Two charges Q are placed at opposite corners | vedclass

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(C) Let the side of the square be $a$. Place the charges $Q$ at $(0, a)$ and $(a, 0)$,and charges $q$ at $(0, 0)$ and $(a, a)$.Consider the force on c

Vedclass · Accepted Answer

$-2\sqrt{2}$

Vedclass · Answer

(C) Let the side of the square be $a$. Place the charges $Q$ at $(0, a)$ and $(a, 0)$,and charges $q$ at $(0, 0)$ and $(a, a)$.Consider the force on charge $Q$ at $(a, 0)$.The forces acting on $Q$ are:$1$. Force due to $q$ at $(0, 0)$: $\vec{F}_1 = \frac{kQq}{a^2} \hat{i}$$2$. Force due to $q$ at $(a, a)$: $\vec{F}_2 = \frac{kQq}{a^2} \hat{j}$$3$. Force due to $Q$ at $(0, a)$: $\vec{F}_3 = \frac{kQ^2}{(\sqrt{2}a)^2} \left( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \right)$For the net force to be zero,the sum of components must be zero:$F_x = \frac{kQq}{a^2} + \frac{kQ^2}{2a^2} \cdot \frac{1}{\sqrt{2}} = 0$$\frac{kQq}{a^2} = -\frac{kQ^2}{2\sqrt{2}a^2}$$q = -\frac{Q}{2\sqrt{2}}$Therefore,$\frac{Q}{q} = -2\sqrt{2}$.

Two charges $Q$ are placed at opposite corners of a square. Two charges $q$ are placed at the other two corners. If the net electric force on $Q$ is zero,then $Q/q$ is equal to:

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