the configuration is

$U = k\cdot\frac{Qq}{a} + k\frac{q^2}{a} + k\frac{qQ}{a\sqrt2} = 0 \Rightarrow Q = \frac{-\sqrt2 q}{\sqrt2 + 1}$

Option 1)

$\frac{-q}{1 +\sqrt2}$

Option 2)

$\frac{-\sqrt2 q}{1 +\sqrt2}$

Option 3)

-2q

Option 4)

+q

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