chapter 24
p45. an electron moving parallel to the $x$ axis has an initial speed of $3.70 \times 10^6$ m/s at the origin. its speed is reduced to $1.40 \times 10^5$ m/s at the point $x = 2.00$ cm. (a) calculate the electric potential difference between the origin and that point. (b) which point is at the higher potential?

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{ "quality": "clear", "question_count": 2, "skills_matched": [ "step_cot", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Electric Potential Difference", "Work in Electrostatics" ], "new_concepts": [], "current_concepts": [ "Electric Potential Difference", "Work in Electrostatics", "Conservation of Energy" ] } ## Identify given values and physical constants Using the *Electric Potential Difference* and *Work in Electrostatics* knowledge points $$ \begin{aligned} &m_e = 9.109 \times 10^{-31}\text{ kg}\ &q = -e = -1.602 \times 10^{-19}\text{ C}\ &v_i = 3.70 \times 10^6\text{ m/s}\ &v_f = 1.40 \times 10^5\text{ m/s}\ &x_i = 0\text{ m},\quad x_f = 0.0200\text{ m} \end{aligned} $$ ## Apply conservation of energy Using the *Work in Electrostatics* knowledge point $$ \begin{aligned} &\Delta K + \Delta U = 0\ &\frac{1}{2}m_e(v_f^2 - v_i^2) + q\Delta V = 0\ &q\Delta V = -\frac{1}{2}m_e(v_f^2 - v_i^2) \end{aligned} $$ ## Calculate the electric potential difference Using the *Electric Potential Difference* knowledge point $$ \begin{aligned} &\Delta V = V_f - V_i = -\frac{m_e(v_f^2 - v_i^2)}{2q}\ &\Delta V = -\frac{(9.109 \times 10^{-31}\text{ kg})((1.40 \times 10^5\text{ m/s})^2 - (3.70 \times 10^6\text{ m/s})^2)}{2(-1.602 \times 10^{-19}\text{ C})}\ &\Delta V = -\frac{(9.109 \times 10^{-31})(1.96 \times 10^{10} - 1.369 \times 10^{13})}{-3.204 \times 10^{-19}}\ &\Delta V = \frac{(9.109 \times 10^{-31})(-1.367 \times 10^{13})}{-3.204 \times 10^{-19}}\ &\Delta V \approx -38.9\text{ V} \end{aligned} $$ ## Determine which point is at the higher potential Using the *Electric Potential Difference* knowledge point $$ \begin{aligned} &\Delta V = V_f - V_i < 0 \implies V_i > V_f\ &\text{Origin } (x = 0) \text{ is at the higher potential.} \end{aligned} $$ ## Question 1 (a) The electric potential difference between the origin and that point is $-38.9\text{ V}$ (or $38.9\text{ V}$ in magnitude decrease). ## Question 2 (b) The origin ($x = 0$) is at the higher potential. { "subject": "Natural Science", "question_type": "Multi-part", "knowledge_point": [ "Natural Science", "Physics", "Electric Potential Difference" ] }

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QUESTION IMAGE

Question

Explanation:

Identify given values and physical constants

Apply conservation of energy

Calculate the electric potential difference

Determine which point is at the higher potential

Answer:

Question 1

Question 2