Saved Bookmarks
| 1. |
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. |
|
Answer» `(a)/(SQRT(3))`  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))` |
|