A problem of practical interest is to make a beam of electrons turn a $90°$corner. This can be done with the parallel-plate capacitor shown in FIGURE. An electron with kinetic energy $3.0\times {10}^{-17}\mathrm{J}$enters through a small hole in the bottom plate of the capacitor.

a. Should the bottom plate be charged positive or negative relative to the top plate if you want the electron to turn to the right? Explain.

b. What strength electric field is needed if the electron is to emerge from an exit hole $1.0cm$away from the entrance hole, traveling at right angles to its original direction?

Hint: The difficulty of this problem depends on how you choose your coordinate system.

c. What minimum separation $d_{\min }$must the capacitor plates have?

Step by step solution

01

Explanation (part a)

A moving electron in a magnetic field

The force on the $(-ve)$electron is downward since the electrical field points upwards.

The electron will travel in a parabolic path.

02

Find Strength electrical field (part b)

Using the equation,

$V_f^2=u_i^2-2aH$

$(atmaxH{V}_f=0)$

$$\begin{gathered} a=\frac{u_i^2}{2H} \\ =\frac{V_o^2\sin 45°}{2H} \end{gathered}$$

$a=\frac{V_o^2}{4H}$

By,

$eE=ma$

$$\begin{gathered} E=\frac{ma}{e} \\ =\frac{1}{2}\times \frac{m{V}_o^2}{2H_e} \end{gathered}$$

Expand it,

localid="1650219707878" $$\begin{gathered} E=\frac{kE}{2H_e} \\ =\frac{3\times {10}^{-17}J}{2\times 0.5\times {10}^{-2}m\times 1.6\times {10}^{-19}C} \end{gathered}$$

$E=1.87\times {10}^4\mathrm{N}/\mathrm{C}$

03

Find dmin (part c)

By Trigonometry,

$\frac{OB}{\sin 45°}=\frac{OA}{\sin 45°}$

$OA=OB$

$A{B}^2=O{A}^2+O{B}^2$

$$\begin{gathered} OA=OB \\ =\frac{AB}{\sqrt{2}}=\frac{1}{\sqrt{2}}\mathrm{cm} \end{gathered}$$

$OP$is the maximum height of the capacitor plate,

$OP=OA\sin 45°$

$$\begin{gathered} d_{\min }=H \\ =\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}} \\ =\frac{1}{2}=0.5\mathrm{cm} \end{gathered}$$

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