# A charge of -1 C is at the origin. How much energy would be applied to or released from a  3 C charge if it is moved from  (3 ,1 )  to (5 ,-6 ) ?

Dec 26, 2017

The energy released is $= 5.08 \cdot {10}^{9} J$

#### Explanation:

Calculate the potential energy at each point.

${E}_{1} = k \frac{{q}_{1} {q}_{2}}{d} _ 1$

${E}_{2} = k \frac{{q}_{1} {q}_{2}}{d} _ 2$

The distance ${d}_{1} = \sqrt{{3}^{2} + {\left(1\right)}^{2}} = \sqrt{10}$

The distance ${d}_{2} = \sqrt{{5}^{2} + {\left(- 6\right)}^{2}} = \sqrt{61}$

The charges are ${q}_{1} = - 1 C$ and ${q}_{2} = 3 C$

${E}_{1} = 9 \cdot {10}^{9} \left(- 1\right) \frac{3}{\sqrt{10}} = - \frac{27}{\sqrt{10}} \cdot {10}^{9} = - {8.5410}^{9} J$

${E}_{2} = 9 \cdot {10}^{9} \left(- 1\right) \frac{3}{\sqrt{61}} = - \frac{27}{\sqrt{61}} \cdot {10}^{9} = - {3.4610}^{9} J$

The work done is

$= {E}_{2} - {E}_{1}$

$= \left(- 3.46 + 8.54\right) \cdot {10}^{9} J$

$= 5.08 \cdot {10}^{9} J$