Question

The electric potential varies in field as $V=3x+4y$. A particle of mass $0.1\text{kg}$ starts from rest from a point $(2,3.2)$ under the influence of this field. If the charge on the particle is $1\mu \text{C}$ and $V(x,y)$ are in SI units, then the velocity of the particle when it crosses the $x-$ axis is

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

Answer 1

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

Given:
$m=0.1\text{kg};q={10}^{-6}\text{C}$

$V=3x+4y$

Using relation between potential and electric field

$E_x=\frac{-\partial V}{\partial x}=-3\text{V/m};$
and $E_y=\frac{-\partial V}{\partial y}=-4\text{V/m}$

Now we know that force $\left(F\right)$ experienced by charge $\left(q\right)$ is given by

$F=qE\Rightarrow ma=qE$

$\Rightarrow a=\frac{qE}{m}$

So acceleration along $x$ and $y$ axis is

$a_x=\frac{q{E}_x}{m}=\frac{{10}^{-6}\times (-3)}{0.1}=-3\times {10}^{-5}{\text{m/s}}^2$

$a_y=\frac{q{E}_y}{m}=\frac{{10}^{-6}\times (-4)}{0.1}=-4\times {10}^{-5}{\text{m/s}}^2$

For the particle to cross the $x-$ axis it has to cover a distance of $3.2\text{m}$.
As initial velocity, $u_y=0$

$s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$\Rightarrow s_y=\frac{1}{2}\times 4\times {10}^{-5}\times t^2$

$\Rightarrow t^2=\frac{3.2}{2}\times {10}^5=1.6\times {10}^5=16\times {10}^4$

$\Rightarrow t=400\text{s}$

So,
velocity along $x-$axis, $v_x=a_{x}t=-12\times {10}^{-3}\text{m/s}$

$v_y=a_{y}t=-16\times {10}^{-3}\text{m/s}$

$v=\sqrt{v_x^2+v_y^2}=\sqrt{{16}^2+{12}^2}\times {10}^{-3}$

$\Rightarrow v=\sqrt{256+144}\times {10}^{-3}\text{m/s}$

$\Rightarrow v=20\times {10}^{-3}\text{m/s}$

So, option $\left(a\right)$ is correct.

Why this question ? Concept: This questions links the study of kinematics and electrostatics.