Question

At some instant the velocity component of an electron moving between two charged parallel plates are $u_x=1.5\times {10}^5\text{m/s}$ and $u_y=3\times {10}^6\text{m/s}$. Suppose that the electric field between the plates is given by $\overrightarrow{E}=120\hat{j}\text{N/C}$. What will be the velocity of the electron as its $x-$coordinate changes by $2\text{cm}$?
(Neglect gravity)

• A. $(1.5\hat{i}+2\hat{j})\times {10}^5\text{m/s}$
• B. $(2\hat{i}+1.5\hat{j})\times {10}^5\text{m/s}$
• C. $(1.5\hat{i}-2\hat{j})\times {10}^5\text{m/s}$
• D. $(2\hat{i}-1.5\hat{j})\times {10}^5\text{m/s}$

Answer 1

The correct option is A $(1.5\hat{i}+2\hat{j})\times {10}^5\text{m/s}$

As there is force only in $y-$direction, velocity along $y-$axis changes.
$F_y=qE\Rightarrow F_y=-1.6\times {10}^{-19}\times 120=-1.92\times {10}^{-17}\text{N}$

$\Rightarrow a_y=\frac{F_y}{m}\Rightarrow a_y=\frac{-1.92\times {10}^{-17}}{9.1\times {10}^{-31}}=-2.1\times {10}^{13}{\text{m/s}}^2$

Now, time taken by electron to move by $2\text{cm}$ along $x-$axis is
$\Rightarrow t=\frac{x}{u_x}\Rightarrow t=\frac{2\times {10}^{-2}}{1.5\times {10}^5}=1.33\times {10}^{-7}\text{s}$
Now, along $y-axis$,
$\Rightarrow v_y=u_y+at$
$\Rightarrow v=3\times {10}^6-2.1\times {10}^{13}\times 1.33\times {10}^{-7}=2\times {10}^5\text{m/s}$
Therefore, final velocity of the particle is $(1.5\hat{i}+2\hat{j})\times {10}^5\text{m/s}$