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 time taken to cross the $x$ - axis is

• A. $20s$
• B. $40s$
• C. $20s$
• D. $400s$

Answer 1

The correct option is D $400s$

$V=3x+4y$
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
$S_y=ut+\frac{1}{2}a{t}^2$
$⟹S_y=\frac{1}{2}\frac{q{E}_y}{m}{t}^2$
$S_y=-3.2m$ (given)
On putting all values from above we will get $t=400s$