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.

Answer

A, B, C

Explanation

`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`