InterviewSolution
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InterviewSolution
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A particle of mass m and charge `-q` is projected from the origin with a horizontal speed v into an electric field of intensity E directed downward. Choose the wrong statement. Neglect gravity A. The kinetic energy after a displacement y is `qEy`B. The horizontal and vertical components of acceleration of `a_x=0,a_y=(qE)/m`C. the equation of trajectory is `y=1/2((qEx^2)/(mv^2))`D. The horizontall and verticasl displacements x and y after a time t x=vt and `y=1/2a_yt^2` |
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Answer» Correct Answer - A `F_x=0` `:. a_=0` `F_y=qE` `:. a_y=(qE)/m` `x=vt` and `y=1/2a_yt^2=1/2((qE)/m)t^2` Substituting `t=x/v` in expression of `y` we get `y=1/2((qEx^2)/(mv^2))` `KE=1/2 m(v_x^2+v_y^2)` where `v_x=v` and `v_h=a_yt=(qE)/mt` |
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