# 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  (6, -1 )  to (-9 ,-2 ) ?

Aug 11, 2017

Energy is released, $E \approx 1.7619134 \cdot {10}^{10} J$

#### Explanation:

Let charge at origin be called ${q}_{1}$ and other is ${q}_{2}$. The initial position vector of charge ${q}_{2}$ is $\vec{{r}_{1}}$ and its final position vector is $\vec{{r}_{2}}$.

So,

$| \vec{{r}_{1}} | = \sqrt{{6}^{2} + {\left(- 1\right)}^{2}} = \sqrt{37}$

$| \vec{{r}_{2}} | = \sqrt{{\left(- 9\right)}^{2} + {\left(- 2\right)}^{2}} = \sqrt{85}$

Since,

$\Delta W = {q}_{2} \cdot \Delta V = E$

$\implies E = {q}_{2} \cdot \left({V}_{\left\{- 9 , - 2\right\}} - {V}_{\left\{6 , - 1\right\}}\right) = {q}_{2} \cdot \left(\frac{{k}_{e} {q}_{1}}{|} \vec{{r}_{2}} | - \frac{{k}_{e} {q}_{1}}{|} \vec{{r}_{1}} |\right)$

$\implies E = \frac{{k}_{e} {q}_{1} {q}_{2}}{| \vec{{r}_{1}} | \cdot | \vec{{r}_{2}} |} \cdot \left(| \vec{{r}_{1}} | - | \vec{{r}_{2}} |\right)$

$\implies E = \frac{9 \cdot {10}^{9} \cdot 7 \cdot 5}{\sqrt{37 \cdot 85}} \cdot \left(\sqrt{37} - \sqrt{85}\right)$

$\implies E \approx - 1.7619134 \cdot {10}^{10} J$