Two 2.0-cm-diameter disks spaced 2.0 mm apart form a

parallel-plate capacitor. The electric field between the disks is

$5.0\times {10}^{5}V/m$.

a. What is the voltage across the capacitor?

b. An electron is launched from the negative plate. It strikes the

positive plate at a speed of $2.0\times {10}^{7}m/s$.

What was the electron’s speed as it left the negative plate?

The electric potential inside a parallel plate

$∆V_c=E.d$

Now;

$$\begin{gathered} E=5.0\times {10}^{5}V/m \\ d=2.0\times {10}^{-3}m \end{gathered}$$

Here, E=Electrical field strength d= distance between space

Hence the Potential energy of the field

role="math" localid="1648306659509" $\begin{array}{rcl}U& =& qEd\\ r& =& \frac{\left(2.0cm\right)\left({10}^{-2}m\right)}{2\times 1cm}\\ & =& 1.0\times {10}^{-2}m\\ d& =& \left(2mm\right)\left(\frac{{10}^{-3}}{1mm}\right)\\ & =& 2.0\times {10}^{-3}m\end{array}$

The electric potential inside a parallel plate

$\begin{array}{rcl}∆V_c& =& 5.0\times {10}^{5}V/m\times 2.0\times {10}^{-3}m\\ & =& 1.0\times {10}^{-3}V\end{array}$

Hence the value of capacitor;

$1.0\times {10}^{-3}V$

The kinetic energy of a particle where m is the mass and v is the velocity

$E_k=\frac{1}{2}m{v}^2$

Initial speed on the negative electric field $V_i$

Final speed on the negative electric field $V_f$

The kinetic energy of a changing electron is equal to the accelerating charge of the particle

$\begin{array}{rcl}qV& =& \frac{1}{2}m{V^2}_f-\frac{1}{2}m{V^2}_i\\ \frac{1}{2}m{V^2}_i& =& \frac{1}{2}m{V^2}_f-qV\\ {V^2}_i& =& {V^2}_f-\frac{2qV}{m}\\ V_i& =& \sqrt{{V^2}_f-\frac{2qV}{m}}\end{array}$

Now placing the values of $V_i,V_f,m,qandV$

$\begin{array}{rcl}{V}_f& =& 2.0\times {10}^{7}m/s\\ m& =& 9.11\times {10}^{-31}m\\ q& =& 1.6\times {10}^{-19}C\\ V& =& 1.0\times {10}^{-3}V\\ & & \\ V_i& =& \sqrt{\left(2.0\times {10}^{7}m/s\right)-\frac{2(1.6\times {10}^{-19}C)(1.0\times {10}^{-3}V)}{9.11\times {10}^{-31}m}}\\ & =& 6.98\times {10}^{6}m/s\end{array}$

Hence the calculated speed is$6.98\times {10}^{6}m/s$