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Two fixed insulating rings A and B carry ahanges with uniform linear aharge density `+lambda` and `-lambda`, respectively, as shown in the adjacent figure. The planes of the rings are parallel to each other and their axes are coinciding. A particle of change "q" and mass "m" is released with zero velocity from centre P of the positively charged ring. The kinetic energy of the particle when it reaches centre Q of the negatively charge ring will be A. `sqrt((2lambdaq)/(m in_(0))(1 - (1)/(sqrt2))`B. `(lambda)/(in_(0))(1 - (1)/(sqrt2))q`C. `(lambda)/(2 in_(0))(1 - (1)/(sqrt2))q`D. none of these |
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Answer» Correct Answer - B Potential at P due to rings `V_(P)=(lambda2piR)/(4piin_(0)R)-(lambda2piR)/(4piin_(0)sqrt2R)=(lambda)/(2in_(0))(1-(1)/(sqrt2))` Potential at Q due to rings is `V_(Q)=-(lambda)/(2in_(0))(1-(1)/(sqrt2))` `Delta(K.E.)+(V_(Q)-V_(P))q=0` `rArr (1)/(2)mv^(2)=(lambda)/(in_(0))(1-(1)/(sqrt2))q` |
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