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

Options

A. `-Q/4(1+2sqrt2)`
B. `Q/4(1+2sqrt2)`
C. `-Q/2(1+2sqrt2)`
D. `Q/2(1+2sqrt2)`

Answer

B

Explanation

The system is in equilibrium net force on any charge will be zero. Consider any corner’s charge
Resultant of all three forces is `(sqrt2 kQ^2)/a^2+(kQ^2)/(sqrt2 a)^2`
The must be balanced by attraction of charge at centre
`(kQq)/((1/2 sqrt2a))= (kQq)/(a^2//2)`
On solving we get `(2kQq)/a^2= (sqrt2 kQ^2)/a^2+(kQ^2)/(2a^2)`
`2q=sqrt2 Q+Q/2 implies q=Q/sqrt2 +Q/4 (1/sqrt2+1/4)=Q/4(1+2sqrt2)`