Question

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?

Short answer

1. Show that this is not sufficient the speed of photon to reach the positive plate

2. The proton’s speed that it is collided with the negative plate is $2.96\times {10}^{5}m/s$

Step-by-step solution

 Explanation of the  insufficient speed to reach the positive 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.

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