Q²/(2√2πε₀)
Q²/(4πε₀)(1+1/√5)
Q²/(4πε₀)(1+1/√3)
Q²/(4πε₀)
The correct option is B
Q²/(4πε₀)(1+1/√5)
potential at origin =kQ/2+kQ/2+kQ/√20+kQ/√20 (potential at ∞=0)=kQ(1+1/√5)
work required to put a fifth charge Q at origin is equal to Q²/(4πε₀)(1+1/√5)