Charges Q,+q,and +q are placed at the vertice | vedclass

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(B) The total electrostatic potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $

Vedclass · Accepted Answer

$\frac{-\sqrt{2}q}{1 + \sqrt{2}}$

Vedclass · Answer

(B) The total 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,the distances between the charges are $a$,$a$,and $a\sqrt{2}$.The potential energy is $U = \frac{kQq}{a} + \frac{kQq}{a} + \frac{kq^2}{a\sqrt{2}} = 0$.Dividing by $k/a$,we get $2Qq + \frac{q^2}{\sqrt{2}} = 0$.$2Qq = -\frac{q^2}{\sqrt{2}}$.$Q = -\frac{q^2}{2\sqrt{2}q} = -\frac{q}{2\sqrt{2}}$.Wait,re-evaluating the geometry: The charges are $Q$ at the top vertex,and $+q, +q$ at the base vertices separated by distance $a$. The distance from $Q$ to each $+q$ is $a$. The distance between the two $+q$ charges is $a\sqrt{2}$.$U = \frac{kQq}{a} + \frac{kQq}{a} + \frac{kq^2}{a\sqrt{2}} = 0$.$2Qq + \frac{q^2}{\sqrt{2}} = 0 \implies 2Q = -\frac{q}{\sqrt{2}} \implies Q = -\frac{q}{2\sqrt{2}}$.However,looking at the options,the intended calculation likely assumes the distance from $Q$ to one $+q$ is $a$ and to the other is $a\sqrt{2}$ (if $Q$ is at a vertex of the right angle). Let's assume the vertices are $(0,a), (0,0), (a,0)$. $Q$ at $(0,a)$,$+q$ at $(0,0)$,$+q$ at $(a,0)$. Distances are $a, a, a\sqrt{2}$.$U = \frac{kQq}{a} + \frac{kQq}{a\sqrt{2}} + \frac{kq^2}{a} = 0$.$Q(1 + \frac{1}{\sqrt{2}}) + q = 0 \implies Q(\frac{\sqrt{2}+1}{\sqrt{2}}) = -q \implies Q = -\frac{\sqrt{2}q}{\sqrt{2}+1}$.This matches option $B$.

Charges $Q$,$+q$,and $+q$ are placed at the vertices of an isosceles right-angled triangle as shown in the figure. If the total electrostatic potential energy of the system is zero,then $Q = $ . . . . . . .

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