Problem

An electron with an initial speed of $29 \times 10^5 \, \mathrm{m/s}$ is fired in the same direction as a uniform electric field of magnitude $80 \, \mathrm{N/C}$. How far does the electron travel before being momentarily brought to rest?

Data

• Initial speed: $v_i = 29 \times 10^5 \, \mathrm{m/s}$
• Final speed: $v_f = 0 \, \mathrm{m/s}$
• Electric field: $E = 80 \, \mathrm{N/C}$
• Charge on electron: $q_e = -1.6 \times 10^{-19} \, \mathrm{C}$
• Mass of electron: $m_e = 9.11 \times 10^{-31} \, \mathrm{kg}$

Prerequisite Concepts

• Equation of Motion:

$$v_f^2 = v_i^2 + 2aS$$

Rearrange for $S$:

$$S = \frac{v_f^2 - v_i^2}{2a}$$

• Acceleration:

$$a = \frac{F}{m_e}, \quad F = q_e E$$

Answer

Step 1: Calculate Force

$$F = q_e E$$

Substitute values:

$$F = (-1.6 \times 10^{-19})(80) = -1.28 \times 10^{-17} \, \mathrm{N}$$

Step 2: Calculate Acceleration

$$a = \frac{F}{m_e}$$ $$a = \frac{-1.28 \times 10^{-17}}{9.11 \times 10^{-31}} = -1.4 \times 10^{13} \, \mathrm{m/s^2}$$

Step 3: Solve for Distance $S$

Using the equation of motion:

$$S = \frac{v_f^2 - v_i^2}{2a}$$

Substitute values:

$$S = \frac{0 - (29 \times 10^5)^2}{2(-1.4 \times 10^{13})}$$ $$S = \frac{-8.41 \times 10^{11}}{-2.8 \times 10^{13}} = 0.03 \, \mathrm{m}$$

Final Answer

The electron travels $0.03 \, \mathrm{m}$ (or $3 \, \mathrm{cm}$) before being brought to rest.

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