Question 3.5: In a cathode ray tube of a TV set, an electron with initial ......
In a cathode ray tube of a TV set, an electron with initial velocity v_{o} = 2×10^{4} m/s enters a region 2 cm long (see Fig. 3.11) where it is electrically accelerated in a straight line. The electron emerges from this region with a velocity v = 3×10^{5} m/s. (a) What was its acceleration, assuming it was constant? (b) How long will the electron be in this region?
(a) Taking the motion to be along the x-axis, and using v_{o} = 2 × 10^{4} m/s, v = 3 × 10^{5} m/s, and x−x_{o} = 2 cm = 2×10^{−2} m, we can find the accel-eration a from the relation v_{2} = v^{2}_{o} + 2 a (x − x_{o}) as follows:
a=\frac{v^{2}-v_{\mathrm{o}}^{2}}{2\,\left(x-x_{\mathrm{o}}\right)}=\frac{(3\times10^{5}\operatorname{m}/s)^{2}-(2\times10^{4}\operatorname{m}/s)^{2}}{2×2\,\times10^{-2}\,\operatorname{m}}=2.24\times10^{12}\,\mathrm{m}/s^{2}(b) Since the displacement and velocities are known, we can use x − x_{o} = \frac{1}{2} (v_{o} + v) t to find the time t that the electron will be electrically accelerated as follows:
t={\frac{2(x-x_{\circ})}{v_{\circ}+v}}={\frac{2\times2\times10^{-2}{\mathrm{m}}}{2\times10^{4}\,{\mathrm{m}}/{\mathrm{s}}+3\times10^{5}\,{\mathrm{m}}/{\mathrm{s}}}}=1.25\times10^{-7}\,{\mathrm{s}}=0.125\,{\mathrm{\mu s}}Another way to find t is to use equation v = v_{o}+a t. In this case, v = 3×10^{5} m/s, and a = 2.24 × 10^{12} m/s^{2}. Thus:
t={\frac{v-v_{o}}{a}}={\frac{3\times10^{5}~\mathrm{m}/s-2\times10^{4}~\mathrm{m}/s}{2.24\times10^{12}~\mathrm{m}/s}}=1.25\times10^{-7}\,\mathrm{s}=0.125\mathrm{\ \mu s}Even though a is very high in this example, but such an acceleration occurs over a very short time interval which is a typical value for such an electrically accelerated charged particle.