# A charge of 7 C is at the origin. How much energy would be applied to or released from a  5 C charge if it is moved from  (-5, 1 )  to (2 ,-6 ) ?

Feb 27, 2018

The energy released is $= 11.97 \cdot {10}^{9} J$

#### Explanation:

The potential energy is

$U = k \frac{{q}_{1} {q}_{2}}{r}$

The charge ${q}_{1} = 7 C$

The charge ${q}_{2} = 5 C$

The Coulomb's constant is $k = 9 \cdot {10}^{9} N {m}^{2} {C}^{-} 2$

The distance

${r}_{1} = \sqrt{{\left(- 5\right)}^{2} + {\left(1\right)}^{2}} = \sqrt{26} m$

The distance

${r}_{2} = \sqrt{{\left(2\right)}^{2} + {\left(- 6\right)}^{2}} = \sqrt{40}$

Therefore,

${U}_{1} = k \frac{{q}_{1} {q}_{2}}{r} _ 1$

${U}_{2} = k \frac{{q}_{1} {q}_{2}}{r} _ 2$

$\Delta U = {U}_{2} - {U}_{1} = k \frac{{q}_{1} {q}_{2}}{r} _ 2 - k \frac{{q}_{1} {q}_{2}}{r} _ 1$

$= k \left({q}_{1} {q}_{2}\right) \left(\frac{1}{r} _ 2 - \frac{1}{r} _ 1\right)$

$= 9 \cdot {10}^{9} \cdot \left(\left(7\right) \cdot \left(5\right)\right) \left(\frac{1}{\sqrt{40}} - \frac{1}{\sqrt{26}}\right)$

$= - 11.97 \cdot {10}^{9} J$

The energy released is $= 11.97 \cdot {10}^{9} J$