In the figure,the value of Q so that the elec | vedclass

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(C) The electrostatic potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = \s

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$\frac{2 q}{2-\sqrt{2}}$

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(C) The electrostatic potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = \sum \frac{k q_i q_j}{r_{ij}}$.For the given system with charges $-q$,$+q$,and $Q$ at the vertices of a right-angled triangle with sides $x, x$,and hypotenuse $\sqrt{x^2 + x^2} = \sqrt{2}x$,the total potential energy is:$U = k \left[ \frac{(-q)(q)}{x} + \frac{(q)(Q)}{x} + \frac{(-q)(Q)}{\sqrt{2}x} ight]$For the system to have zero potential energy,we set $U = 0$:$0 = k \left[ -\frac{q^2}{x} + \frac{qQ}{x} - \frac{qQ}{\sqrt{2}x} ight]$Dividing by $\frac{kq}{x}$ (assuming $q, x 
eq 0$):$0 = -q + Q - \frac{Q}{\sqrt{2}}$$q = Q \left( 1 - \frac{1}{\sqrt{2}} ight)$$q = Q \left( \frac{\sqrt{2} - 1}{\sqrt{2}} ight)$$Q = \frac{\sqrt{2} q}{\sqrt{2} - 1}$To match the options,multiply the numerator and denominator by $(\sqrt{2} + 1)$ or simplify differently:$Q = \frac{\sqrt{2} q (\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{(2 + \sqrt{2})q}{2 - 1} = (2 + \sqrt{2})q$Alternatively,looking at option $C$: $\frac{2q}{2-\sqrt{2}} = \frac{2q(2+\sqrt{2})}{4-2} = \frac{2q(2+\sqrt{2})}{2} = (2+\sqrt{2})q$. Thus,option $C$ is correct.

In the figure,the value of $Q$ so that the electrostatic potential energy of the system becomes zero is

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