Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,
![v = \dfrac{e_p}{q}](/tex.php?f=v = \dfrac{e_p}{q})
then the following gives us the potential difference,
![\begin{aligned} \delta v & = \dfrac{e_p_b}{q} - \dfrac{e_p_a}{q} \\ & = \dfrac{\delta e_p}{q} \end{aligned}](/tex.php?f=\begin{aligned} \delta v & = \dfrac{e_p_b}{q} - \dfrac{e_p_a}{q} \\ & = \dfrac{\delta e_p}{q} \end{aligned})
the formula
![\delta v = \delta e_p / q](/tex.php?f=\delta v = \delta e_p / q)
may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get
![\delta e_p = q\delta v](/tex.php?f=\delta e_p = q\delta v)
using q = 8.00 × 10⁻¹⁹ c and δv = 2.00 v, we get
![\begin{aligned} \delta e_p & = (8.00\times10^{-19} \text{ c})(2.00\text{ v}) \\ & = 1.60\times 10^{-18}\text{ j} \end{aligned}](/tex.php?f=\begin{aligned} \delta e_p & = (8.00\times10^{-19} \text{ c})(2.00\text{ v}) \\ & = 1.60\times 10^{-18}\text{ j} \end{aligned})
the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.