Electric Potential Energy Video Lessons

Concept

# Problem: A point charge with a charge q1 = 3.10 μC is held stationary at the origin. A second point charge with a charge q2 = -4.10 μC moves from the point x = 0.130 m , y = 0 to the point x = 0.230 m , y = 0.290 m.How much work is done by the electric force on q2?

###### FREE Expert Solution

The work done:

$\overline{)\begin{array}{rcl}{\mathbf{W}}& {\mathbf{=}}& \mathbf{∆}\mathbf{E}\\ & {\mathbf{=}}& {\mathbf{E}}_{\mathbf{f}}\mathbf{-}{\mathbf{E}}_{\mathbf{i}}\end{array}}$

Electric potential energy:

$\overline{){\mathbf{E}}{\mathbf{=}}{\mathbf{-}}{\mathbf{k}}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{r}}}$

Initial electric potential energy:

${\mathbit{E}}_{\mathbf{i}}\mathbf{=}\mathbf{-}\mathbf{\left(}\mathbf{8}\mathbf{.}\mathbf{99}\mathbf{×}{\mathbf{10}}^{\mathbf{9}}\mathbf{\right)}\mathbf{\left[}\frac{\mathbf{\left(}\mathbf{3}\mathbf{.}\mathbf{10}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{\right)}\mathbf{\left(}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{10}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{\right)}}{\mathbf{0}\mathbf{.}\mathbf{130}}\mathbf{\right]}$

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###### Problem Details

A point charge with a charge q1 = 3.10 μC is held stationary at the origin. A second point charge with a charge q2 = -4.10 μC moves from the point x = 0.130 m , y = 0 to the point x = 0.230 m , y = 0.290 m.

How much work is done by the electric force on q2?