A proton is released from rest at the positive plate of a parallelplate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 50,000 m/s. What will be the final speed of an electron released from rest at the negative plate?

Given information:

A proton speed reaches the negative place is $50,000m/s$.

Find:

The objective is to find final speed of proton.

We can estimate the changes in electric field seen between favourable and unfavourable plates for the electron using alternative energy:

$\mathrm{\Delta }{U}_{\text{e.p}}=-\mathrm{\Delta }{K}_p$

$=-\left(\frac{1}{2}{m}_p{v}_{p,f}^2-\frac{1}{2}{m}_p{v}_{p,i}^2\right)$

$∆U_{ep}=-\frac{1}{2}{m}_p{v}_{p,f}^2$

We could obtain the $qEd$quantity by situating the minus panel at $x=0$and therfore the optimistic plate at $x=d$using equation $(28.9)$:

$\begin{array}{r}\mathrm{\Delta }{U}_{e,e}=U_{\mathrm{f}}-U_{\mathrm{i}}\end{array}$

$=\left(\mathrm{\Delta }{U}_0+(-q)Ed\right)-\left(\mathrm{\Delta }{U}_0+(-q)E\times 0\right)$

$∆U_{ee}=-\alpha Ed$

The gravitational potential shift has been converted into energy.

From the angle of reducing emissions:

$$\begin{gathered} \mathrm{\Delta }{U}_{\mathrm{e},\mathrm{e}}=-\mathrm{\Delta }{K}_{\mathrm{e}} \\ =-\left(\frac{1}{2}{m}_{\mathrm{e}}{v}_{\mathrm{e},\mathrm{f}}^2-\frac{1}{2}{m}_{\mathrm{e}}{v}_{\mathrm{e},\mathrm{i}}^2\right) \end{gathered}$$

$qEd=\frac{1}{2}{m}_{\mathrm{e}}{v}_{\mathrm{e},\mathrm{f}}^2$

Substituting the determined $qEd$value from either the proton case:

$\frac{1}{2}{m}_{\mathrm{p}}{v}_{\mathrm{p},\mathrm{f}}^2=\frac{1}{2}{m}_{\mathrm{e}}{v}_{\mathrm{e},\mathrm{f}}^2$

localid="1649457305020" $$\begin{gathered} v_{\mathrm{e},\mathrm{f}}=v_{\mathrm{p},\mathrm{f}}\sqrt{\frac{m_{\mathrm{p}}}{m_{\mathrm{e}}}} \\ =50000\frac{\mathrm{m}}{\mathrm{s}}\times \sqrt{\frac{1.6726\times {10}^{-27\mathrm{kg}}}{9.109\times {10}^{-31}\mathrm{kg}}} \end{gathered}$$

localid="1649457587623" $v_{\mathrm{e},\mathrm{f}}=2.14\times {10}^6\frac{\mathrm{m}}{\mathrm{s}}$