The electric potential varies in space according to the relation `V = 3 x + 4 y`. A particle of mass `0.1 kg` starts from rest from point `(2,3.2)` under the influence of this field. The charge on the particle is `+1 mu C`. Assume `V` and (x,y) are in `S I` units.
The velocity of the particle when it crosses the x - axis is

Options

A. `20 xx 10^-3 ms^-1`
B. `40 xx 10^-3 ms^-1`
C. `30 xx 10^-3 ms^-1`
D. `50 xx 10^-3 ms^-1`

Answer

A

Explanation

`E_x = - (delta V)/(delta x) = -3 Vm^-1, E_y = (delta V)/(delta y) = -4 Vm^-1`
`a_x = (q E_x)/(m) = (1 xx 10^-6 xx 3)/(0.1) = -3 xx 10^-5 ms^-2`
`a_y = (q E_y)/(m) = (1 xx 10^-6 xx 4)/(0.1) = -4 xx 10^-5 ms^-2`
Time taken to cross the x - axis
Using `s = ut + (1)/(2) at^2` we get
`3.2 = (1)/(2) xx 4 xx 10^-5 xx t^2`
`t = 400 s`
`v_x = a_x t = -3 xx 10^-5 xx 400`
=`12 xx 10^-3 ms^-1`
`v_y = a_y t = -4 xx 10^-5 xx 400`
=`16 xx 10^-3 ms^-1`
`v = sqrt(v_x^2 + v_y^2) = 20 xx 10^-3 ms^-1`.