Question 32: Electrostatics & Capacitance

Question

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, then the value of $$-$$q is

Accepted Answer

Consider the equilibrium of charge Q at corner A
$${F_B} = {F_D} = {1 \over {4\pi {\varepsilon _0}}}{{{Q^2}} \over {{a^2}}} = {{k{Q^2}} \over {{a^2}}}$$ [$$\because$$ $${1 \over {4\pi {\varepsilon _0}}} = k$$]
$${F_C} = {{k{Q^2}} \over {{{(\sqrt 2 a)}^2}}}$$ [$$\because$$ $$AC = \sqrt 2 a$$]
$$ = {{k{Q^2}} \over {2{a^2}}}$$
$${F_O} = {{k\,Qq} \over {{{\left( {{a \over {\sqrt 2 }}} ight)}^2}}} = {{k\,2Qq} \over {{a^2}}}$$ [$$\because$$ $$OC = {a \over {\sqrt 2 }}$$]
For equilibrium of charge Q at corner A, the net force on this charge along AC must be zero.
$$	herefore$$ $${F_O} = {F_C} + {F_B}\cos 45^\circ  + {F_D}\cos 45^\circ $$
$$ = {F_C} + 2{F_B}\cos 45^\circ $$ [$$\because$$ $${F_B} = {F_D}$$]
$$ = {F_C} + 2{F_B} 	imes {1 \over {\sqrt 2 }}$$ [$$\because$$ $$\cos 45^\circ  = {1 \over {\sqrt 2 }}$$]
$$ \Rightarrow {{k\,2Qq} \over {{a^2}}} = {{k\,{Q^2}} \over {2{a^2}}} + {{2 	imes k{Q^2}} \over {{a^2}}} 	imes {1 \over {\sqrt 2 }}$$
$$ \Rightarrow 2Qq = {{{Q^2}} \over 2} + {{{Q^2}} \over {\sqrt 2 }} 	imes 2$$
$$ \Rightarrow 2q = {Q \over 2} + Q\sqrt 2 $$
$$ \Rightarrow q = {Q \over 4}(1 + 2\sqrt 2 )$$

Answer

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

Answer

$${Q \over 2}(2\sqrt 2  + 1)$$

Answer

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

Answer

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

Electrostatics & Capacitance

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

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