Question

The electric field strength is $50,000\mathrm{N}/\mathrm{C}$inside a parallel-plate capacitor with a $2.0mm$ spacing. A proton is released from rest at the positive plate. What is the proton’s speed when it reaches the negative plate?

Short answer

The proton's final velocity at negative plate is$v_{\mathrm{f}}\approx 1.38\times {10}^5\frac{\mathrm{m}}{\mathrm{s}}$

Step-by-step solution

Step :1  Angular momentum

Given information:

Electric field strength is $50,000\mathrm{N}/\mathrm{C}$.

Spacing is $2.0\mathrm{mm}$.

Find:

The objective is to find proton's speed reaches the negative plate.

The particle travels along the field, more toward the capacitor's deleterious anode, in a static electric field. At the expenditure of electrostatic field, it gains angular momentum.

The variation in electromotive force can be computed using equation $(28.10)$:

$$\begin{gathered} \mathrm{\Delta }{U}_{\mathrm{e}}=U_{\mathrm{e},\mathrm{f}}-U_{\mathrm{e},\mathrm{i}} \\ =\left(U_0+qE\cdot 0\right)+\left(U_0+qEd\right) \end{gathered}$$

$=-qEd$

Step :2 Explanation

We can compute the proton's mobility when it approaches the negatively pole by transforming the difference in electromotive force energy into kinetic energy.

From energy conversation

$(\mathrm{\Delta }K+\mathrm{\Delta }U=0)$

$\mathrm{\Delta }K=-\mathrm{\Delta }{U}_{\mathrm{e}}$

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

$v_f=\sqrt{\frac{2qEd}{m_p}}$

$=\sqrt{\frac{2\times 1.602\times {10}^{-19}\mathrm{C}\times 50000\frac{\mathrm{N}}{\mathrm{C}}\times 2\times {10}^{-3}\mathrm{m}}{1.673\times {10}^{-27}\mathrm{kg}}}$

localid="1649456505449" $v_{\mathrm{f}}\approx 1.38\times {10}^5\frac{\mathrm{m}}{\mathrm{s}}$