Answer to Question #171640 in Electric Circuits for Pranali Patil

Question

You are building a cathode ray tube and you need to ensure that the electron gun emits electrons at 4500000m/s; if the field in the electron gun is created by two plates separated by 5.5cm, what voltage should you use?

Answer 1

Explanations & Calculations

• A simple way of the electron gun can be sketched as shown

• What happens is that a force is generated on the electron ($\small F=Eq$) which accelerates it towards the anode with the final velocity as described in the description.
• Since the field is uniform, the acceleration is constant & to get a connection between the data, the four equations of motion can be employed.
• Field between a plates is given by $\small E = \Large\frac{\Delta V}{d}$
• Then,

$\qquad\qquad \begin{aligned} \small \rightarrow v^2&=\small u^2+2as\\ \small v^2 &=\small 0+2\cdot\frac{F}{m}\cdot d\\ &=\small 2\cdot\frac{\Delta Vq}{md}\cdot d\\ \small \Delta V&=\small \frac{mv^2}{2q} \end{aligned}$

• The same result can be gained by considering the work transferred on to the electron by the fiels

$\qquad\qquad \begin{aligned} \small \frac{1}{2}mv^2&=\small q\Delta V \end{aligned}$

• Mass of en electron is $\small 9.11\times10^{-31}kg$ and its charge is $\small -1.6\times10^{-19}C$
• Therefore,

$\qquad\qquad \begin{aligned} \small \Delta V &=\small \frac{9.11\times10^{-31}kg\times(4.5\times10^6ms^{-1})^2}{2\times1.6\times10^{-19}C}\\ &=\small \bold{57.649\,V} \end{aligned}$