Question

The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.2 cm apart with a 25 kV potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential.

a. What is the electric field strength between the plates?

b. With what speed does an electron exit the electron gun if its entry speed is close to zero?

Note: The exit speed is so fast that we really need to use the theory of relativity to compute an accurate value.

Your answer to part b is in the right range but a little too big.

Short answer

a. The electric field strength between the plates $2.1\times {10}^{6}V/m$

b. At $9.4\times {10}^{7}m/s$speed an electron exit the electron gun if its entry speed is close to zero

Step-by-step solution

Definition of Electric Potential and Electric field strength

the equation of electric potential inside an parallel plate capacitor is ;$\Delta V_c=Ed$

Here in this equation E is the electric field strength, and d is the distance between two plates;

$E=\frac{\Delta V_C}{d}$

Given Data

Here the given data is

$\begin{array}{rcl}\Delta V_C& =& 25kV\\ d& =& 1.2cm\\ & & \end{array}$

Substitution of data within Equation

$\begin{array}{rcl}qV& =& \frac{1}{2}m{v_f}^2-\frac{1}{2}m{v_i}^2\\ or,qV& =& \frac{1}{2}m{v_f}^2-\frac{1}{2}m{\left(0\right)}^2\\ Thus;qV& =& \frac{1}{2}m{v_f}^2\\ & & \end{array}$

However this can be Written as;

${v}_f=\sqrt{\frac{2qV}{m}}$

As;

localid="1648287412324" $\begin{array}{rcl}E& =& \frac{\Delta V_C}{d}\\ E& =& \frac{25kV}{1.2cm}\\ & =& \frac{25\times 1000V}{1.2\times {10}^{-2}m}\\ & =& 2.083\times {10}^{6}V/m\end{array}$

a. The electric field strength between the plates

In this scenario the initial speed of electron can be considered as Zero. The more the electron gets accelerated the more change has seen in the kinetic energy.

localid="1648287430301" role="math" $\begin{array}{rcl}qV& =& \frac{1}{2}m{v_f}^2-\frac{1}{2}m{v_i}^2\\ qV& =& \frac{1}{2}m{v_f}^2-\frac{1}{2}m\left(0\right)\\ qV& =& \frac{1}{2}m{v_f}^2\\ v_f& =& \sqrt{\frac{2qV}{m}}\\ & & \end{array}$

b) hence the calculated speed of electron will be;

localid="1648287440340" $$\begin{gathered} v_f=\sqrt{\frac{2\left(1.6\times {10}^{-19}\right)\left(25000\right)V}{9.11\times {10}^{-31}kg}} \\ =9.37\times {10}^{7}m/s \end{gathered}$$