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

Answer

B

Explanation

On establishing equilibrium on charge at A
`vec(F)_(AC)=(kQ^(2))/((sqrt(2)r)^(2))[cos 45^(@)(-hat(i))+sin 45^(@)hat(j)]`
`vec(F)_(AB)=(kQ^(2))/r^(2)(-hat(i)): vec(F)_(AD)=(kQ^(2))/r^(2) hat(j)`
`vec(F)_(AE)=(kQq)/((r//sqrt(2))^(2))[cos 45^(@) hat(i)+sin 45^(@)(-hat(j))]`
On solving we get `q=+Q/4[1+2sqrt(2)]`