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

Options

A. `(-Q)/(4)(1+2sqrt(2))`
B. `(Q)/(4)(1+2sqrt(2))`
C. `(-Q)/(2)(1+2sqrt(2))`
D. `(Q)/(2)(1+2sqrt(2))`

Answer

b

Explanation

Let a be side of the square ABCD.
Diagonal `AC=sqrt(a^(2)+a^(2))=asqrt(2)`
`thereforeOA=OC=(asqrt(2))/(2)=(a)/(sqrt(2))`
The systen is in equilibrium means the force experienced by each charge is zero . It is clear that charge placed at centre would be at equilibrium for any value of q, so we are considering the equilibrium of charge placed at any comer (sayA).

Hence, `vecF_(A)=vecF_(AB)+vecF_(AD)+vecF_(AC)+vecF_(AO)=vec0`
`vecF_(AB)=(1)/(4piepsilon_(0))(Q^(2))/(a^(2))` along `BA,vecF_(AC)=(1)/(4piepsilon_(0))(Q^(2))/(2a^(2))` along CA
`vecF_(AD)=(1)/(4piepsilon_(0))(Q^(2))/(a^(2))` along `DA,vecF_(AO)=-(1)/(4piepsilon_(0))(2Qq)/(a^(2))` along AO
Resultant of `vecF_(AB)andvecF_(AD)=(1)/(4piepsilon_(0))(Q^(2))/(a^(2))-(1)/(4piepsilon_(0))(2Qq)/(a^(2))=0`
or `sqrt(2)Q+(Q)/(2)-2q=0`
`2q=(Q)/(2)[2sqrt(2)+1]rArrq=(Q)/(4)[2sqrt(2)+1]`