NEET Physics Electrostatics & Capacitance Question Bank NEET PYQ - Electrostatics and Capacitance

  • question_answer
    A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is :               [AIPMT 1998]

    A) \[qE{{y}^{2}}\]

    B) \[q{{E}^{2}}y\]

    C) \[qEy\]

    D) \[{{q}^{2}}Ey\]

    Correct Answer: C

    Solution :

    [c] Key Idea: Kinetic energy obtained by the particle is equal to the work done in moving a distance y.
    Electric force on charged particle
                            \[F=qE\]
    Kinetic energy attained by particle
                            = work done
                            = force \[\times \] displacement
                            \[=qE\times y\]
                Alternative:
                Force on charged particle in a uniform electric field is
                            \[F=ma=Eq\]
                or         \[a=\frac{Eq}{m}\]                                …(i)
                From the equation of motion, we have
                            \[{{v}^{2}}={{u}^{2}}+2ay\]
                            \[=0+2\times \frac{Eq}{m}\times y\]
                            \[=\frac{2Eqy}{m}\]     
                Now kinetic energy of the particle
                            \[K=\frac{1}{2}m{{v}^{2}}\]
                            \[=\frac{m}{2}\times \frac{2\,E\,qy}{m}=qEy\]


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