Question

an electron with speed $v_0 = 5.75 \times 10^6 \\ m/s$ is traveling parallel to an electric field of magnitude $e = 8850 \\ n/c$
express your answer to three significant figures and include the appropriate units.
$\delta x=$
part b
how much time will elapse before it returns to its starting point?
express your answer to three significant figures and include the appropriate units.
$t=$

Explanation:

Step1: Calculate the acceleration of the electron

The force on an electron in an electric field is \(F = eE\), where \(e = 1.6\times10^{-19}\ C\) and \(E = 8850\ N/C\). According to Newton's second law \(F = ma\), and the mass of an electron \(m = 9.11\times10^{-31}\ kg\). So, \(a=\frac{eE}{m}\).
Substitute the values: \(a=\frac{1.6\times 10^{-19}\times8850}{9.11\times 10^{-31}}\)

$$a=\frac{1.6\times8850\times10^{-19}}{9.11\times10^{-31}}=\frac{14160\times10^{-19}}{9.11\times10^{-31}} = 1.554\times10^{15}\ m/s^{2}$$

Step2: Find the distance (for part A)

When the electron stops (final velocity \(v = 0\)), we use the kinematic equation \(v^{2}-v_{0}^{2}=2a\Delta x\). Since \(v = 0\), \(\Delta x=\frac{-v_{0}^{2}}{2a}\). Given \(v_{0}=5.75\times 10^{6}\ m/s\) and \(a=- 1.554\times10^{15}\ m/s^{2}\) (negative because it's decelerating).
\(\Delta x=\frac{-(5.75\times 10^{6})^{2}}{2\times(-1.554\times 10^{15})}\)

$$=\frac{-33.0625\times10^{12}}{-3.108\times10^{15}}=\frac{33.0625}{3.108}\times10^{-3}$$

\(\Delta x = 10.6\ mm\)

Step3: Find the time (for part B)

Using the kinematic equation \(v = v_{0}+at\). When \(v = 0\), \(t=\frac{-v_{0}}{a}\).
\(t=\frac{-5.75\times 10^{6}}{-1.554\times 10^{15}}\)

$$t = 3.70\times10^{-9}\ s$$

Since the time to go and come back is considered (symmetric motion), the total time \(t_{total}=2t\)
\(t_{total}=2\times3.70\times 10^{-9}=7.40\times10^{-9}\ s\)

Answer:

For part A: \(\Delta x = 10.6\ mm\)
For part B: \(t = 7.40\times10^{-9}\ s\)