Two $2.0-cm$diameter disks face each other, $1.0mm$apart. They are charged to $\pm 10nC$.

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

b. A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk?

Given :

Two disks of diameter each : $2.0-cm-\mathrm{diameter}$

Distance between disks : $1.0mm$

They are charged to : $\pm 10nC.$

Theory used :

$E=\frac{Q}{{\epsilon }_{0}A}$ gives the electric field inside the capacitor.

Where$A$denotes the disk's surface area. The charge on the disk is $Q$The conservation of energy law states that

$K_1+U_1=K_2+U_2$, where

$K_1=$ initial kinetic energy

$U_1$= initial potential energy

$K_2$= final kinetic energy

$U_2=$ final potential energy

(a) Using the equation $E=\frac{Q}{{\epsilon }_{0}A}$, we have :

$$\begin{gathered} E=\frac{10\times {10}^{-9}C}{(8.85\times {10}^{-12}{C}^2/N{m}^2)\left(\pi {(0.01m)}^2\right)} \\ =\boxed{\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathit{N}\mathbf{/}\mathit{C}} \end{gathered}$$

(b) The proton's velocity when it hits the capacitor's positive disk is$0m/s$.

So applying conservation of energy to find the initial velocity of proton :

role="math" localid="1649090929584" $$\begin{gathered} \frac{1}{2}mv²+0=QEd \\ \Rightarrow v²=\frac{2QEd}{m} \\ =\frac{2(10\times {10}^{-9}C)(1.6\times {10}^{-19}N/C)(0.001m)}{1.67\times {10}^{-27}kg} \\ \Rightarrow v=\boxed{\mathbf{}\mathbf{8}\mathbf{.}\mathbf{2}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}\mathbf{}} \end{gathered}$$