1.

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

Answer»

`(-Q)/(4)(1+2sqrt(2))`
`(Q)/(4)(1+2sqrt(2))`
`(-Q)/(2)(1+2sqrt(2))`
`(Q)/(2)(1+2sqrt(2))`

Solution :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]`


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