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

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Question

(b) If all charges are in equilibrium , system is also in equilibrium.
Charge at centre : charge $q$ is in equilibrium because no net force acting on

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) If all charges are in equilibrium , system is also in equilibrium.
Charge at centre : charge $q$ is in equilibrium because no net force acting on it corner 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}}}{\rm{and}}\,{F_O} = \frac{{KQq}}{{{{(a\sqrt 2 )}^2}}}$
Force at $B$ away from the centre = ${F_{AC}} + {F_D}$
$ = \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{{2kQq}}{{{a^2}}}$
For equilibrium of charge at $B$, ${F_{AC}} + {F_D} = {F_O}$
$==>$ $\frac{{K{Q^2}}}{{{a^2}}}\left( {\sqrt 2 + \frac{1}{2}} \right) = \frac{{2KQq}}{{{a^2}}}$ $==>$ $q = \frac{Q}{4}\left( {1 + 2\sqrt 2 } \right)$

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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