A charge Q is placed at each of the opposite | vedclass

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Question

If is given that $\overrightarrow{\mathrm{F}}_{1}+\overrightarrow{\mathrm{F}}_{2}+\overrightarrow{\mathrm{F}}_{\mathrm{D}}=0$

where $\overrightarrow{

Vedclass · Accepted Answer

$-2\sqrt 2$

Vedclass · Answer

If is given that $\overrightarrow{\mathrm{F}}_{1}+\overrightarrow{\mathrm{F}}_{2}+\overrightarrow{\mathrm{F}}_{\mathrm{D}}=0$

where $\overrightarrow{F}_{\mathrm{A}}+\overrightarrow{\mathrm{F}}_{\mathrm{B}}+\overrightarrow{\mathrm{F}}_{\mathrm{D}}=0$ are the forces applied by charges placed at $A, B$ and $D$ on the charge placed at $C$

$\Rightarrow \overrightarrow{\mathrm{F}}_{\mathrm{B}}+\overrightarrow{\mathrm{F}}_{\mathrm{D}}=-\overrightarrow{\mathrm{F}}_{\mathrm{A}}$

$\left|\overrightarrow{\mathrm{F}}_{\mathrm{B}}+\overrightarrow{\mathrm{F}}_{\mathrm{D}}ight|=\sqrt{2} \frac{\mathrm{kqQ}}{\mathrm{a}^{2}}$ and $\left|\overrightarrow{\mathrm{F}}_{\mathrm{A}}ight|=\frac{\mathrm{kQ}^{2}}{2 \mathrm{a}^{2}}$

$	herefore $ $\sqrt{2} \frac{\mathrm{k} \mathrm{qQ}}{\mathrm{a}^{2}}=\frac{\mathrm{k} \mathrm{Q}^{2}}{2 \mathrm{a}^{2}} \Rightarrow \frac{\mathrm{Q}}{\mathrm{q}}=-2 \sqrt{2}$

A charge $Q$ is placed at each of the opposite corners of a square. A charge $q$ is placed at each of the other two corners. If the electrical force on $Q$ is zero, then $Q/q$ equals

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