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?
The text passage contains a physics problem with multiple questions related to the motion of an electron near a charged sphere, and it provides a detailed solution for the first question.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
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?
user
The text passage contains a physics problem with multiple questions related to the motion of an electron near a charged sphere, and it provides a detailed solution for the first question.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS