A charge +Q is placed at each of the diagonal | vedclass

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(D) Let the side length of the square be $a$. Consider the charge $+Q$ at one corner. The forces acting on it are:$1$. Force due to the other $+Q$ at

Vedclass · Accepted Answer

$-2 \sqrt{2}$

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(D) Let the side length of the square be $a$. Consider the charge $+Q$ at one corner. The forces acting on it are:$1$. Force due to the other $+Q$ at the diagonally opposite corner: $F_1 = \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{(\sqrt{2}a)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{2a^2}$ (repulsive,directed away from the opposite corner).$2$. Forces due to the two $-q$ charges at the adjacent corners: $F_2 = \frac{1}{4 \pi \varepsilon_0} \frac{Qq}{a^2}$ (attractive,directed towards each $-q$ charge).Since the two $-q$ charges are at equal distances $a$,the resultant force $F_2'$ of these two attractive forces is $F_2' = \sqrt{F_2^2 + F_2^2} = \sqrt{2} F_2 = \sqrt{2} \frac{1}{4 \pi \varepsilon_0} \frac{Qq}{a^2}$.For the net force on $+Q$ to be zero,the magnitude of the repulsive force $F_1$ must equal the magnitude of the resultant attractive force $F_2'$.$\frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{2a^2} = \sqrt{2} \frac{1}{4 \pi \varepsilon_0} \frac{Qq}{a^2}$$\frac{Q}{2} = \sqrt{2} q$$Q = 2\sqrt{2} q$Therefore,$\frac{Q}{-q} = -2\sqrt{2}$.However,the question asks for the ratio $\frac{+Q}{-q}$,which is $-2\sqrt{2}$.

$A$ charge $+Q$ is placed at each of the diagonally opposite corners of a square. $A$ charge $-q$ is placed at each of the other diagonally opposite corners as shown. If the net electrical force on $+Q$ is zero,then $\frac{+Q}{-q}$ is equal to

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