Question

The electric potential varies in space according to the relation $V=3x+4y.$ A particle of mass $0.1kg$ 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

• A. $20\times {10}^{-3}m{s}^{-1}$
• B. $40\times {10}^{-3}m{s}^{-1}$
• C. $30\times {10}^{-3}m{s}^{-1}$
• D. $50\times {10}^{-3}m{s}^{-1}$

Answer 1

The correct option is A $20\times {10}^{-3}m{s}^{-1}$

$V=3x+4y$
We know that the relation between $V$ and $E$ is given by
$E=-[\frac{\delta V}{\delta x}\hat{i}+\frac{\delta V}{\delta y}\hat{j}+\frac{\delta V}{\delta z}\hat{k}]$
$⟹E_x=-\frac{\delta V}{\delta x}\hat{i}$
$\therefore E_x=-3V{m}^{-1}$
Similarly, $E_y=-\frac{\delta V}{\delta y}\hat{j}$
$\therefore E_y=-4V{m}^{-1}$ (downward)
Only this y component of electric field is responsible for the particle to cross x-axis.
The acceleration of particle along y- axis will be
$a_y=\frac{q{E}_y}{m}$
We know
$S_y=ut+\frac{1}{2}{a}_y{t}^2$
$⟹S_y=\frac{1}{2}\frac{q{E}_y}{m}{t}^2$........(i)
$S_y=-3.2m$ (given)
On putting all values from above in (i) we will get $t=400s$
Velocity of the when it crosses x-axis is given by
$V=\sqrt{{V_x}^2+{V_y}^2}$...........(ii)
$V_x=u_x+a_{x}t$
As $u_x=0$ then,
$V_x=\frac{q{E}_y}{m}t$
Ily, $V_y=\frac{q{E}_x}{m}t$
On solving we will get $V_x=-12\times {10}^{-3}$ $m{s}^{-1}$ and $V_y=-16\times {10}^{-3}$ $m{s}^{-1}$
Putting these values in (ii) we will get
$V=20\times {10}^{-3}$ $m{s}^{-1}$