Two charges `q` and `-3q` are placed fixed on x-axis separated by distance `d`. Where should a third charge `2q` be placed such that it will not experience any force ?

Options

A. `(d)/(2)(1+sqrt(3))` to the left of q
B. `(d)/(2)(1+sqrt(3))` tothe right of q
C. `(d)/(2)(1+sqrt(5))` to the left of q
D. `(d)/(2)(1+sqrt(5))` to the right of q

Answer

A

Explanation

If on 2q , force due to q is to the left and that due to `-3q` is to the right `therefore(2Q^(2))/(4piepsilon_(0)X^(2))=(6q^(2))/(4piepsilon_(0)(d+x)^(2))`
`therefore(d+x)^(2)=3x^(2)`
`therefore2x^(2)-2dx-d^(2)=0`

`rArrx=(d)/(2)+-(sqrt(3)d)/(2)`
(-ve sign would be between and `q+3q` unacceptable )
`x=(d)/(2)+(sqrt(3)d)/(2)=(d)/(2)(1+sqrt(3))` to the left of q