Saved Bookmarks
| 1. |
Two point charges +Q_(1) and -Q_(2) are placed at A and B respectively. A line of force emanates from Q_(1) at an angle theta with the line joining A and B. At what angle will it terminate at B ? |
|
Answer» Solution :We know that number of lines of force emerge is proportional to magnitude of the CHARGE. The field lines emanating from `Q_(1)`, spread out equally in all directions. The number of field lines or FLUX through cone of HALF ANGLE `THETA` is `(Q_(1))/(4pi) 2pi (1- cos theta)`. Similarly the number of lines of force terminating on `-Q_(2)` at an angle `phi " is " (Q_(2))/(4pi) 2pi (1- cos phi)`. The total lines of force emanating from `Q_(1)` is equal to the total lines of force terminating on `Q_(2)` `rArr (Q_(1))/(4pi) 2pi (1- cos theta) = (Q_(2))/(4pi) 2pi (1- cos phi) or (Q_(1))/(2) (1- cos theta) = (Q_(2))/(2) (1- cos phi)` `Q_(1) sin^(2) theta//2 = Q_(2) sin^(2) phi//2` `sin phi//2= sqrt((Q_(1))/(Q_(2))) sin theta//2 rArr phi= 2 sin^(-1) {sqrt((Q_(1))/(Q_(2))) sin theta//2}` |
|