text: Electric Potential Energy Video Lessons
  
  Concept
  
  # Problem: A 2.5 mm -diameter sphere is charged to -4.6 nC . An electron fired directly at the sphere from far away comes to within 0.35 mm of the surface of the target before being reflected.What was the electron's initial speed?At what distance from the surface of the sphere is the electron's speed half of its initial value?What is the acceleration of the electron at its turning point?
  
  ###### FREE Expert Solution
  
  Kinetic energy,
  
  $\overline{)\begin{array}{rcl}\mathbf{K}\mathbf{.}\mathbf{E}& {\mathbf{=}}& \frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m}}_{\mathbf{e}}{\mathbf{v}}^{\mathbf{2}}\end{array}}$
  
  Potential Energy,
  
  $\overline{)\begin{array}{rcl}\mathbf{∆}\mathbf{U}& {\mathbf{=}}& \frac{{\mathbf{kq}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{r}}\end{array}}$
  
  Force,
  
  r = d /2 = 2.5 mm/ 2 ( 1 m/ 1000 mm) = 1.25 × 10-3 m
  
  qs = - 4.6 nC = - 4.6 × 10-9  C
  
  qs = - 1.6 nC = - 1.6 × 10-9  C
  
  me = 9.1 × 10-31 kg
  
  d = 0.35 mm = 0.35 × 10-3 m
  
  a)
  
  From the law of conservation of energy;
  
  $\begin{array}{rcl}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m}}_{\mathbf{e}}{\mathbf{v}}^{\mathbf{2}}& \mathbf{=}& \frac{\mathbf{k}{\mathbf{q}}_{\mathbf{s}}{\mathbf{q}}_{\mathbf{e}}}{\mathbf{\left(}\mathbf{r}\mathbf{+}\mathbf{d}\mathbf{\right)}}\\ \mathbf{v}& \mathbf{=}& \sqrt{\frac{\mathbf{2}\mathbf{k}{\mathbf{q}}_{\mathbf{s}}{\mathbf{q}}_{\mathbf{e}}}{{\mathbf{m}}_{\mathbf{e}}\mathbf{\left(}\mathbf{r}\mathbf{+}\mathbf{d}\mathbf{\right)}}}\\ & \mathbf{=}& \sqrt{\frac{\mathbf{2}\mathbf{\left(}\mathbf{9}\mathbf{×}{\mathbf{10}}^{\mathbf{9}}\mathbf{\right)}\mathbf{\left(}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{6}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{\right)}\mathbf{\left(}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{\right)}}{\mathbf{\left(}\mathbf{9}\mathbf{.}\mathbf{1}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{31}}\mathbf{\right)}\mathbf{\left(}\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{\right)}}}\end{array}$
  
  v = 9.54  × 107 m/s
  
  90% (274 ratings)
  ###### Problem Details
  
  A 2.5 mm -diameter sphere is charged to -4.6 nC . An electron fired directly at the sphere from far away comes to within 0.35 mm of the surface of the target before being reflected.
  
  What was the electron's initial speed?
  
  At what distance from the surface of the sphere is the electron's speed half of its initial value?
  
  What is the acceleration of the electron at its turning point?
explanation: The document is a physics problem involving electric potential energy and the motion of an electron near a charged sphere. It contains a detailed solution to the first part of the problem, which calculates the electron's initial speed. The document also lists two additional questions that are not answered in the provided text.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: Electric Potential Energy Video Lessons
  
  Concept
  
  # Problem: A 2.5 mm -diameter sphere is charged to -4.6 nC . An electron fired directly at the sphere from far away comes to within 0.35 mm of the surface of the target before being reflected.What was the electron's initial speed?At what distance from the surface of the sphere is the electron's speed half of its initial value?What is the acceleration of the electron at its turning point?
  
  ###### FREE Expert Solution
  
  Kinetic energy,
  
  $\overline{)\begin{array}{rcl}\mathbf{K}\mathbf{.}\mathbf{E}& {\mathbf{=}}& \frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m}}_{\mathbf{e}}{\mathbf{v}}^{\mathbf{2}}\end{array}}$
  
  Potential Energy,
  
  $\overline{)\begin{array}{rcl}\mathbf{∆}\mathbf{U}& {\mathbf{=}}& \frac{{\mathbf{kq}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{r}}\end{array}}$
  
  Force,
  
  r = d /2 = 2.5 mm/ 2 ( 1 m/ 1000 mm) = 1.25 × 10-3 m
  
  qs = - 4.6 nC = - 4.6 × 10-9  C
  
  qs = - 1.6 nC = - 1.6 × 10-9  C
  
  me = 9.1 × 10-31 kg
  
  d = 0.35 mm = 0.35 × 10-3 m
  
  a)
  
  From the law of conservation of energy;
  
  $\begin{array}{rcl}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m}}_{\mathbf{e}}{\mathbf{v}}^{\mathbf{2}}& \mathbf{=}& \frac{\mathbf{k}{\mathbf{q}}_{\mathbf{s}}{\mathbf{q}}_{\mathbf{e}}}{\mathbf{\left(}\mathbf{r}\mathbf{+}\mathbf{d}\mathbf{\right)}}\\ \mathbf{v}& \mathbf{=}& \sqrt{\frac{\mathbf{2}\mathbf{k}{\mathbf{q}}_{\mathbf{s}}{\mathbf{q}}_{\mathbf{e}}}{{\mathbf{m}}_{\mathbf{e}}\mathbf{\left(}\mathbf{r}\mathbf{+}\mathbf{d}\mathbf{\right)}}}\\ & \mathbf{=}& \sqrt{\frac{\mathbf{2}\mathbf{\left(}\mathbf{9}\mathbf{×}{\mathbf{10}}^{\mathbf{9}}\mathbf{\right)}\mathbf{\left(}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{6}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{\right)}\mathbf{\left(}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{\right)}}{\mathbf{\left(}\mathbf{9}\mathbf{.}\mathbf{1}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{31}}\mathbf{\right)}\mathbf{\left(}\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{\right)}}}\end{array}$
  
  v = 9.54  × 107 m/s
  
  90% (274 ratings)
  ###### Problem Details
  
  A 2.5 mm -diameter sphere is charged to -4.6 nC . An electron fired directly at the sphere from far away comes to within 0.35 mm of the surface of the target before being reflected.
  
  What was the electron's initial speed?
  
  At what distance from the surface of the sphere is the electron's speed half of its initial value?
  
  What is the acceleration of the electron at its turning point?
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a physics problem involving electric potential energy and the motion of an electron near a charged sphere. It contains a detailed solution to the first part of the problem, which calculates the electron's initial speed. The document also lists two additional questions that are not answered in the provided text.
  
  Label: multiple-question-and-answer