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 $\overrightarrow{E}=\left(120N/C\right)\hat{j}$. The acceleration of the electron is approximately $2.1\times {10}^{(10+x)}$. Find the value of $x$. Take charge on electron to be $1.6\times {10}^{-19}C$ and mass of electron to be $9.1\times {10}^{-31}kg$.

Answer 1

Given:

The velocity component in X-direction$v_x=1.5\times {10}^{5}m/s$.

The velocity in y-direction $v_y=3.0\times {10}^{6}m/s$.

The electric field present in the region is $\overrightarrow{E}=120N/C\hat{j}$.

The net electric force on the charged particle accelerates the particle through the electric field.

So, the force balancing equation for the charged particle is:

$m\times a=q\times \overrightarrow{E}$

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

Substituting the values in the above equation:

$a=\frac{1.6\times {10}^{-19}\times 120}{9.1\times {10}^{-31}}$

$a=2.1\times {10}^{13}m/s^2$

So, the value of $x$ will be $3$.