InterviewSolution
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InterviewSolution
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A particle of mass 40 mg and carrying a charge `5 xx 10^(-9) C` is moving directly towards a fixed positive point charge on magnitude `10^(-8) C`. When it is at a distance of 10 cm from the fixed positive point charge it has a velocity of `50 cm s^(-1)` at what distance from the fixed point charge will the particle come momentarily to rest ? Is the acceleration constant during motion? |
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Answer» If the particle comes to rest momentarily at a distance r from the fixed charge, then from conservation of energy, we have ? `1/2 m u^(2) +1/(4pi epsi_(0)) (Qq)/a=1/(4pi epsi_(0)) (Qq)/r` Substituting the given data, we get : `1/2xx40xx10^(-6)xx1/2xx1/2=9xx10^(9)xx5xx10^(-8)xx10^(-9) [1/r-10]` or, `1/r-10=(5xx10^(-6))/(9xx5xx10^(-8))=100/9 implies" "1/r=190/9" "implies" "r=9/190 m` or, i.e. `r=4.7xx10^(-2)m`. As here, `F=1/(4pi epsi_(0)) (Qq)/r^(2)` So, acc. `=F/m prop 1/r^(2)` i.e., Acceleration is not constant during the motion. |
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