Two points +ve charges q each are placed at (-a, 0) and (a,0), A third +ve charge `q_(0)` is placed at (0, y). Find the value of y for which the force at `q_(0)` is maximum.

Options

A. `(a)/(sqrt(3))`
B. `(a)/(sqrt(2))`
C. a
D. 2a

Answer

B

Explanation

Force on charge `+q_(0)` due to charge +q at A is
`F_(1) = (1)/(4pi epsilon_(0))(q_(0)q)/((a^(2) + y^(2)))`
Force on charge `+q_(0)` due to charge +q at B is
`F_(2) = (1)/(4pi epsilon_(0))(q_(0)q)/((a^(2) + y^(2))) :. F_(1) = F_(2)`
By symmetry, the components of force on charge `q_(0)` due to charges at A and B along X-axis will conact each other while along Y-axis will add up.
`:.` The resultant force on charge `+q_(0)` is
`F = 2F_(1) cos theta " "(because F_(1) = F_(2))`
`F = (2)/(4pi epsilon_(0))(q_(0)q)/(a^(2) + y^(2))(y)/(sqrt(a^(2) + y^(2))) = (2)/(4pi epsilon_(0))(q_(0)qy)/((a^(2) + y^(2))^(3//2))`
Force on charge `q_(0)` will be maximum, when `(dF)/(dy) = 0`
`:. (2q_(0)q)/(4pi epsilon_(0)) [(1)/((a^(2) + y^(2))^(3//2)) - (((3)/(2)y)(2y))/((a^(2) + y^(2))^(5//2))] = 0`
or `(1)/((a^(2) + y^(2))^(3//2)) - (3y^(2))/((a^(2) + y^(2))^(5//2)) = 0` or `1 = (3y^(2))/(a^(2) + y^(2))`
or `a^(2) + y^(2) = 3y^(2)` or `2y^(2) = a^(2)` or `y^(2) = (a^(2))/(2)` or `y = (a)/(sqrt(2))`