Question

In a certain region, uniform electric field exists as $\overrightarrow{E}=E_0\hat{j}.$ A proton and an electron are projected from origin at time $t=0$ with certain velocities along $+x$-axis direction. Due to the electric field, they experience force and move in the $xy$-plane along different trajectories.

$\left(ii\right)$ If they have same kinetic energy then for the same displacement along $+x$-direction, the deflection is

• A. more for proton
• B. more for electron
• C. equal for both
• D. independent of kinetic energy

Answer 1

The correct option is C equal for both

As electric field exists along the $y$-axis,

$a_x=0$ and $a_y=\frac{qE}{m}$

Let initial velocity along the $x$-axis be $u_x=v_0$

Displacement along the $x$-axis is,
$x=v_{0}t\Rightarrow t=\frac{x}{v_0}$

Displacement along the $y$-axis is,

$y=\frac{1}{2}a{t}^2=\frac{qE{t}^2}{2m}$

$\Rightarrow y=\frac{qE{x}^2}{2m(v_0{)}^2}[\because K.E=\frac{1}{2}m(v_0{)}^2\Rightarrow 2K.E=m\left(v_0{)}^2\right]$

$\Rightarrow y=\frac{qE{x}^2}{2(2K.E)}$

Hence, for same value of $x$ and initial kinetic energy, $y$ is same for the both.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.