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

Jul 17, 2017

We will use the formula to find the potential energy of the charge $U = K \frac{Q \cdot q}{r}$

#### Explanation:

Let's denote the points
${A}_{1} \left(8 , - 3\right)$
${A}_{2} \left(- 2 , 7\right)$
the distanses of the points from the center are
${r}_{1} = \sqrt{64 + 9} = \sqrt{73} m$
${r}_{2} = \sqrt{4 + 49} = \sqrt{53} m$
because ${r}_{1} > {r}_{2}$ the object comes closer to the center and beacuse the charges are opposite the energy will be released and will be equal to the difference between the potential energies of the object at the two points
$E = {U}_{1} - {U}_{2} = K \frac{Q \cdot q}{r} _ 1 - K \frac{Q \cdot q}{r} _ 2 =$
$9 \cdot {10}^{9} \cdot \frac{6}{\sqrt{73}} - 9 \cdot {10}^{9} \cdot \frac{6}{\sqrt{53}} = 9 \cdot {10}^{9} \left(\frac{6}{\sqrt{73}} - \frac{6}{\sqrt{53}}\right)$ Joules