# 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  (-3 , 7 )  to (-2 ,-2 ) ?

Dec 26, 2017

The energy to be applied is $= 73 \cdot {10}^{9} J$

#### 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 ${d}_{1} = \sqrt{{\left(- 3\right)}^{2} + {\left(7\right)}^{2}} = \sqrt{58}$

The distance ${d}_{2} = \sqrt{{\left(- 2\right)}^{2} + {\left(- 2\right)}^{2}} = \sqrt{8}$

The charges are ${q}_{1} = 7 C$ and ${q}_{2} = 5 C$

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

${E}_{1} = 9 \cdot {10}^{9} \cdot \left(7\right) \frac{5}{\sqrt{58}} = 41.4 \cdot {10}^{9} J$

${E}_{2} = 9 \cdot {10}^{9} \cdot \left(7\right) \frac{5}{\sqrt{8}} = 111.4 \cdot {10}^{9} J$

The work done is

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

$= \left(114.4 - 41.4\right) \cdot {10}^{9} J$

$= 73 \cdot {10}^{9} J$