Question

problem 16.36
an electron (mass ( m = 9.11 \times 10^{-31} ) kg) is accelerated from rest in the uniform field ( vec{e} ) (( e = 1.45 \times 10^4 ) n/c) between two thin parallel charged plates. the separation of the plates is 1.90 cm. the electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate, see the figure (figure 1).

figure
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part a
with what speed does it leave the hole?
express your answer to three significant figures and include the appropriate units.

part b
complete previous part(s)

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Explanation:

Step1: Identify the force on the electron

The force on an electron in an electric field is given by \( F = eE \), where \( e = 1.60\times10^{-19}\ C \) (charge of electron) and \( E = 1.45\times 10^{4}\ N/C \).
\( F=(1.60\times 10^{-19}\ C)(1.45\times 10^{4}\ N/C)=2.32\times 10^{-15}\ N \)

Step2: Calculate acceleration

Using Newton's second law \( F = ma \), we can find acceleration \( a=\frac{F}{m} \), where \( m = 9.11\times 10^{-31}\ kg \)
\( a=\frac{2.32\times 10^{-15}\ N}{9.11\times 10^{-31}\ kg}\approx 2.546\times 10^{15}\ m/s^{2} \)

Step3: Use kinematic equation

The electron starts from rest (\( u = 0 \)), travels a distance \( s=1.90\ cm = 0.019\ m \). The kinematic equation \( v^{2}=u^{2}+2as \)
Since \( u = 0 \), \( v^{2}=2as \)
\( v=\sqrt{2as}=\sqrt{2\times(2.546\times 10^{15}\ m/s^{2})\times0.019\ m} \)
\( v=\sqrt{9.72\times 10^{13}\ m^{2}/s^{2}}\approx 9.86\times 10^{6}\ m/s \)

Answer:

The speed of the electron when it leaves the hole is \(\boldsymbol{9.86\times 10^{6}\ m/s}\) (or in appropriate units, the value is approximately \(9.86\times 10^{6}\) with unit \(m/s\))