# 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  ( -4, 0 )  to ( 2 , 4 ) ?

Mar 5, 2018

The energy released is $= 1.43 \cdot {10}^{9} J$

#### Explanation:

The potential energy is

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

The charge ${q}_{1} = 2 C$

The charge ${q}_{2} = 3 C$

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

The distance

${r}_{1} = \sqrt{{\left(- 4\right)}^{2} + {\left(0\right)}^{2}} = \sqrt{16} = 4 m$

The distance

${r}_{2} = \sqrt{{\left(2\right)}^{2} + {\left(4\right)}^{2}} = \sqrt{20}$

Therefore,

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

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

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

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

$= 9 \cdot {10}^{9} \cdot \left(\left(2\right) \cdot \left(3\right)\right) \left(\frac{1}{\sqrt{20}} - \frac{1}{4}\right)$

$= - 1.43 \cdot {10}^{9} J$

The energy released is $= 1.43 \cdot {10}^{9} J$