In FIGURE EX25.26, a proton is fired with a speed of 200,000 m/s from the midpoint of the capacitor toward the positive plate.

a. Show that this is insufficient speed to reach the positive plate.

b. What is the proton’s speed as it collides with the negative plate?

Initial kinetic energy of the photon KE

$KE=\frac{1}{2}m{v_0}^2$

Here, m is the mass of the photon and v₀ is the speed

$\begin{array}{rcl}m& =& 1.67\times {10}^{-27}\\ v_0& =& 200000m/s^2\\ KE& =& \frac{1}{2}(.67\times {10}^{-27})({200000m/s^2)}^2\\ & =& 3.34\times {10}^{-17}J\\ & & \end{array}$

As increase the poetical energy of the photon, then it reaches the positive plate

$\begin{array}{rcl}∆U& =& e\left(\frac{∆V}{2}\right)\\ e& =& 1.6\times {10}^{-19}C\\ ∆V& =& 500V\\ ∆U& =& 1.6\times {10}^{-19}C\left(\frac{500V}{2}\right)\\ & =& 4.0\times {10}^{-17}J\\ & & \\ & & \end{array}$

KE is not sufficient to increase photon energy to reach the positive plate. Due to this reason, the photon cannot reach the positive plate.

The sum of the K.E and P.E are equal to the total energy of the photon.

$$\begin{gathered} \frac{1}{2}m{V^2}_{mid}+e{V}_{mid}=\frac{1}{2}m{V^2}_{negative}+e{V}_{neagtive} \\ \frac{1}{2}m{V^2}_{negative}=\frac{1}{2}m{V^2}_{mid}+e{V}_{mid}-e{V}_{neagtive} \\ {V}_{negative}=\sqrt{\left(\frac{2}{m}\right)}\left(\frac{1}{2}m{V^2}_{mid}+e{V}_{mid}-e{V}_{neagtive}\right) \\ {V}_{mid}=200000m/s \\ m=1.67\times {10}^{-17} \\ e=1.6\times {10}^{-19}C \\ {V}_{negative}=0 \\ when{V}_{mid}=250 \\ {V}_{negative}=\sqrt{\left(\frac{2}{1.67\times {10}^{-17}}\right)}\left(\frac{1}{2}1.67\times {10}^{-17}{(200000m/s)}^2+1.6\times {10}^{-19}C(250-0\right) \\ {V}_{negative}=2.96\times {10}^{5}m/s \end{gathered}$$

The proton’s speed as it collides with the negative plate is$2.96\times {10}^{5}m/s$