Four charges equal to `-Q` are placed at the four corners of a square and a charge q is at its center. If the system is in equilibrium the value of q is

Options

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

Answer

B

Explanation

Let us consider charge at `A` to be in equilibrium

Now `F_(AB) = F_(DA) = (KQ^(2))/(a^(2))`
and `F_(CA) = (KQ^(2))/((sqrt(2)a//2)^(2)) = (KQ^(2))/(2a^(2))`
Now `F_(AO) = (KQq)/((sqrt(2)a//2)^(2)) = (K 2Q q)/(a^(2))`
for equilibrium all forces along AO should be equal to forces along `CA`,
`F_(AO) = F_(CA) + F_(BA) cos 45^(@) + F_(DA) cos 45^(@)`
`F_(AO) = F_(CA) + 2 F_(BA) cos 45^(@)`
`(K 2Q q)/(a^(2)) = (KQ^(2))/(2a^(2)) + (2 KQ^(2))/(a^(2)) xx *(1)/(sqrt(2))`
`:. q = (Q)/(4) (1 + 2 sqrt(2))`