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  (1, -1)  to ( 7 , -4 ) ?

Dec 24, 2017

The energy released is $= 15.74 \cdot {10}^{9} J$

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 ${d}_{1} = \sqrt{{1}^{2} + {\left(- 1\right)}^{2}} = \sqrt{2}$

The distance ${d}_{2} = \sqrt{{7}^{2} + {\left(- 4\right)}^{2}} = \sqrt{65}$

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

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

${E}_{2} = 9 \cdot {10}^{9} \left(- 1\right) \frac{3}{\sqrt{65}} = - \frac{27}{\sqrt{65}} \cdot {10}^{9} J$

The work done is

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

$= \left(- 3.35 + 19.09\right) \cdot {10}^{9} J$

$= 15.74 \cdot {10}^{9} J$