Question

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?

Short answer

a. The electric field strength between the disks is$\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathit{N}\mathbf{/}\mathit{C}$

b. The proton has launch speed of $\mathbf{}\mathbf{8}\mathbf{.}\mathbf{2}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}\mathbf{}$to reach the positive disk

Step-by-step solution

Given information and formula used

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

Calculating the electric field strength between the disks

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

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

Calculating the launch speed of the proton

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