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Charge q_(1)is fixed andanotherpoint charge q_(2) is placedat a distance r_(0) from q_(1) on a frictionless horizontal surface. Find the velocity of q_(2) as a function of seperation r between them (treat as point charges and mass of q_(2) is m) |
Answer» Solution : When the seperationbetween the CHARGES is r, the force between them is `F=(1)/(4pi in_(0))(q_(1)q_(2))/( r^(2))` Accelerationof `q_(2) = (F)/(m)= (q_(1)q_(2))/( 4 pi in_(0) MR^(2))` `(DV)/(dt) = (dv)/(dt) ((dr)/(dt)) = (vdv)/(dr)` `rArr ( vdv)/(dr) = (q_(1)q_(2))/( 4 pi in_(0)mr^(2))` `int_(0)^(v) vdv=(q_(1)q_(2))/( 4pi in_(0) m) int_(r_(0)) ^( r)r^(-2) dr ` or`(v^(2))/(2) = (q_(1)q_(2))/( 4 pi in_(0) m ) |(-1)/(r ) |_(r_(0))^(r )` `rArr v = SQRT((q_(1)q_(2))/( 2pi in_(0) m) {(1)/(r_(0)) - (1)/(r )})` |
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