Two equal charges are separated by a distance `d`. A third charge placed on a perpendicular bisector at `x` distance will experience maximum coulomb force when

Options

A. `x=(d)/(sqrt(2))`
B. `x=(d)/(2)`
C. `x=(d)/(2sqrt(2))`
D. `x=(d)/(2sqrt(3))`

Answer

c

Explanation

Suppose third charge is similar to `Q` and it is `q`

So net force on it
`F_(n et)= 2F cos theta`
Where `F=(1)/(4pi epsilon_(0)).(Qq)/((x^(2)+(d^(2))/(4)))` and `cos theta=(x)/(sqrt(x^(2)+(d^(2))/(4)))`
`:. F_(n et)= 2xx(1)/(4pi epsilon_(0)).(Qq)/((x^(2)+(d^(2))/(4)))xx(x)/((x^(2)+(d^(2))/(4))^(1//2))`
`=(2Qqx)/(4pi epsilon_(0)(x^(2)+(d^(2))/(4))^(3//2))`
For `F_(n et)` to be maximum `(dF_(n et))/(dx)=0`
`e.(d)/(dx)[(2Qqx)/(4pi epsilon_(0)(x^(2)+(d^(2))/(4))^(3//2))]=0`
or `[(x^(2)+(d^(2))/(4))^(-3//2)-3x^(2)(x^(2)+(d^(2))/(4))^(-5//2)]=0`
i.e., `x=+-(d)/(2sqrt(2))`