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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Self Evaluation Test - Electric Chages and Fields

  • question_answer
    A particle of charge q and mass m moves rectilinearly under the action of electric field \[E=A-Bx,\]where A and B are positive constants and x is distance from the point where particle was initially at rest then the distance traveled by the particle before coming to rest and acceleration of particle at that moment are respectively:

    A) \[\frac{2A}{B},0\]         

    B)        \[0,-\frac{qA}{m}\]

    C) \[\frac{2A}{B},-\frac{qA}{m}\]  

    D)        \[\frac{-2A}{B},-\frac{qA}{m}\]

    Correct Answer: C

    Solution :

    [c] \[F=qE=q\left( A-Bx \right)\] \[ma=q\left( A-Bx \right)\Rightarrow a=\frac{q}{m}\left( A-Bx \right)\]         ?.(i) \[\frac{vdv}{dx}=\frac{q}{m}\left( A-Bx \right);\,\,\,vdv=\frac{q}{m}q\left( A-Bx \right)dx\] \[\int\limits_{0}^{0}{vdv=\frac{q}{m}}\int\limits_{0}^{x}{(A-Bx)dx}\,\,\,\,;\,\,\,\,Ax-\frac{B{{x}^{2}}}{2}=0\] \[x=0,x=\frac{2A}{B}\]                         ?(ii) From eqs. (i) and (ii) \[\frac{q}{m}(A-Bx)=\frac{q}{m}\,\left( A-B\times \frac{2A}{B} \right)\,=\frac{-\,aA}{m}\]


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