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A charge +q_(0) is fixed at a position in space. From a large distance another chaged particle of charge -q and mass m is thrown towards +q_(0) with an impact parameter L as shown. The initial speed of the projected particle is v. Find the distance of closet approach of the two particles. |
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Answer» Solution :As -Q movestowards `+q_(0)` an attractiveforce acts on -q towards `+q_(0)`. No torque acts on -q relative to `+q_(0)`. In other WORDS angular MOMENTUM of -q must remain constant. When -q is closet to +q, it will be moving perpendicularly to the line joining the two charges as shown.  Let r be the closed seperation between the charges and `v_(c )` be the velocity of -q at the instant. From conservation of angular momentum, `mvL=mv_(c )r` From conservation of mechanical energy `(1)/(2) mv^(2)=(1)/(2)mv_(c )^(2)-(1)/(4pi in_(0)) (qq_(0))/(r )` On solving the above equation, we can get r. |
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