InterviewSolution
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InterviewSolution
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The electric potential varies in space according to the relation `V = 3x + 4y`. A particle of mass 10 kg starts from rest from point `(2, 3.2)` m under the influence of this field. Find the velocityr,f the particle when it crosses the x-axis. The charge on the particle is `+1muC`. Assume V (x, y)are in SI units. |
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Answer» Correct Answer - A::B::C `E=-((delV)/(delx)hati+(delV)/(dely)hatj)=(-3hati+4hatj)N//C` `a=(qE)/m` `=10^-6/10(-3hati-4hatj)` `=(-3xx10^-7hati-4xx10^-7hatj)m//s^2` When particle crosses `x` axis `y=0` Initial `y`-coordinate was `3.2m` and `a_y=-4xx10^-7m//s^2` `:.y=0` at time `t=sqrt((2xx3.2)/(4xx10^-7))=4000s` At this instant `x`-coordinate will be `x=x_i+1/2a_xt^2` `=2+1/2(-3xx10^-7)(4000)^2=-0.4m` Now `V_i=(3xx2)+(4xx3.2)=18.8V` `V_f=(3)(-0.4)=-1.2V` `/_V=20V` `:. ` Speed `v=sqrt((2q/_V)/m)` `=sqrt((2xx10^-6xx20)/10)` `2.0xx10^-3m//s` |
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