Point charges g, -q, 2Q and Q are placed in order at the corners A, B, Cand D of a square of side 2b. If the field at the midpoint of CD is zero, then `(q)/(Q)` is

Options

A. 1
B. 2
C. `(2 sqrt2)/(5)`
D. `(5 sqrt5)/(2)`

Answer

D

Explanation

`E = E _(C) - E _(D) - E _(A) cos theta - E _(B) cos theta `
`( k ( 2Q))/( b ^(2)) - ( k Q)/(b ^(2)) = 2 xx ( k q b)/( (b ^(2) + 4 b ^(2) ) (3)/(2))`
`implies Q = ( 2 q)/( 5 sqrt5)`