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?