# A charge of 2 C is at the origin. How much energy would be applied to or released from a  -1 C charge if it is moved from  (3 ,7)  to (8,2 ) ?

Feb 9, 2016

$\Delta E \cong - 4 , 452 \cdot {10}^{9} \text{ J}$

#### Explanation:

$\text{distance between origin and the point(3,7) is:}$
${r}_{1} = \sqrt{{\left(3 - 0\right)}^{2} + {\left(7 - 0\right)}^{2}}$
${r}_{1} = \sqrt{58}$
$\text{distance between origin and the point (8,2) is:}$
${r}_{2} = \sqrt{{\left(8 - 0\right)}^{2} + {\left(2 - 0\right)}^{2}}$
${r}_{2} = \sqrt{68}$
${E}_{1} = \frac{k \cdot 2 \cdot \left(- 1\right)}{\sqrt{58}}$
${E}_{1} = \frac{- 2 \cdot 9 \cdot {10}^{9}}{\sqrt{58}}$
${E}_{1} \cong - 2 , 362 \cdot {10}^{9} \text{ J}$
${E}_{2} = \frac{k \cdot 2 \cdot \left(- 1\right)}{\sqrt{68}}$
${E}_{2} = \frac{- 2 \cdot 9 \cdot {10}^{9}}{\sqrt{68}}$
${E}_{2} \cong - 2 , 18 \cdot {10}^{9} \text{ J}$
$\Delta E = {E}_{2} - {E}_{1}$
$\Delta E = - 2 , 18 \cdot {10}^{9} - 2 , 362 \cdot {10}^{-} 9$
$\Delta E \cong - 4 , 452 \cdot {10}^{9} \text{ J}$