Example 5.2: Motion of an electron in an electric field

Question: An electron in a television set is accelerated from the cathode to the screen through a potential difference of +1000 V. The screen is 35 mm from the cathode. What is the net change in the potential energy of the electron during the acceleration process? How much work is done by the electric field in accelerating the electron? What is the speed of the electron when it strikes the screen?
 
Solution: Let call the cathode point $A$ and the screen point $B$. We are told that the potential difference between points $B$ and $A$ is +1000 V, so

$$V_B - V_A = 1000\,{\rm V}.$$

By definition, the difference in electric potential energy of some charge $q$ at points $B$ and $A$ is the product of the charge and the difference in electric potential between these points. Thus,

$$P_B- P_A = q\,(V_B - V_A)= (-1.6\times 10^{-19})\,(1000) = -1.6\times 10^{-16}\,{\rm J},$$

since $q=-1.6\times 10^{-19}\,{\rm C}$ for an electron. Note that the potential energy of the electron decreases as it is accelerated towards the screen. As we have seen, the electric potential energy of a charge is actually held in the surrounding electric field. Thus, a decrease in the potential energy of the charge corresponds to a reduction in the energy of the field. In this case, the energy of the field decreases because it does work $W'$ on the charge. Clearly, the work done (i.e., energy lost) by the field equals the decrease in potential energy of the charge,

$$W' = - {\mit\Delta}P.$$

Thus,

$$W' = 1.6\times 10^{-16}\,{\rm J}.$$

The total energy $E$ of the electron is made up of two components--the electric potential energy $P$, and the kinetic energy $K$. Thus,

$$E = P + K.$$

Of course,

$$K = \frac{1}{2} \,m\, v^2,$$

where $m=9.11\times 10^{-31}$ kg is the mass of the electron, and $v$ its speed. By conservation of energy, $E$ is a constant of the motion, so

$$K_B - K_A = {\mit\Delta} K = - {\mit\Delta}P.$$

In other words, the decrease in electric potential energy of the electron, as it is accelerated towards the screen, is offset by a corresponding increase in its kinetic energy. Assuming that the electron starts from rest (i.e. $v_A=0$), it follows that

$$\frac{1}{2}\,m\,v_B^{~2} = - {\mit\Delta}P,$$

or

$$v_B = \sqrt{\frac{-2\,{\mit\Delta}P}{m}} = \sqrt{ \frac{-2\,... ...6})}{9.11\times 10^{-31}}} = 1.87\times 10^7\,{\rm m\,s^{-1}}.$$

Note that the distance between the cathode and the screen is immaterial in this problem. The final speed of the electron is entirely determined by its charge, its initial velocity, and the potential difference through which it is accelerated.