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(B) For the system to be in equilibrium,the net force on every charge must be zero.Consider a charge $-Q$ at corner $B$ of the square with side length

Vedclass · Accepted Answer

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

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(B) For the system to be in equilibrium,the net force on every charge must be zero.Consider a charge $-Q$ at corner $B$ of the square with side length $a$.The forces acting on this charge due to the other three charges at the corners are:$1$. Force due to charge at $A$: $F_A = \frac{kQ^2}{a^2}$ (along $AB$)$2$. Force due to charge at $C$: $F_C = \frac{kQ^2}{a^2}$ (along $CB$)$3$. Force due to charge at $D$: $F_D = \frac{kQ^2}{(a\sqrt{2})^2} = \frac{kQ^2}{2a^2}$ (along $DB$)The resultant force of $F_A$ and $F_C$ is $F_{AC} = \sqrt{F_A^2 + F_C^2} = \sqrt{2} \frac{kQ^2}{a^2}$.The total force away from the center is $F_{net} = F_{AC} + F_D = \sqrt{2} \frac{kQ^2}{a^2} + \frac{kQ^2}{2a^2} = \frac{kQ^2}{a^2} (\sqrt{2} + 0.5)$.For equilibrium,the force due to the central charge $q$ must balance this force:$F_O = \frac{k |Q| |q|}{(a/\sqrt{2})^2} = \frac{2kQq}{a^2}$.Equating the forces: $\frac{2kQq}{a^2} = \frac{kQ^2}{a^2} (\sqrt{2} + 0.5)$.$2q = Q (\frac{2\sqrt{2} + 1}{2})$.$q = \frac{Q}{4} (1 + 2\sqrt{2})$.Since the corner charges are negative,the central charge $q$ must be positive to provide an attractive force.

List $I$	List $II$
$(A)$ Fleming's left hand rule	$(i)$ Direction of induced current
$(B)$ Right hand thumb rule	(ii) Magnitude and direction of magnetic induction
$(C)$ Biot-Savart law	(iii) Direction of force due to magnetic induction
$(D)$ Fleming's right hand rule	(iv) Direction of magnetic lines due to current

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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