Question

At some instant the velocity components of an electron moving between two charged parallel plates are $v_x=1.5\times {10}^5$ m/s and $v_y=3.0\times {10}^6$ m/s. Suppose that the electric field between the plates is given by $\overrightarrow{E}=\left(120N/C\right)\hat{j}$.

What will be the velocity of the electron after its $x$-coordinate has changed by $2.0cm$?

• A. $(3\hat{i}+4.0\hat{j})\times {10}^{5}m/s$
• B. $(1.5\hat{i}+2.0\hat{j})\times {10}^{5}m/s$
• C. $(1.5\hat{i}-2.0\hat{j})\times {10}^{5}m/s$
• D. $(-1.5\hat{i}+2.0\hat{j})\times {10}^{5}m/s$

Answer 1

The correct option is B $(1.5\hat{i}+2.0\hat{j})\times {10}^{5}m/s$

Acceleration of the electron = $\frac{E{q}_e}{m_e}=2.1\times {10}^{13}m/s^2$ in the direction opposite of the field.

$2$ cm is covered in $x$ direction in $t=\frac{2\times {10}^{-2}}{1.5\times {10}^5}=1.333\times {10}^{-7}s$
$v_{x}att=1.5\times {10}^{5}m/s$
$v_y=3\times {10}^6-2.1\times {10}^{13}\times 1.333\times {10}^{-7}=2\times {10}^{5}m/s$