Question

In Exercise 21.29, what is the speed of the electron as it emerges from the field?

Short answer

Speed of electron is $v=1.79\times {10}^{6}m/s$.

Step-by-step solution

Step 1:

As the velocity vector is made up of two components x and y.

Therefore,$v_x=1.6\times {10}^{6}m/s$

As no acceleration is along x-direction, the velocity remains unchanged from initial velocity.

As the particle has no initial velocity so$v_{y0}=0$

$$\begin{gathered} v_y=v_{y0}+at \\ {v}_y=at=\left(\frac{E_{0}q}{m}\right)t \\ =\frac{364\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}\times \left(1.25\times {10}^{-8}\right) \\ =8\times {10}^5\mathrm{m}/\mathrm{s} \end{gathered}$$

Step 2:

$$\begin{gathered} \left|\overrightarrow{v}\right|=\sqrt{v_x^2+v_y^2} \\ =\sqrt{{\left(1.6\times {10}^6\right)}^2+{\left(8\times {10}^5\right)}^2} \\ \approx 1.79\times {10}^6\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, Speed of electron is$v=1.79\times {10}^6\mathrm{m}/\mathrm{s}$.