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.

$\left(iii\right)$ If they have the same initial velocity then for the same displacement along $x$ axis, displacement along $y$ axis is

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

Answer 1

The correct option is B more for electron

As, electric field exist 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,

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

$\Rightarrow y=\frac{qE{x}^2}{2m(v_0{)}^2}$

$\Rightarrow y\propto \frac{1}{m}$

$\because$mass of the proton $\left(m_p\right)$>mass of the proton$\left(m_e\right)$

Hence, the deflection of the electron is greater than the deflection of the proton.

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