Question

An electron in a vacuum chamber is fired with a speed of $8300km/s$toward a large, uniformly charged plate $75cm$ away. The electron reaches a closest distance of $15cm$ before being repelled. What is the plate’s surface charge density?

Short answer

The plate’s surface charge density is$\mathbf{2}\mathbf{.}\mathbf{8}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{}\mathit{C}\mathbf{/}\mathit{m}\mathbf{²}\mathbf{}$

Step-by-step solution

Given information and formula used

Given :

The electron is fired with a speed of : $8300km/s$

Distance from the uniformly charged plate : $75cm$

The electron reaches a closest distance of : $15cm$

Theory used :

The electric field owing to an infinite plane is calculated as follows:

$E=\frac{\sigma }{2{\epsilon }_0}$

The electron's force

$F=eE=\frac{e\sigma }{{\epsilon }_0}$

The electron's work consists of the following:

$$\begin{gathered} W=Fd=\frac{de\sigma }{{\epsilon }_0} \\ d=\frac{1}{2}m{v}^2 \\ \Rightarrow \underline{\frac{de\sigma }{{\epsilon }_0}=\frac{1}{2}m{v}^2} \end{gathered}$$

Calculating the plate’s surface charge density 

Finding $d$:

$$\begin{gathered} d=d₁-d₂ \\ =75cm-15cm \\ =60cm \\ =\underline{0.6m} \end{gathered}$$

Finally, we locate $a$:

$$\begin{gathered} \frac{de\sigma }{{\epsilon }_0}=\frac{1}{2}m{v}^2 \\ \Rightarrow \sigma =\frac{m{v}^2{\epsilon }_0}{2de} \\ =\frac{(9.11\times {10}^{-31})\times (8.3\times {10}^6)²\times (8.85\times {10}^{-12})}{2x0.6x(1.6x{10}^{-19})} \\ =\boxed{\mathbf{2}\mathbf{.}\mathbf{8}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{}\mathit{C}\mathbf{/}\mathit{m}\mathbf{²}\mathbf{}} \end{gathered}$$