A charge Q is place at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then `Q//q` equals:

Options

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

Answer

4

Explanation

`q (at B)`:
Due to `Q (at A) : F_(1) = k.(Qq)/(d^(2))` , along `AB`
Due to `q(at D) : F_(2) = k .(q.q)/((sqrt(2) d)^(2))`, along `DB`
Due to `Q (at C) : F_(3) = k.(qQ)/(d^(2))` , along `CB`
If a force is zero, its component are also zero.
`F_(x) = F_(1) + F_(2) cos 45^(@) = 0`
`k.(Qq)/(d^(2)) + k .(q.q)/((sqrt(2) d)^(2)) .(1)/(sqrt(2)) = 0`
`Q + (q)/(2sqrt(2)) = 0`
`(q)/(Q) = - 2 sqrt(2)`