Question 20.7: In Fig. 20.23, assume that the horizontal length L of the pl......
In Fig. 20.23, assume that the horizontal length L of the plates is 5 cm, and assume that the separation D between the plates and the screen is 50 cm. If the uniform electric field has E = 250 N/C, and the electron’s initial speed v_0 is 2 × 10^{6} m/s, then; (a) What is the acceleration of the electron between the two plates? (b) Find the time when the electron leaves the two plates. (c) Find the electron’s vertical position before leaving the field region. (d) Find the electron’s vertical distance before hitting the screen.
(a) Using the magnitude of the electronic charge e = 1.6 × 10^{−19} C and the electronic mass m = 9.11 × 10^{−31} kg in Eq. 20.62, we get:
a_{x}=0\ \ a_{y}={\frac{e E}{m}} (Upwards) (20.62)
a_{x}=0 \text{and} a_{y}={\frac{e E}{m}} ={\frac{(1.6\times10^{-19}\,{\mathrm{C}})(250\,{\mathrm{N}}/{\mathrm{C}})}{9.11\times10^{-31}\,{\mathrm{kg}}}}=4.391\times10^{13}\,{\mathrm{m/s}}^{2}(b) Using Eq. 20.65 for the horizontal motion, we get:
t_{1}={\frac{L}{\operatorname{v_{\circ}}}} (20.65)
={\frac{0.05\,\mathrm{m}}{2\times10^{6}\,\mathrm{m}/\mathrm{s}}}=2.5\times10^{-8}\,\mathrm{s}(c) Using Eq. 20.66 for the vertical motion, we get:
y_{1}={\frac{e E L^{2}}{2m v_{\circ}^{2}}} (20.66)
={\frac{(1.6\times10^{-19}\,\mathrm{C})(250\,\mathrm{N}/\mathrm{C})(0.05\,\mathrm{m})^{2}}{2(9.11\times10^{-31}\,\mathrm{Kg})(2\times10^{6}\,\mathrm{m}/\mathrm{s})^{2}}}=0.0137\,\mathrm{m}=1.37\,\mathrm{cm}Alternatively, we can use Eq. 20.64 to find y_{1} as follows:
\begin{array}{c c}{{A l o n g~x~~~x=v_{o}t~~}}&{{}}\\ {{A l o n g~y~~~y=\frac12a_{y}t^{2}=\frac{e E}{2m}t^{2}}}\end{array} (20.64)
y_{1}={\textstyle{\frac{1}{2}}}a_{y}\,t_{1}^{2}={\textstyle{\frac{1}{2}}}(4.391\times10^{13}\,\mathrm{m/s^{2}})(2.5\times10^{-8}\,\mathrm{s})^{2}=0.0137\,\mathrm{m}=1.37\,\mathrm{cm}(d) We calculate y_{2} from Eq. 20.69 as follows:
y_{2} = D tan α = D\frac{eEL}{m^{2}v^2_{o}} (20.69)
={\frac{(0.5\operatorname{m})(1.6\times10^{-19}\,{\mathrm{C}})(250{\mathrm{N}}/{\mathrm{C}})(0.05\operatorname{m})}{(9.11\times10^{-31}\,{\mathrm{kg}})(2\times10^{6}\,{\mathrm{m}}/{\mathrm{s}})^{2}}}=0.274\,{\mathrm{m}}=27.4\,{\mathrm{cm}}Therefore, the total vertical distance moved by the electron is:
h=y_{1}+y_{2}=0.0137\,\mathrm{m}+0.274\,\mathrm{m}=0.2877\,\mathrm{m}=28.77\,\mathrm{cm}