91影视

1. Initial speed of electron,$v_0=2.00x{10}^6\text{m/s}$
2. The angle with x axis,${胃}_0={40.0}^o$
3. The uniform electric field,$\stackrel{鈫�}{E}=(5.00\text{N/C})\widehat{j}$
4. Screen distance parallel to the x-axis,$x=3.00\text{m}$

Since the electron is negatively charged, then (as a consequence of the magnitude of the electrostatic force on a point charge of magnitude q (given by F = qE, where E is the magnitude of the electric field at the location of the particle) with Newton鈥檚 second law) the field E pointing in the +y direction (which we will call 鈥渦pward鈥�) leads to a downward acceleration. This is exactly like a projectile motion problem (but with g replaced with, acceleration

$\begin{array}{c}\mathbf{a}\mathbf{=}\mathbf{e}\mathbf{E}\mathbf{/}\mathbf{m}\\ \mathbf{=}\mathbf{8}\mathbf{.78}\mathbf{x}{\mathbf{10}}^{\mathbf{11}}\mathbf{\text{m/s}}\end{array}$).

Formulae:

Using the concept of projectile motion, the time of travel by the electron,

$t=\frac{x}{v_{o}cos{胃}_o}$(i)

Now, the velocity of the electron using projectile motion,$v_y=v_{o}sin{胃}_o鈭�a{t}^{}$ (ii)

$\begin{array}{c}t=\frac{3.00\text{m}}{(2.00脳{10}^6\text{m/s})cos{40.0}^o}\\ 鈭�=1.96脳{10}^{鈭�6}\text{s}\end{array}$

The required velocity component using equation (ii) is given as:

$\begin{array}{c}{v}_y=(2.00脳{10}^6\text{m/s})sin{40.0}^o鈭�(8.78脳{10}^{11}{\text{m/s}}^{\text{2}})(1.96脳{10}^{鈭�6}\text{s})\\ =鈭�4.34脳{10}^5\text{m/s}\end{array}$

Since the x component of velocity does not change, then the final velocity is.

$(1.53脳{10}^6\text{m/s})\widehat{i}鈭�(4.34脳{10}^5\text{m/s})\widehat{j}$