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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 ? |
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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)/(4 pi) 2 pi (1 - cos phi)`. The total lines of force emanating from `Q_1` is equal to the total lines of force terminating on `Q_2` `implies (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 implies phi = 2 sin^(-1) {sqrt((Q_1)/(Q2)) sin theta//2}`. |
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