# A charge of -1 C is at the origin. How much energy would be applied to or released from a  6 C charge if it is moved from  (-3, 1 )  to (2 ,-1 ) ?

Mar 18, 2016

#### Answer:

$W = {E}_{p} = - 54 \cdot {10}^{9} \left(\frac{\sqrt{10} + \sqrt{5}}{\sqrt{50}}\right) \text{ } J$

#### Explanation:

${q}_{1} = - 1 C \text{ } {q}_{2} = 6 C$
${r}_{1} : \text{distance between origin and the point of (-3,1)}$
${r}_{1} = \sqrt{{\left(- 3 - 0\right)}^{2} + {\left(1 - 0\right)}^{2}} = \sqrt{{\left(- 3\right)}^{2} + {1}^{2}} = \sqrt{10}$
${r}_{2} : \text{distance between origin and the point of (2,-1)}$
${r}_{2} = \sqrt{{\left(2 - 0\right)}^{2} + {\left(- 1 - 0\right)}^{2}} = \sqrt{{2}^{2} + {\left(- 1\right)}^{2}} = \sqrt{5}$

$W = {E}_{p} = \frac{k \cdot {q}_{1} \cdot {q}_{2}}{r} _ 1 + \frac{k \cdot {q}_{1} \cdot {q}_{2}}{r} _ 2$

$W = {E}_{p} = k \cdot {q}_{1} \cdot {q}_{2} \left(\frac{1}{r} _ 1 + \frac{1}{r} _ 2\right)$

$W = {E}_{p} = k \cdot {q}_{1} \cdot {q}_{2} \left(\frac{{r}_{1} + {r}_{2}}{{r}_{1} \cdot {r}_{2}}\right)$

$W = {E}_{p} = 9 \cdot {10}^{9} \cdot \left(- 1\right) \cdot 6 \left(\frac{\sqrt{10} + \sqrt{5}}{\sqrt{10} \cdot \sqrt{5}}\right)$

$W = {E}_{p} = - 54 \cdot {10}^{9} \left(\frac{\sqrt{10} + \sqrt{5}}{\sqrt{50}}\right) \text{ } J$
$\text{The energy of system has decreased}$