Four charges equal to -Q are placed at the fo | vedclass

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Question

If all charges in equilibrium, system is also in equilibrium.

Charge at centre : charge $q$ is in equilibrium because no net force acting on it com

Vedclass · Accepted Answer

$\frac{Q}{4} (1 + 2 \sqrt 2 )$

Vedclass · Answer

If all charges in equilibrium, system is also in equilibrium.

Charge at centre : charge $q$ is in equilibrium because no net force acting on it comer charge$:$

If we consider the charge at corner $B$. This charge will experience following forces

$F_{A}=k \frac{Q^{2}}{a^{2}}, F_{C}=\frac{K Q^{2}}{a^{2}} F_{D}=\frac{k Q^{2}}{(a \sqrt{2})^{2}}$ and $F_{o}=\frac{k Q q}{(a \sqrt{2})^{2}}$

Force at $B$ away from the centre $=F_{A C}+F_{D}$

$=\sqrt{F_{A}^{2}+F_{C}^{2}}+F_{D}$$=\sqrt{2} \frac{k Q^{2}}{a^{2}}+\frac{k Q^{2}}{2 a^{2}}=\frac{k Q^{2}}{a^{2}}\left(\sqrt{2}+\frac{1}{2}\right)$

Force at $B$ towards the centre $=F_{o}=\frac{2 k Q q}{a^{2}}$

For equilibrium of charges at $B, F_{A C}+F_{D}=F_{o}$

$\Rightarrow \frac{K Q^{2}}{a^{2}}\left(\sqrt{2}+\frac{1}{2}\right)$$=\frac{2 K Q q}{a^{2}}=q=\frac{Q}{4}(1+2 \sqrt{2})$

Four charges equal to $-Q$ are placed at the four corners of a square and a charge $q$ is at its centre. If the system is in equilibrium, the value of $q$ is

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